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**Unformatted text preview: **Book of Proof
Richard Hammack
Virginia Commonwealth University Mathematics Textbook Series. Editor: Lon Mitchell
1. Book of Proof by Richard Hammack
2. Linear Algebra by Jim Heﬀeron
3. Abstract Algebra: Theory and Applications by Thomas Judson
4. Ordinary and Partial Diﬀerential Equations by John W. Cain and Angela Reynolds
5. Introduction to Modern Set Theory by Judith Roitman Department of Mathematics & Applied Mathematics
Virginia Commonwealth University
Richmond, Virginia, 23284 Book of Proof
Edition 1.1
© 2009 by Richard Hammack
This work is licensed under the Creative Commons Attribution-No Derivative Works 3.0
License Cover art by Eleni Kanakis, © 2009, all rights reserved, used by permission
A
Typeset in 11pt TEX Gyre Schola using PDFL TEX To my students Contents vii Preface
Introduction I viii Fundamentals
3 1. Sets
1.1.
1.2.
1.3.
1.4.
1.5.
1.6.
1.7.
1.8. Introduction to Sets
The Cartesian Product
Subsets
Power Sets
Union, Intersection, Diﬀerence
Complement
Venn Diagrams
Indexed Sets 28 2. Logic
2.1. Statements
2.2. And, Or, Not
2.3. Conditional Statements
2.4. Biconditional Statements
2.5. Truth Tables for Statements
2.6. Logical Equivalence
2.7. Quantiﬁers
2.8. More on Conditional Statements
2.9. Translating English to Symbolic Logic
2.10. Negating Statements
2.11. Logical Inference
2.12. An Important Note 3. Counting
3.1.
3.2.
3.3.
3.4.
3.5. 3
8
11
14
17
19
21
24 Counting Lists
Factorials
Counting Subsets
Pascal’s Triangle and the Binomial Theorem
Inclusion-Exclusion 29
33
36
39
41
44
46
48
50
52
55
56 57
57
64
67
72
75 v II How to Prove Conditional Statements
4. Direct Proof 81 4.1.
4.2.
4.3.
4.4.
4.5. 81
83
85
90
92 Theorems
Deﬁnitions
Direct Proof
Using Cases
Treating Similar Cases 5. Contrapositive Proof
5.1. Contrapositive Proof
5.2. Congruence of Integers
5.3. Mathematical Writing 6. Proof by Contradiction
6.1.
6.2.
6.3.
6.4. III Proving Statements with Contradiction
Proving Conditional Statements by Contradiction
Combining Techniques
Some Words of Advice 94
94
97
99 103
104
107
108
109 More on Proof
7. Proving Non-conditional Statements
7.1. If-And-Only-If Proof
7.2. Equivalent Statements
7.3. Existence Proofs 8. Proofs Involving Sets
8.1.
8.2.
8.3.
8.4. How to Prove a ∈ A
How to Prove A ⊆ B
How to Prove A = B
Examples: Perfect Numbers 9. Disproof
9.1. Counterexamples
9.2. Disproving Existence Statements
9.3. Disproof by Contradiction 10. Mathematical Induction
10.1. Proof by Strong Induction
10.2. Proof by Smallest Counterexample
10.3. Fibonacci Numbers 113
113
115
116 119
119
121
124
127 134
136
138
139 142
148
152
153 vi IV Relations, Functions and Cardinality
11. Relations
11.1.
11.2.
11.3.
11.4.
11.5. Properties of Relations
Equivalence Relations
Equivalence Classes and Partitions
The Integers Modulo n
Relations Between Sets 12. Functions
12.1.
12.2.
12.3.
12.4.
12.5.
12.6. Functions
Injective and Surjective Functions
The Pigeonhole Principle
Composition
Inverse Functions
Image and Preimage 13. Cardinality of Sets
13.1. Sets With Equal Cardinalities
13.2. Countable and Uncountable Sets
13.3. Comparing Cardinalities 161
164
169
173
176
179 181
181
186
190
193
196
199 202
202
206
211 Conclusion 215 Solutions 216 Index 274 Preface n writing this book I have been motivated by the desire to create a
high-quality textbook that costs almost nothing.
The book is available on my web page for free, and the paperback
versions (produced through an on-demand press) cost considerably less
than comparable traditional textbooks. Any new editions of this text will
be issued solely for the purpose of correcting mistakes and clarifying the
exposition. New exercises may be added, but the existing ones will not be
unnecessarily changed or renumbered.
This text is an expansion and reﬁnement of lecture notes I developed
while teaching proofs courses over the past ten years. It is written for an
audience of mathematics majors at Virginia Commonwealth University,
a large state university. It is catered to our program, and is intended
to prepare our students for our more advanced courses. However, I am
mindful of a larger audience. I believe this book is suitable for almost any
undergraduate mathematics program. I Richmond, Virginia
June 12, 2010 R H Introduction T his is a book about how to prove theorems. Until this point in your education, you may have regarded mathematics
as being a primarily computational discipline. You have learned to solve
equations, compute derivatives and integrals, multiply matrices and ﬁnd
determinants; and you have seen how these things can answer practical
questions about the real world. In this setting, your primary goal in using
mathematics has been to compute answers.
But there is another approach to mathematics that is more theoretical
than computational. In this approach, the primary goal is to understand
mathematical structures, to prove mathematical statements, and even
to invent or discover new mathematical theorems and theories. The
mathematical techniques and procedures that you have learned and used
up until now have their origins in this theoretical side of mathematics. For
example, in computing the area under a curve, you use the Fundamental
Theorem of Calculus. It is because this theorem is true that your answer
is correct. However, in your calculus class you were probably far more
concerned with how that theorem could be applied than in understanding
why it is true. But how do we know it is true? How can we convince
ourselves or others of its validity? Questions of this nature belong to the
theoretical realm of mathematics. This book is an introduction to that
realm.
This book will initiate you into an esoteric world. You will learn to
understand and apply the methods of thought that mathematicians use to
verify theorems, explore mathematical truth and create new mathematical
theories. This will prepare you for advanced mathematics courses, for you
will be better able to understand proofs, write your own proofs and think
critically and inquisitively about mathematics. ix
The book is organized into four parts, as outlined below.
PART I
• Chapter 1: Sets
• Chapter 2: Logic
• Chapter 3: Counting Chapters 1 and 2 lay out the language and conventions used in all advanced
mathematics. Sets are fundamental because every mathematical structure,
object or entity can be described as a set. Logic is fundamental because it
allows us to understand the meanings of statements, to deduce information
about mathematical structures and to uncover further structures. All
subsequent chapters will build on these ﬁrst two chapters. Chapter 3
is included partly because its topics are central to many branches of
mathematics, but also because it is a source of many examples and exercises
that occur throughout the book. (However, the course instructor may choose
to skip Chapter 3.)
PART II
• Chapter 4: Direct Proof
• Chapter 5: Contrapositive Proof
• Chapter 6: Proof by Contradiction Chapters 4 through 6 are concerned with three main techniques used for
proving theorems that have the “conditional” form “If P , then Q”.
PART III
•
•
•
• Chapter 7: Proving Non-Conditional Statements
Chapter 8: Proofs Involving Sets
Chapter 9: Disproof
Chapter 10: Mathematical Induction
These chapters deal with useful variations, embellishments and consequences of the proof techniques introduced in chapters 4 through 6.
PART IV
• Chapter 11: Relations
• Chapter 12: Functions
• Chapter 13: Cardinality of Sets These ﬁnal chapters are mainly concerned with the idea of functions, which
are central to all of mathematics. Upon mastering this material you will
be ready for many advanced mathematics courses, such as combinatorics,
abstract algebra, analysis and topology. Introduction x The book is designed to be covered in a fourteen-week semester. Here
is a possible timetable.
Week Monday Wednesday Friday 1 Section 1.1 Section 1.2 Sections 1.3, 1.4 2 Sections 1.5, 1.6, 1.7 Section 1.8 Section 2.1 3 Section 2.2 Sections 2.3, 2.4 Sections 2.5, 2.6 2.8∗ , Sections 2.10, 2.11∗ , 2.12∗
4 Section 2.7 Sections 5 Sections 3.1, 3.2 Section 3.3 Sections 3.4, 3.5∗ 6 EXAM Sections 4.1, 4.2, 4.3 Sections 4.3, 4.4, 4.5∗ 7 Sections 5.1, 5.2, 5.3∗ Section 6.1 Sections 6.2 6.3∗ 8 Sections 7.1, 7.2∗ , 7.3 Sections 8.1, 8.2 9 Section 8.4 2.9 Sections 9.1, 9.2,
10.3∗ Section 8.3
9.3∗ Section 10.0 10 Sections 10.0, Sections 10.1, 10.2 EXAM 11 Sections 11.0, 11.1 Sections 11.2, 11.3 Sections 11.4, 11.5 12 Section 12.1 Section 12.2 Section 12.2 Section 12.5 Sections 12.5, 12.6∗ Section 13.2 Section 13.3 13 Sections 12.3, 14 Section 13.1 12.4∗ Sections marked with ∗ may require only the briefest mention in class, or
may be best left for the students to digest on their own. Some instructors
may prefer to omit Chapter 3.
Acknowledgments. I thank my students in VCU’s MATH 300 courses
for oﬀering feedback as they read this book. Thanks especially to Cory
Colbert and Lauren Pace for rooting out many typographical mistakes and
inconsistencies. I am especially indebted to Cory for reading early drafts
of each chapter and catching numerous mistakes before I posted the ﬁnal
draft on my web page. Cory also created the index, suggested some of the
more interesting exercises, and wrote many of the solutions. Thanks also
to Micol Hammack for proofreading the entire text, and to Andy Lewis for
suggesting many improvements while teaching from it in Fall 2008 and
2009. I am grateful to Eleni Kanakis for creating the cover art.
I am indebted to VCU’s Center For Teaching Excellence for awarding
Lon Mitchell and me a CTE Small Grant for open-source textbook publishing. It was through this grant that we were able to buy the ISBN and
professional fonts. Thanks also to series editor Lon Mitchell, whose expertise with typesetting and on-demand publishing made the print version of
this book a reality. Part I
Fundamentals CHAPTER 1 Sets ll of mathematics can be described with sets. This becomes more and
more apparent the deeper into mathematics you go. It will be apparent
in most of your upper-level courses, and certainly in this course. The theory
of sets is a language that is perfectly suited to describing and explaining
all types of mathematical structures. A 1.1 Introduction to Sets
A set is a collection of things. The things in the collection are called
elements of the set. We are mainly concerned with sets whose elements
are mathematical entities, such as numbers, points, functions, etc.
A set is often expressed by listing its elements between commas, enclosed by braces. For example, the collection 2, 4, 6, 8 is a set which has
four elements, the numbers 2, 4, 6 and 8. Some sets have inﬁnitely many
elements. For example, consider the collection of all integers,
. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . . . Here the dots indicate a pattern of numbers that continues forever in both
the positive and negative directions. A set is called an inﬁnite set if it
has inﬁnitely many elements; otherwise it is called a ﬁnite set.
Two sets are equal if they contain exactly the same elements. Thus
2, 4, 6, 8 = 4, 2, 8, 6 because even though they are listed in a diﬀerent
order, the elements are identical; but 2, 4, 6, 8 = 2, 4, 6, 7 . Also
. . . − 4, −3, −2, −1, 0, 1, 2, 3, 4 . . . = 0, −1, 1, −2, 2, −3, 3, −4, 4, . . . . We often let upper-case letters stand for sets. In discussing the set
2, 4, 6, 8 we might declare A = 2, 4, 6, 8 and then use A to stand for
2, 4, 6, 8 . To express that 2 is an element of the set A , we write 2 ∈ A , and
read this as “2 is an element of A ” or “2 is in A ” or just “2 in A .” We also
have 4 ∈ A , 6 ∈ A and 8 ∈ A , but 5 ∉ A . We read this last expression as “5 is
not an element of A ,” or “5 not in A .” Expressions like 6, 2 ∈ A or 2, 4, 8 ∈ A
are commonly used for indicating that several things are in a set. Sets 4 Some sets are so signiﬁcant and prevalent that we reserve special
symbols for them. The set of natural numbers (i.e. the positive whole
numbers) is denoted by N, that is,
N = 1, 2, 3, 4, 5, 6, 7, . . . . The set of integers
Z = . . . , −3, −2, −1, 0, 1, 2, 3, 4, . . . is another fundamental set. The symbol R stands for the set of all real
numbers, a set that is undoubtedly familiar to you from calculus. Other
special sets will be listed later in this section.
Sets need not have just numbers as elements. The set B = T , F consists
of two letters, perhaps representing the values “true” and “false.” The set
C = a, e, i , o, u consists of the lower-case vowels in the English alphabet.
The set D = (0, 0), (1, 0), (0, 1), (1, 1) has as elements the four corner points
of a square on the x- y coordinate plane. Thus (0, 0) ∈ D , (1, 0) ∈ D , etc., but
(1, 2) ∉ D (for instance). It is even possible for a set to have other sets
as elements. Consider E = 1, 2, 3 , 2, 4 , which has three elements: the
number 1, the set 2, 3 and the set 2, 4 . Thus 1 ∈ E and 2, 3 ∈ E and
2, 4 ∈ E . But note that 2 ∉ E , 3 ∉ E , and 4 ∉ E .
For yet another example, consider the set M = 0 0 , 1 0 , 1 0 of three
00
01
11
two-by-two matrices. We have 0 0 ∈ M , but 1 0 ∉ M .
00
21
If X is a ﬁnite set, its cardinality or size is the number of elements
it has, and this number is denoted as | X |. Thus for the sets above, | A | = 4,
|B| = 2, |C | = 5, |D | = 4, |E | = 3 and | M | = 3.
There is a special set that, although small, plays a big role. The
empty set is the set that has no elements. We denote it as , so = .
Whenever you see the symbol , it stands for . Observe that | | = 0. The
empty set is the only set whose cardinality is zero.
Be very careful how you write the empty set. Don’t write
when
you mean . These sets can’t be equal because contains nothing while
contains one thing, namely the empty set. If this is confusing, think
of a set as a box with things in it, so, for example, 2, 4, 6, 8 is a “box”
containing four numbers. Thus the empty set = is an empty box. By
contrast,
is a box with an empty box inside it. Obviously, there’s a
diﬀerence: An empty box is not the same as a box with an empty box
=1
inside it. Thus =
. (You might also observe that | | = 0 and
as additional evidence that =
.) Introduction to Sets 5 This box analogy can help you think about sets. The set F = ,
,
may look strange but it is really very simple. Think of it as a box containing three things: an empty box, a box containing an empty box, and a box
containing a box containing an empty box. Thus |F | = 3. The set G = N, Z
is a box containing two boxes, the box of natural numbers and the box of
integers. Thus |G | = 2.
A special notation called set-builder notation is used to describe sets
that are too big or complex to list between braces. Consider the inﬁnite
set of even integers E = . . . , −6, −4, −2, 0, 2, 4, 6, . . . . In set-builder notation
this set is written as
E = 2n : n ∈ Z . We read the ﬁrst brace as “the set of all things of form,” and the colon as
“such that.” Thus the entire expression E = 2n : n ∈ Z is read as “E equals
the set of all things of form 2n, such that n is an element of Z.” The idea
is that E consists of all possible values of 2n, where n is allowed to take
on all values in Z.
In general, a set X written with set-builder notation has the syntax
X = expression : rule , where the elements of X are understood to be all values of “expression”
that are speciﬁed by “rule.” For example, the set E above is the set
of all values the expression 2n that satisfy the rule n ∈ Z. There can
be many ways to express the same set. For example E = 2n : n ∈ Z =
n : n is an even integer = n : n = 2 k, k ∈ Z . Another common way of
writing it is
E = n ∈ Z : n is even
which we read as “E is the set of all n in Z such that n is even.”
Example 1.1 Here are some further illustrations of set-builder notation. 1. n : n is a prime number = 2, 3, 5, 7, 11, 13, 17, . . .
2. n ∈ N : n is prime = 2, 3, 5, 7, 11, 13, 17, . . .
3. n2 : n ∈ Z = 0, 1, 4, 9, 16, 25, . . .
4. x ∈ R : x2 − 2 = 0 = 2, − 2 5. x ∈ Z : x2 − 2 = 0 =
6. x ∈ Z : | x| < 4 = − 3, −2, −1, 0, 1, 2, 3
7. 2 x : x ∈ Z, | x| < 4 = − 6, −4, −2, 0, 2, 4, 6
8. x ∈ Z : |2 x| < 4 = − 1, 0, 1 Sets 6 These last three examples highlight a conﬂict of notation that we must
always be alert to. The expression | X | means absolute value if X is a
number and cardinality if X is a set. The distinction should always be
clear from context. In the | x| in Example 6 above, we have x ∈ Z, so x is
a number (not a set) and thus the bars in | x| must mean absolute value,
not cardinality. On the other hand, consider A = 1, 2 , 3, 4, 5, 6 , 7 and
B = X ∈ A : | X | < 3 . The elements of A are sets (not numbers) so the | X |
in the expression for B must mean cardinality. Therefore B = 1, 2 , 7 .
We close this section with a summary of special sets. These are sets or
types of sets that come up so often that they are given special names and
symbols.
• The empty set: = • The natural numbers: N = 1, 2, 3, 4, 5, . . .
• The integers: Z = . . . , −3, −2, −1, 0, 1, 2, 3, 4, 5, . . .
• The rational numbers: Q = x : x =
• The real numbers: R m
, where m, n ∈ Z and n = 0
n (the set of all real numbers on the number line) Notice Q is the set of all numbers that can be expressed as a fraction of
two integers. You are surely aware that Q = R, for 2 ∉ Q but 2 ∈ R.
There are some other special sets that you will recall from your study
of calculus. Given two numbers a, b ∈ R with a < b, we can form various
intervals on the number line.
• Closed interval: [a, b] = x ∈ R : a ≤ x ≤ b
• Half open interval: (a, b] = x ∈ R : a < x ≤ b
• Half open interval: [a, b) = x ∈ R : a ≤ x < b
• Open interval: (a, b) = x ∈ R : a < x < b
• Inﬁnite interval: (a, ∞) = x ∈ R : a < x
• Inﬁnite interval: [a, ∞) = x ∈ R : a ≤ x
• Inﬁnite interval: (−∞, b) = x ∈ R : x < b
• Inﬁnite interval: (−∞, b] = x ∈ R : x ≤ b
Remember that these are intervals on the number line, so they have inﬁnitely many elements. The set (0.1, 0.2) contains inﬁnitely many numbers,
even though the end points may be close together. It is an unfortunate
notational accident that (a, b) can denote both an interval on the line and
a point on the plane. The diﬀerence is usually clear from context. In the
next section we will see still another meaning of (a, b). Introduction to Sets 7 Exercises for Section 1.1
A. Write each of the following sets by listing their elements between braces.
1. 5 x − 1 : x ∈ Z 9. 2. 3 x + 2 : x ∈ Z 10. x ∈ R : cos x = 1 x ∈ R : sin π x = 0 3. x ∈ Z : −2 ≤ x < 7 11. x ∈ Z : | x| < 5 4. x ∈ N : −2 < x ≤ 7 12. x ∈ Z : |2 x | < 5 5. 2 13. x ∈ Z : |6 x | < 5 2 14. 5 x : x ∈ Z, |2 x| ≤ 8 6.
7.
8. x∈R:x =3
x∈R:x =9
2 x ∈ R : x + 5 x = −6
3 15. 5a + 2b : a, b ∈ Z 2 x ∈ R : x + 5 x = −6 x 16. 6a + 2b : a, b ∈ Z B. Write each of the following sets in set-builder notation.
17. 2, 4, 8, 16, 32, 64 . . . 23. 3, 4, 5, 6, 7, 8 18. 0, 4, 16, 36, 64, 100, . . . 24. − 4, −3, −2, −1, 0, 1, 2 19. . . . , −6, −3, 0, 3, 6, 9, 12, 15, . . . 25. 1
1
. . . , 8 , 1 , 2 , 1, 2, 4, 8, . . .
4 20. . . . , −8, −3, 2, 7, 12, 17, . . . 26. 1 11
. . . , 27 , 9 , 3 , 1, 3, 9, 27, . . . 21. 0, 1, 4, 9, 16, 25, 36, . . . 27. π
π
. . . , −π, − π , 0, π , π, 32 , 2π, 52 , . . .
2
2 22. 3, 6, 11, 18, 27, 38, . . . 28. 3
3
9
. . . , − 3 , − 4 , 0, 3 , 2 , 9 , 3, 15 , 2 , . . .
2
4
4
4 C. Find the following cardinalities.
29. 1 , 2, 3, 4 , 34. x ∈ N : | x| < 10 30. 1, 4 , a, b, 3, 4 , 35. x ∈ Z : x2 < 10 31. 1 , 2, 3, 4 , 36. x ∈ N : x2 < 10 32. 1, 4 , a , b , 3, 4 , 37. x ∈ N : x2 < 0 38. x ∈ N : 5 x ≤ 20 33. x ∈ Z : | x| < 10 D. Sketch the following sets of points in the x- y plane.
39. ( x, y) : x ∈ [1, 2], y ∈ [1, 2] 46. ( x, y) : x, y ∈ R, x2 + y2 ≤ 1 40. ( x, y) : x ∈ [0, 1], y ∈ [1, 2] 47. ( x, y) : x, y ∈ R, y ≥ x2 − 1 41. ( x, y) : x ∈ [−1, 1], y = 1 48. ( x, y) : x, y ∈ R, x > 1 42. ( x, y) : x = 2, y ∈ [0, 1] 49. ( x, x + y) : x ∈ R, y ∈ Z 43. ( x, y) : | x| = 2, y ∈ [0, 1] 50. ( x, xy ) : x ∈ R, y ∈ N 44. ( x, x2 ) : x ∈ R 51. ( x, y) ∈ R2 : ( y − x)( y + x) = 0
2 2 45. ( x, y) : x, y ∈ R, x + y = 1 2 52. ( x, y) ∈ R2 : ( y − x2 )( y + x2 ) = 0 Sets 8
1.2 The Cartesian Product Given two sets A and B, it is possible to “multiply” them to produce a new
set denoted as A × B. This operation is called the Cartesian product. To
understand it, we must ﬁrst understand the idea of an ordered pair.
Deﬁnition 1.1 An ordered pair is a list ( x, y) of two things x and y,
enclosed in parentheses and separated by a comma.
For example (2, 4) is an ordered pair, as is (4, 2). These ordered pairs
are diﬀerent because even though they have the same things in them,
the order is diﬀerent. We write (2, 4) = (4, 2). Right away you can see that
ordered pairs can be used to describe points on the plane, as was done in
calculus, but they are not limited to just that. The things in an ordered
pair don’t have to be numbers. You can have ordered pairs of letters, such
as ( m, ), ordered pairs of sets such as ( 2, 2 , 3, 2 ), even ordered pairs
of ordered pairs like ((2, 4), (4, 2)). The following are also ordered pairs:
(2, 1, 2, 3 ), (R, (0, 0)). Any list of two things enclosed by parentheses is an
ordered pair. Now we are ready to deﬁne the Cartesian product.
Deﬁnition 1.2 The Cartesian product of two sets A and B is another
set, denoted as A × B and deﬁned as A × B = (a, b) : a ∈ A , b ∈ B .
Thus A × B is a set of ordered pairs of elements from A and B. For
example, if A = k, , m and B = q, r , then
A × B = ( k, q), ( k, r ), ( , q), ( , r ), ( m, q), ( m, r ) . Figure 1.1 shows how to make a schematic diagram of A × B. Line up the
elements of A horizontally and line up the elements of B vertically, as if A
and B form an x- and y-axis. Then ﬁll in the ordered pairs so that each
element ( x, y) is in the column headed by x and the row headed by y.
B
r
q ( k, r )
( k, q) k A×B
( , r ) ( m, r )
( , q ) ( m, q )
m A Figure 1.1. A diagram of a Cartesian product The Cartesian Product 9 For another example, 0, 1 × 2, 1 = (0, 2), (0, 1), (1, 2), (1, 1) . If you are
a visual thinker, you may wish to draw a diagram similar to Figure 1.1.
The rectangular array of such diagrams give us the following general fact.
Fact 1.1 If A and B are ﬁnite sets, then | A × B| = | A | · |B|. The set R × R = ( x, y) : x, y ∈ R should be very familiar. It can be viewed
as the set of points on the Cartesian plane, and is drawn in Figure 1.2(a).
The set R × N = ( x, y) : x ∈ R, y ∈ N can be regarded as all of the points on
the Cartesian plane whose second coordinate is a natural number. This
is illustrated in Figure 1.2(b), which shows that R × N looks like inﬁnitely
many horizontal lines at integer heights above the x axis. The set N × N
can be visualized as the set of all points on the Cartesian plane whose
coordinates are both natural numbers. It looks like a grid of dots in the
ﬁrst quadrant, as illustrated in Figure 1.2(c).
y y R×R (a) x y R×N (b) x N×N x (c) Figure 1.2. Drawings of some Cartesian products
It is even possible for one factor of a Cartesian product to be a Cartesian
product itself, as in R × (N × Z) = ( x, ( y, z)) : x ∈ R, ( y, z) ∈ N × Z .
We can also deﬁne Cartesian products of three or more sets by moving
beyond ordered pairs. An ordered triple is a list ( x, y, z). The Cartesian
product of the three sets R, N and Z is R × N × Z = ( x, y, z) : x ∈ R, y ∈ N, z ∈ Z .
Of course there is no reason to stop with ordered triples. In general,
A 1 × A 2 × · · · × A n = ( x1 , x2 , . . . , xn ) : x i ∈ A i for each i = 1, 2, . . . , n . But we should always be mindful of parentheses. There is a slight
diﬀerence between R × (N × Z) and R × N × Z. The ﬁrst is a Cartesian product
of two sets. Its elements are ordered pairs ( x, ( y, z)). The second is a
Cartesian product of three sets, and its elements look like ( x, y, z). Sets 10 We can also take Cartesian powers of sets. For any set A and positive
integer n, the power A n is the Cartesian product of A with itself n times.
A n = A × A × · · · × A = ( x1 , x2 , . . . , xn ) : x1 , x2 , . . . , xn ∈ A In this way, R2 is the familiar Cartesian plane and R3 is three-dimensional
space. You can visualize how, if R2 is the plane, then Z2 = ( m, n) : m, n ∈ Z
is a grid of dots on the plane. Likewise, as R3 is 3-dimensional space,
Z3 = ( m, n, p) : m, n, p ∈ Z is a grid of dots in space.
In other courses you may encounter sets that are very similar to Rn ,
but yet have slightly diﬀerent shades of meaning. Consider, for example,
the set of all two-by-three matrices with entries from R:
M= uvw
xyz : u , v , w , x , y, z ∈ R . This is not really all that diﬀerent from the set
R 6 = ( u , v , w , x , y, z ) : u , v , w , x , y, z ∈ R . The elements of these sets are merely certain arrangements of six real
numbers. Despite their similarity, we maintain that M = R6 , for a two-bythree matrix is not the same thing as an ordered sequence of six numbers.
Exercises for Section 1.2
A. Write out the indicated sets by listing their elements between braces.
1. Suppose A = 1, 2, 3, 4 and B = a, c .
(a) A × B
(b) B × A (c) A × A
(d) B × B (e) × B
(f) ( A × B) × B (g) A × (B × B)
(h) B3 (e) A ×
(f) ( A × B) × B (g) A × (B × B)
(h) A × B × B 9. Suppose A = π, e, 0 and B = 0, 1 .
(a) A × B
(b) B × A (c) A × A
(d) B × B 9. x ∈ R : x2 = 2 × a, c, e
10. n ∈ Z : 2 < n < 5 × n ∈ Z : | n| = 5
11. x ∈ R : x2 = 2 × x ∈ R : | x| = 2 12. x ∈ R : x2 = x × x ∈ N : x2 = x
13.
× 0, × 0, 1
4
14. 0, 1
B. Sketch these Cartesian products on the x- y plane R2 (or R3 for the last two).
15.
16.
17.
18.
19.
20. 1, 2, 3 × − 1, 0, 1
− 1, 0, 1 × 1, 2, 3
[0, 1] × [0, 1]
[−1, 1] × [1, 2]
1, 1.5, 2 × [1, 2]
[1, 2] × 1, 1.5, 2 21.
22.
23.
24.
25.
26. 1 × [0, 1]
[0, 1] × 1
N×Z
Z×Z
[0, 1] × [0, 1] × [0, 1]
( x, y) ∈ R2 : x2 + y2 ≤ 1 × [0, 1] Subsets 11 1.3 Subsets
It can happen that every element of some set A is also an element of
another set B. For example, each element of A = 0, 2, 4 is also an element
of B = 0, 1, 2, 3, 4 . When A and B are related this way we say that A is a
subset of B.
Deﬁnition 1.3 Suppose A and B are sets. If every element of A is also
an element of B, then we say A is a subset of B, and we denote this as
A ⊆ B. We write A ⊆ B if A is not a subset of B, that is if it is not true that
every element of A is also an element of B. Thus A ⊆ B means that there
is at least one element of A that is not an element of B.
Example 1.2 Be sure you understand why each of the following is true. 1. 2, 3, 7 ⊆ 2, 3, 4, 5, 6, 7
2. 2, 3, 7 ⊆ 2, 4, 5, 6, 7
3. 2, 3, 7 ⊆ 2, 3, 7
4. 2n : n ∈ Z ⊆ Z
5. ( x, sin( x)) : x ∈ R ⊆ R2
6. 2, 3, 5, 7, 11, 13, 17, . . . ⊆ N
7. N ⊆ Z ⊆ Q ⊆ R
8. R × N ⊆ R × R
This brings us to a particularly important fact: If B is any set whatsoever, then ⊆ B. To see why this is true, look at the deﬁnition of ⊆.
If ⊆ B were false, there would be an element in that was not in B.
But there can be no such element because contains no elements! The
inescapable conclusion is that ⊆ B.
Fact 1.2 The empty set is a subset of every set, that is ⊆ B for any set B. Here is another way to look at it. Imagine a subset of B as something
you make by starting with braces , then ﬁlling them with selections from
B. For instance, suppose B = a, b, c . To make one particular subset of B,
start with , select b and c from B and insert them into
to form the
subset b, c . Alternatively, you could have chosen a and b to make a, b ,
and so on. But one option is to simply make no selections from B. This
leaves you with the subset . Thus ⊆ B. More often we write it as ⊆ B. Sets 12 This idea of “making” a subset can help us list out all the subsets of
a given set B. As an example, let B = a, b, c . Let’s list all of its subsets.
One way of approaching this is to make a tree-like structure. Begin with
the subset , which is shown on the left of Figure 1.3. Considering the
element a of B, we have a choice; insert it or not. The lines from point
to what we get depending whether or not we insert a, either or a . Now
move on to the element b of B. For each of the sets just formed we can
either insert or not insert b, and the lines on the diagram point to the
resulting sets , b , a , or a, b . Finally, to each of these sets, we can
either insert c or not insert it, and this gives us, on the far right-hand
column, the sets , c , b , b, c , a , a, c , a, b and a, b, c . These are
the eight subsets of B = a, b, c .
Insert a ? Insert b ? Insert c ?
No
Yes No b Yes b No c No
Yes b, c
a a Yes No
Yes No a, c a
a, b Yes a, b No
Yes a, b , c Figure 1.3. A “tree” for listing subsets
We can see from the way this tree branches out that if it happened that
B = a , then B would have just two subsets, those in the second column
of the diagram. If it happened that B = a, b , then B would have four subsets, those listed in the third column, and so on. At each branching of
the tree, the number of subsets doubles. Thus in general, if |B| = n, then
B must have 2n subsets.
Fact 1.3 If a ﬁnite set has n elements, then it has 2n subsets. Subsets 13 For a slightly more complex example, consider listing the subsets of
B = 1, 2, 1, 3 . This B has just three elements: 1, 2 and 1, 3 . At this
point you probably don’t even have to draw a tree to list out B’s subsets.
You just make all the possible selections from B and put them between braces to get
, 1, 2, 1, 3 , 1, 2 , 1, 1, 3 , 2, 1, 3 , 1, 2, 1, 3 . These are the eight subsets of B. Exercises like this help you identify what
is and isn’t a subset. You know immediately that a set such as 1, 3 is not
a subset of B because it can’t be made by selecting elements from B, as
the 3 is not an element of B and thus is not a valid selection.
Example 1.3 Be sure you understand why the following statements are
true. Each illustrates an aspect of set theory that you’ve learned so far.
1. 1 ∈ 1, 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 is the ﬁrst element listed in 1, 1
2. 1 ⊆ 1, 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because 1 is not a set 3. 1 ∈ 1, 1 . . . . . . . . . . . . . . . . . . . . . . . . . 1 is the second element listed in 1, 1 4. 1 ⊆ 1, 1 . . . . . . . . . . . . . . . . . . . . . . . make subset 1 by selecting 1 from 1, 1 5. 1 ∉ 1, 1 . . . . . . . . . . . because 1, 1 6. 1 ⊆ 1, 1 . . . . . . . . . . . . . . . . . .make subset contains only 1 and 1 , and not
1 1 by selecting 1 from 1, 1 7. N ∉ N . . . . . . . . .because N is a set (not a number) and N contains only numbers
8. N ⊆ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because X ⊆ X for every set X
9.
10. ∉ N . . . . . . . . . . . . . . . . . . . . because the set N contains only numbers and no sets
⊆ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because is a subset of every set 11. N ∈ N . . . . . . . . . . . . . . . . . . . . . . . . . . .because N has just one element, the set N
12. N ⊆ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because, for instance, 1 ∈ N but 1 ∉ N
13. ∉ N . . . . . . . . . . . . . . . . . . . . . note that the only element of N is N, and N = 14. ⊆ N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because 15. ∈ ,N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16. ⊆ , N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .because is a subset of every set is the ﬁrst element listed in ,N is a subset of every set 17. N ⊆ , N . . . . . . . . . . . . . . . . . . . . . . . make subset N by selecting N from ,N 18. N ⊆ ,N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . because N ∉ ,N 19. N ∈ , N . . . . . . . . . . . . . . . . . . . . . . N is the second element listed in ,N 20. (1, 2), (2, 2), (7, 1) ⊆ N × N . . . . . . . . . . . . . . . . . . . each of (1, 2), (2, 2), (7, 1) is in N × N Though they should help you understand the concept of subset, the
above examples are somewhat artiﬁcial. But subsets arise very naturally
in mathematics. Consider the unit circle C = ( x, y) ∈ R2 : x2 + y2 = 1 . This is Sets 14 a subset C ⊆ R2 . Likewise the graph of a function y = f ( x) is a set of points
G = ( x, f ( x)) : x ∈ R , and G ⊆ R2 . You will surely agree that sets such as C
and G are more easily understood or visualized when regarded as subsets
of R2 . Mathematics is ﬁlled with such instances where it is important to
regard one set as a subset of another.
Exercises for Section 1.3
A. List all the subsets of the following sets.
1. 1, 2, 3, 4 5. 2. 1, 2, 6. R, Q, N 3. 7. R, Q, N R 8. 4. 0, 1 , 0, 1, 2 , 0 B. Write out the following sets by listing their elements between braces.
9.
10. X ⊆ N : |X | ≤ 1 11. X : X ⊆ 3, 2, a and | X | = 4 12. X : X ⊆ 3, 2, a and | X | = 2 X : X ⊆ 3, 2, a and | X | = 1 C. Decide if the following statements are true or false. Explain.
13. R3 ⊆ R3 15. ( x, y) : x − 1 = 0 ⊆ ( x, y) : x2 − x = 0 14. R2 ⊆ R3 16. ( x, y) : x2 − x = 0 ⊆ ( x, y) : x − 1 = 0 1.4 Power Sets
Given a set, you can form a new set with the power set operation, deﬁned
as follows.
Deﬁnition 1.4 If A is a set, the power set of A is another set, denoted
as P ( A ) and deﬁned to be the set of all subsets of A . In symbols, P ( A ) =
X:X⊆A .
For example, suppose A = 1, 2, 3 . The power set of A is the set of all
subsets of A . We learned how to ﬁnd these in the previous section, and
they are , 1 , 2 , 3 , 1, 2 , 1, 3 , 2, 3 and 1, 2, 3 . Therefore the power
set of A is
P ( A) = , 1 , 2 , 3 , 1, 2 , 1, 3 , 2, 3 , 1, 2, 3 . As we saw in the previous section, if ﬁnite set A has n elements, then
it has 2n subsets, and thus its power set has 2n elements. Power Sets 15 Fact 1.4 If A is a ﬁnite set, then |P ( A )| = 2| A | .
Example 1.4 You should examine the following statements and make
sure you understand how the answers were obtained. In particular, notice
that in each instance the equation |P ( A )| = 2| A | is true.
1. P 0, 1, 3 = , 0 , 1 , 3 , 0, 1 , 0, 3 , 1, 3 , 0, 1, 3 2. P 1, 2 = , 1 , 2 , 1, 2 3. P 1 = ,1 4. P ( ) =
5. P a = ,a 6. P , = 7. P a × P
8. P P ), ( a , ), ( a , = ( , ), ( , , = 9. P 1, 1, 2 = 10. P Z, N = ,
, 1, , ) , 1, 2 , 1, 1, 2 , Z , N , Z, N Next are some that are wrong. See if you can determine why they are wrong
and make sure you understand the explanation on the right.
11. P (1) = ,1 . . . . . . . . . . . . . . . . . . . . . . . . . . . meaningless because 1 is not a set 12. P 1, 1, 2 = , 1 , 1, 2 , 1, 1, 2 13. P 1, 1, 2 = , 1 , 1, 2 , . . . . . . . . wrong because 1, 2 ⊆ 1, 1, 2 , 1, 2 . . . . . wrong because 1 ⊆ 1, 1, 2 If A is ﬁnite, it is possible (though maybe not practical) to list out P ( A )
between braces as was done in examples 1–10 above. That is not possible
if A is inﬁnite. For example, consider P (N). You can start writing out the
answer, but you quickly realize N has inﬁnitely many subsets, and it’s not
clear how (or if) they could be arranged as a list with a deﬁnite pattern:
P (N) = , 1 , 2 , . . . , 1, 2 , 1, 3 , . . . , 39, 47 ,
. . . , 3, 87, 131 , . . . , 2, 4, 6, 8, . . . , . . . ? . . . . The set P (R2 ) is mind boggling. Think of R2 = ( x, y) : x, y ∈ R as the set
of all points on the Cartesian plane. A subset of R2 (that is, an element
of P (R2 )) is a set of points in the plane. Let’s look at some of these sets.
Since (0, 0), (1, 1) ⊆ R2 , we know that (0, 0), (1, 1) ∈ P (R2 ). We can even
draw a picture of this subset, as in Figure 1.4(a). For another example,
the graph of the equation y = x2 is the set of points G = ( x, x2 ) : x ∈ R and
this is a subset of R2 , so G ∈ P (R2 ). Figure 1.4(b) is a picture of G . Since
this can be done for any function, the graph of every imaginable function
f : R → R can be found inside of P (R2 ). Sets 16
y y y x (a) x x (b) (c) Figure 1.4. Three of the many, many sets in P (R2 )
In fact, any black-and-white image on the plane can be thought of as a
subset of R2 , where the black points belong to the subset and the white
points do not. So the text “INFINITE” in Figure 1.4(c) is a subset of R2
and therefore an element of P (R2 ). By that token, P (R2 ) contains a copy
of the page you are reading now.
Thus in addition to containing every imaginable function and every
imaginable black-and-white image, P (R2 ) also contains the full text of
every book that was ever written, those that are yet to be written and
those that will never be written. Inside of P (R2 ) is a detailed biography of
your life, from beginning to end, as well as the biographies of all of your
unborn descendants. It is startling that the ﬁve symbols used to write
P (R2 ) can express such an incomprehensibly large set.
Homework: Think about P (P (R2 )).
Exercises for Section 1.4
A. Find the indicated sets.
1. P a, b , c 7. P a, b × P 0, 1 2. P 1, 2, 3, 4 8. P 1, 2 × 3 3. P 9. P a, b × 0 ,5 4. P R, Q 10. X ∈P 5. P P 2 11. X ⊆ P ( 1, 2, 3 ) : | X | ≤ 1 6. P 1, 2 × P 3 12. X ∈P 1, 2, 3
1, 2, 3 : |X | ≤ 1
:2∈ X B. Suppose that | A | = m and |B| = n. Find the following cardinalities.
13. |P (P (P ( A )))| 17. X ∈ P ( A) : | X | ≤ 1 14. |P (P ( A ))| 18. |P ( A × P (B))| 15. |P ( A × B)| 19. |P (P (P ( A × )))| 16. |P ( A ) × P (B)| 20. X ⊆ P ( A) : | X | ≤ 1 Union, Intersection, Diﬀerence 17 1.5 Union, Intersection, Diﬀerence
Just as numbers are combined with operations such as addition, subtraction and multiplication, there are various operations that can be applied to
sets. The Cartesian product (deﬁned in Section 1.2) is one such operation;
given sets A and B, we can combine them with × to get a new set A × B.
Here are three new operations called union, intersection and diﬀerence.
Deﬁnition 1.5 Suppose A and B are sets.
The union of A and B is the set
A ∪ B = x : x ∈ A or x ∈ B .
The intersection of A and B is the set A ∩ B = x : x ∈ A and x ∈ B .
The diﬀerence of A and B is the set
A − B = x : x ∈ A and x ∉ B .
In words, the union A ∪ B is the set of all things that are in A or in B
(or in both). The intersection A ∩ B is the set of all things in both A and B.
The diﬀerence A − B is the set of all things that are in A but not in B.
Example 1.5 Suppose A = a, b, c, d , e , B = d , e, f and C = 1, 2, 3 . 1. A ∪ B = a, b, c, d , e, f
2. A ∩ B = d , e
3. A − B = a, b, c
4. B − A = f
5. ( A − B) ∪ (B − A ) = a, b, c, f
6. A ∪ C = a, b, c, d , e, 1, 2, 3
7. A ∩ C =
8. A − C = a, b, c, d , e
9. ( A ∩ C ) ∪ ( A − C ) = a, b, c, d , e
10. ( A ∩ B) × B = ( d , d ), ( d , e), ( d , f ), ( e, d ), ( e, e), ( e, f )
11. ( A × C ) ∩ (B × C ) = ( d , 1), (d , 2), ( d , 3), ( e, 1), ( e, 2), ( e, 3)
Observe that for any sets X and Y it is always true that X ∪ Y = Y ∪ X
and X ∩ Y = Y ∩ X , but in general X − Y = Y − X .
Continuing the example, parts 12–15 below use the interval notation
discussed in Section 1.1, so [2, 5] = x ∈ R : 2 ≤ x ≤ 5 , etc. Sketching these
examples on the number line may help you understand them.
12. [2, 5] ∪ [3, 6] = [2, 6]
13. [2, 5] ∩ [3, 6] = [3, 5]
14. [2, 5] − [3, 6] = [2, 3)
15. [0, 3] − [1, 2] = [0, 1) ∪ (2, 3] Sets 18
A∪B
B A−B
A∩B A (a) (b) (c) (d) Figure 1.5. The union, intersection and diﬀerence of sets A and B
Example 1.6 Let A = ( x, x2 ) : x ∈ R be the graph of the equation y = x2
and let B = ( x, x + 2) : x ∈ R be the graph of the equation y = x + 2. These sets
are subsets of R2 . They are sketched together in Figure 1.5(a). Figure 1.5(b)
shows A ∪ B, the set of all points ( x, y) that are on one (or both) of the two
graphs. Observe that A ∩ B = (−1, 1), (2, 4) consists of just two elements,
the two points where the graphs intersect, as illustrated in Figure 1.5(c).
Figure 1.5(d) shows A − B, which is the set A with “holes” where B crossed it.
In set builder notation, we could write A ∪ B = ( x, y) : x ∈ R, y = x2 or y = x + 2
and A − B = ( x, x2 ) : x ∈ R − − 1, 2 .
Exercises for Section 1.5
1. Suppose A = 4, 3, 6, 7, 1, 9 , B = 5, 6, 8, 4 and C = 5, 8, 4 . Find:
(a) A ∪ B (d) A − C (g) B ∩ C (b) A ∩ B (e) B − A (h) B ∪ C (c) A − B (f) A ∩ C (i) C − B 2. Suppose A = 0, 2, 4, 6, 8 , B = 1, 3, 5, 7 and C = 2, 8, 4 . Find:
(a) A ∪ B (d) A − C (g) B ∩ C (b) A ∩ B (e) B − A (h) C − A (c) A − B (f) A ∩ C (i) C − B 3. Suppose A = 0, 1 and B = 1, 2 . Find:
(a) ( A × B) ∩ (B × B) (d) ( A ∩ B) × A (g) P ( A ) − P (B) (b) ( A × B) ∪ (B × B) (e) ( A × B) ∩ B (h) P ( A ∩ B) (c) ( A × B) − (B × B) (f) P ( A ) ∩ P (B) (i) P ( A × B) 4. Suppose A = b, c, d and B = a, b . Find:
(a) ( A × B) ∩ (B × B) (d) ( A ∩ B) × A (g) P ( A ) − P (B) (b) ( A × B) ∪ (B × B) (e) ( A × B) ∩ B (h) P ( A ∩ B) (c) ( A × B) − (B × B) (f) P ( A ) ∩ P (B) (i) P ( A ) × P (B) Complement 19 5. Sketch the sets X = [1, 3] × [1, 3] and Y = [2, 4] × [2, 4] on the plane R2 . On separate
drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X . (Hint: X and Y are
Cartesian products of intervals. You may wish to review how you drew sets
like [1, 3] × [1, 3] in the exercises for Section 1.2.)
6. Sketch the sets X = [−1, 3] × [0, 2] and Y = [0, 3] × [1, 4] on the plane R2 . On
separate drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X .
7. Sketch the sets X = ( x, y) ∈ R2 : x2 + y2 ≤ 1 and Y = ( x, y) ∈ R2 : x ≥ 0 on R2 . On
separate drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X .
8. Sketch the sets X = ( x, y) ∈ R2 : x2 + y2 ≤ 1 and Y = ( x, y) ∈ R2 : −1 ≤ y ≤ 0 on R2 .
On separate drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X .
9. Is the statement (R × Z) ∩ (Z × R) = Z × Z true or false? What about the statement
(R × Z) ∪ (Z × R) = R × R?
10. Do you think the statement (R − Z) × N = (R × N) − (Z × N) is true, or false? Justify. 1.6 Complement
This section introduces yet another set operation, called the set complement.
The deﬁnition requires the idea of a universal set, which we now discuss.
When dealing with a set, we almost always regard it as a subset
of some larger set. For example, consider the set of prime numbers
P = 2, 3, 5, 7, 11, 13, . . . . If asked to name some things that are not in P , we
might mention some composite numbers like 4 or 6 or 423. It probably
would not occur to us to say that Vladimir Putin is not in P . True, Vladimir
Putin is not in P , but he lies entirely outside of the discussion of what is
a prime number and what is not. We have an unstated assumption that
P ⊆N because N is the most natural setting in which to discuss prime numbers.
In this context, anything not in P should still be in N. This larger set N is
called the universal set or universe for P .
Almost every useful set in mathematics can be regarded as having
some natural universal set. For instance, the unit circle is the set C =
( x, y) ∈ R2 : x2 + y2 = 1 , and since all these points are in the plane R2 it is
natural to regard R2 as the universal set for C . In the absence of speciﬁcs,
if A is a set, then its universal set is often denoted as U . We are now
ready to deﬁne the complement operation.
Deﬁnition 1.6 Suppose A is a set with a universal set U . The complement of A , denoted A , is the set A = U − A . Sets 20
Example 1.7 If P is the set of prime numbers, then
P = N − P = 1, 4, 6, 8, 9, 10, 12, . . . . Thus P is the set of composite numbers.
Example 1.8 Let A = ( x, x2 ) : x ∈ R be the graph of the equation y = x2 .
Figure 1.6(a) shows A in its universal set R2 . The complement of A is A =
R2 − A = ( x, y) R2 : y = x2 , illustrated by the shaded area in Figure 1.6(b).
A (a) A (b) Figure 1.6. A set and its complement
Exercises for Section 1.6
1. Let A = 4, 3, 6, 7, 1, 9 and B = 5, 6, 8, 4 have universal set U = n ∈ Z : 0 ≤ n ≤ 10 .
(a) A (d) A ∪ A (g) A − B (b) B (e) A − A (h) A ∩ B (c) A ∩ A (f) A − B (i) A ∩ B 2. Let A = 0, 2, 4, 6, 8 and B = 1, 3, 5, 7 have universal set U = n ∈ Z : 0 ≤ n ≤ 8 .
(a) A (d) A ∪ A (g) A ∩ B (b) B (e) A − A (h) A ∩ B (c) A ∩ A (f) A ∪ B (i) A × B 3. Sketch the set X = [1, 3] × [1, 2] on the plane R2 . On separate drawings, shade in
the sets X , and X ∩ ([0, 2] × [0, 3]).
4. Sketch the set X = [−1, 3] × [0, 2] on the plane R2 . On separate drawings, shade
in the sets X , and X ∩ ([−2, 4] × [−1, 3]).
5. Sketch the set X = ( x, y) ∈ R2 : 1 ≤ x2 + y2 ≤ 4 on the plane R2 . On a separate
drawing, shade in the set X .
6. Sketch the set X = ( x, y) ∈ R2 : y < x2 on R2 . Shade in the set X . Venn Diagrams 21 1.7 Venn Diagrams
In thinking about sets, it is sometimes helpful to draw informal, schematic
diagrams of them. In doing this we often represent a set with a circle
(or oval), which we regard as enclosing all the elements of the set. Such
diagrams can illustrate how sets combine using various operations. For
example, Figures 1.7(a–c) show two sets A and B which overlap in a
middle region. The sets A ∪ B, A ∩ B and A − B are shaded. Such graphical
representations of sets are called Venn diagrams, after their inventor,
British logician John Venn, 1834–1923. A∪B A∩B
B A A−B
B A (a) B A (b) (c) Figure 1.7. Venn diagrams for two sets
Though you are not likely to draw Venn diagrams as a part of a proof
of any theorem, you will probably ﬁnd them to be useful “scratch work”
devices that help you to understand how sets combine, and to develop
strategies for proving certain theorems or solving certain problems. The
remainder of this section uses Venn diagrams to explore how three sets
can be combined using ∪ and ∩.
Let’s begin with the set A ∪ B ∪ C . Our deﬁnitions suggest this should
consist of all elements which are in one or more of the sets A , B and
C . Figure 1.8(a) shows a Venn diagram for this. Similarly, we think of
A ∩ B ∩ C as all elements common to each of A , B and C , so in Figure 1.8(b)
the region belonging to all three sets is shaded.
C C B A B A A∪B∪C A∩B∩C (a) (b) Figure 1.8. Venn diagrams for three sets Sets 22 We can also think of A ∩ B ∩ C as the two-step operation ( A ∩ B) ∩ C . In
this expression the set A ∩ B is represented by the region common to both
A and B, and when we intersect this with C we get Figure 1.8(b). This is
a visual representation of the fact that A ∩ B ∩ C = ( A ∩ B) ∩ C . Similarly we
have A ∩ B ∩ C = A ∩ (B ∩ C ). Likewise, A ∪ B ∪ C = ( A ∪ B) ∪ C = A ∪ (B ∪ C ).
Notice that in these examples, where the expression either contains
only the symbol ∪ or only the symbol ∩, the placement of the parentheses
is irrelevant, so we are free to drop them. It is analogous to the situations
in algebra involving expressions (a + b) + c = a + ( b + c) or (a · b) · c = a · ( b · c).
We tend to drop the parentheses and write simply a + b + c or a · b · c. By
contrast, in an expression like (a + b) · c the parentheses are absolutely
essential because (a + b) · c and a + ( b · c) are generally not equal.
Now let’s use Venn diagrams to help us understand the expressions
( A ∪ B) ∩ C and A ∪ (B ∩ C ) which use a mix of ∪ and ∩. Figure 1.9 shows
how to draw a Venn diagram for ( A ∪ B) ∩ C . In the drawing on the left, the
set A ∪ B is shaded with horizontal lines while C is shaded with vertical
lines. Thus the set ( A ∪ B) ∩ C is represented by the cross hatched region
where A ∪ B and C overlap. The superﬂuous shadings are omitted in the
drawing on the right showing the set ( A ∪ B) ∩ C .
C C B A B A Figure 1.9. How to make a Venn diagram for ( A ∪ B) ∩ C .
Now think about A ∪ (B ∩ C ). In Figure 1.10 the set A is shaded with
horizontal lines, and B ∩C is shaded with vertical lines. The union A ∪(B ∩C )
is represented by the totality of all shaded regions, as shown on the right.
C A C B A B Figure 1.10. How to make a Venn diagram for A ∪ (B ∩ C ). Venn Diagrams 23 Compare the diagrams for ( A ∪ B) ∩ C and A ∪ (B ∩ C ) in Figures 1.9 and
1.10. The fact that the diagrams are diﬀerent indicates that ( A ∪ B) ∩ C =
A ∪ (B ∩ C ) in general. Thus an expression such as A ∪ B ∩ C is absolutely
meaningless because we can’t tell whether it means ( A ∪ B) ∩ C or A ∪ (B ∩ C ).
In summary, Venn diagrams have helped us understand the following.
Important Points:
• If an expression involving sets uses only ∪, then parentheses are optional.
• If an expression involving sets uses only ∩, then parentheses are optional.
• If an expression uses both ∪ and ∩, then parentheses are essential. In the next section we will study types of expressions that use only ∪
or only ∩. These expressions will not require the use of parentheses. Exercises for Section 1.7
1. Draw a Venn diagram for A .
2. Draw a Venn diagram for B − A .
3. Draw a Venn diagram for ( A − B) ∩ C .
4. Draw a Venn diagram for ( A ∪ B) − C .
5. Draw Venn diagrams for A ∪ (B ∩ C ) and ( A ∪ B) ∩ ( A ∪ C ). Based on your drawings,
do you think A ∪ (B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )?
6. Draw Venn diagrams for A ∩ (B ∪ C ) and ( A ∩ B) ∩ ( A ∪ C ). Based on your drawings,
do you think A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )?
7. Suppose sets A and B are in a universal set U . Draw Venn diagrams for A ∩ B
and A ∪ B. Based on your drawings, do you think it’s true that A ∩ B = A ∪ B?
8. Suppose sets A and B are in a universal set U . Draw Venn diagrams for A ∪ B
and A ∩ B. Based on your drawings, do you think it’s true that A ∪ B = A ∩ B?
9. Draw a Venn diagram for ( A ∩ B) − C .
10. Draw a Venn diagram for ( A − B) ∪ C .
Following are Venn diagrams for expressions involving sets A , B and C . Write the
corresponding expression.
C 11. A C B 12. A C B 13. A C B 14. A B Sets 24
1.8 Indexed Sets When a mathematical problem involves lots of sets it is often convenient to
keep track of them by using subscripts (also called indices). Thus instead
of denoting three sets as A , B and C , we might instead write them as A 1 , A 2
and A 3 . These are called indexed sets.
Although we deﬁned union and intersection to be operations that
combine two sets, you by now have no diﬃculty forming unions and
intersections of three or more sets. (For instance, in the previous section
we drew Venn diagrams for the intersection and union of three sets.)
But let’s take a moment to write down careful deﬁnitions. Given sets
A 1 , A 2 , . . . , A n , the set A 1 ∪ A 2 ∪ A 3 ∪ · · · ∪ A n consists of everything that is
in at least one of the sets A i . Likewise A 1 ∩ A 2 ∩ A 3 ∩ · · · ∩ A n consists of
everything that is common to all of the sets A i . Here is a careful deﬁnition.
Deﬁnition 1.7 Suppose A 1 , A 2 , . . . , A n are sets. Then
A1 ∪ A2 ∪ A3 ∪ · · · ∪ A n = x : x ∈ A i for at least one set A i , for 1 ≤ i ≤ n , A1 ∩ A2 ∩ A3 ∩ · · · ∩ A n = x : x ∈ A i for every set A i , for 1 ≤ i ≤ n . But if the number n of sets is large, these expressions can get messy.
To overcome this, we now develop some notation that is akin to sigma
notation. You already know that sigma notation is a convenient symbolism
for expressing sums of many numbers. Given numbers a 1 , a 2 , a 3 , . . . , a n ,
then
n a i = a1 + a2 + a3 + · · · + a n .
i =1 Even if the list of numbers is inﬁnite, the sum
∞ a i = a1 + a2 + a3 + · · · + a i + · · ·
i =1 is often still meaningful. The notation we are about to introduce is very
similar to this. Given sets A 1 , A 2 , A 3 , . . . , A n , we deﬁne
n n A i = A1 ∪ A2 ∪ A3 ∪ · · · ∪ A n A i = A1 ∩ A2 ∩ A3 ∩ · · · ∩ A n . and i =1 i =1 Example 1.9 Suppose A 1 = 0, 2, 5 , A 2 = 1, 2, 5 and A 3 = 2, 5, 7 . Then
3 3 A i = A 1 ∪ A 2 ∪ A 3 = 0, 1, 2, 5, 7
i =1 A i = A 1 ∩ A 2 ∩ A 3 = 2, 5 . and
i =1 Indexed Sets 25 This notation is also used when the list of sets A 1 , A 2 , A 3 , . . . is inﬁnite:
∞ A i = A1 ∪ A2 ∪ A3 ∪ · · · = x : x ∈ A i for at least one set A i with 1 ≤ i . A i = A1 ∩ A2 ∩ A3 ∩ · · · = x : x ∈ A i for every set A i with 1 ≤ i . i =1
∞
i =1 Example 1.10 This example involves the following inﬁnite list of sets.
A 2 = − 2, 0, 2 , A 1 = − 1, 0, 1 ,
∞ A i = − i , 0, i , · · · ∞ A i = Z, and Observe that A 3 = − 3, 0, 3 , · · · ,
Ai = 0 . i =1 i =1 Here is a useful twist on our new notation. We can write
3 Ai =
i =1 Ai,
i ∈{1,2,3} as this takes the union of the sets A i for i = 1, 2, 3. Likewise:
3 Ai i ∈{1,2,3} Ai = i =1
∞ Ai
i =1 Ai = i =1
∞ = Ai
i ∈N Ai
i ∈N Here we are taking the union or intersection of a collection of sets A i
where i is an element of some set, be it 1, 2, 3 or N. In general, the way
this works is that we will have a collection of sets A i for i ∈ I , where I is
the set of possible subscripts. The set I is called an index set.
It is important to realize that the set I need not even consist of integers.
(We could subscript with letters or real numbers, etc.) Since we are
programmed to think of i as an integer, let’s make a slight notational
change: we use α, not i , to stand for an element of I . Thus we are dealing
with a collection of sets A α for α ∈ I . This leads to the following deﬁnition.
Deﬁnition 1.8 If we have a set A α for every α in some index set I , then
α∈ I
α∈ I Aα = x : x ∈ A α for at least one set A α with α ∈ I Aα = x : x ∈ A α for every set A α with α ∈ I . Sets 26 Here the sets A α will be subsets of R2 . Let I = [0, 2] =
x ∈ R : 0 ≤ x ≤ 2 . For each number α ∈ I , let A α = ( x, α) : x ∈ R, 1 ≤ x ≤ 2 . For
instance, given α = 1 ∈ I the set A 1 = ( x, 1) : x ∈ R, 1 ≤ x ≤ 2 is a horizontal
line segment one unit above the x-axis and stretching between x = 1 and
x = 2, as shown in Figure 1.11(a). Likewise A 2 = ( x, 2) : x ∈ R, 1 ≤ x ≤ 2 is
a horizontal line segment 2 units above the x-axis and stretching between
x = 1 and x = 2. A few other of the A α are shown in Figure 1.11(a) but they
can’t all be drawn because there is one A α for each of the inﬁnitely many
numbers α ∈ [0, 2]. The totality of them covers the shaded region in Figure
1.11(b), so this region is the union of all the A α . Since the shaded region
is the set ( x, y) ∈ R2 : 1 ≤ x ≤ 2, 0 ≤ y ≤ 2 = [1, 2] × [0, 2], it follows that
Example 1.11 α∈[0,2] A α = [1, 2] × [0, 2]. Likewise, since there is no point ( x, y) that is in every set A α , we have
α∈[0,2] Aα = . y y
A2 2 A 2
2 A1 1 1 2 1 A 0. 5
A 0.25
x α∈[0,2] 1 (a) Aα 2 x (b) Figure 1.11. The union of an indexed collection of sets One ﬁnal comment. Observe that A α = [1, 2] × α , so the above expressions can be written as
α∈[0,2] [1, 2] × α = [1, 2] × [0, 2] and [1, 2] × α = . α∈[0,2] Indexed Sets 27 Exercises for Section 1.8
1. Suppose A 1 = a, b, d , e, g, f , A 2 = a, b, c, d , A 3 = b, d , a and A 4 = a, b, h .
4 4 (a) (b) Ai =
i =1 Ai =
i =1 A1
A
2. Suppose
2
A3 =
=
= 0, 2, 4, 8, 10, 12, 14, 16, 18, 20, 22, 24 ,
0, 3, 6, 9, 12, 15, 18, 21, 24 ,
0, 4, 8, 12, 16, 20, 24 . 3 3 (a) (b) Ai =
i =1 Ai =
i =1 3. For each n ∈ N, let A n = 0, 1, 2, 3, . . . , n .
(a) (b) Ai = i ∈N Ai = i ∈N 4. For each n ∈ N, let A n = − 2n, 0, 2n .
(a) Ai = (b) [ i , i + 1] = (b) [0, i + 1] = (b) R × [ i , i + 1] = (b) α × [0, 1] = (b) i ∈N 5. (a)
i ∈N 6. (a)
i ∈N 7. (a)
i ∈N 8. (a)
α∈R 9. (a) i ∈N 12. If α∈ I (b) α × [0, 1] = (b)
X ∈P (N) X= [ x, 1] × [0, x2 ] =
x∈[0,1] Aα ⊆
Aα = R × [ i , i + 1] = α∈R [ x, 1] × [0, x2 ] = α∈ I [0, i + 1] = i ∈N x∈[0,1] 11. Is [ i , i + 1] = i ∈N X= X ∈P (N) 10. (a) Ai = i ∈N α∈ I α∈ I A α always true for any collection of sets A α with index set I ?
A α , what do you think can be said about the relationships between the sets A α ? 13. If J ⊆ I , does it follow that α∈ J Aα ⊆ α∈ I A α ? What about α∈ J Aα ⊆ α∈ I Aα? CHAPTER 2 Logic ogic is a systematic way of thinking that allows us to deduce new information from old information. You use logic informally in everyday
life, and certainly also in doing mathematics. For example, suppose you
are working with a certain circle, call it “Circle X,” and you have available
the following two pieces of information. L 1. Circle X has radius equal to 3.
2. If any circle has radius r , then its area is π r 2 square units. You have no trouble putting these two facts together to get:
3. Circle X has area 9π square units. In doing this you are using logic to combine existing information to
produce new information. Since a major objective in mathematics is to
deduce new information, logic must play a fundamental role. This chapter
is intended to give you a suﬃcient mastery of logic.
It is important to realize that logic is a process of deducing information
correctly, not just deducing correct information. For example, suppose we
were mistaken and Circle X actually had a radius of 4, not 3. Let’s look at
our exact same argument again.
1. Circle X has radius equal to 3.
2. If any circle has radius r , then its area is π r 2 square units.
3. Circle X has area 9π square units. The sentence “Circle X has radius equal to 3.” is now untrue, and so is our
conclusion “Circle X has area 9π square units.” But the logic is perfectly
correct; the information was combined correctly, even if some of it was
false. This distinction between correct logic and correct information is
signiﬁcant because it is often important to follow the consequences of an
incorrect assumption. Ideally, we want both our logic and our information
to be correct, but the point is that they are diﬀerent things. Statements 29 In proving theorems, we apply logic to information that is considered
obviously true (such as “Any two points determine exactly one line.”) or is
already known to be true (e.g. the Pythagorean theorem). If our logic is
correct, then anything we deduce from such information will also be true.
(Or at least as true as the “obviously true” information we began with.)
2.1 Statements
The study of logic begins with statements. A statement is a sentence
or a mathematical expression that is either deﬁnitely true or deﬁnitely
false. You can think of statements as pieces of information that are either
correct or incorrect. Thus statements are pieces of information that we
might apply logic to in order to produce other pieces of information (which
are also statements).
Example 2.1 Here are some examples of statements. They are all true. If a circle has radius r , then its area is π r 2 square units.
Every even number is divisible by 2.
2∈Z
2∉Z
N⊆Z The set {0, 1, 2} has three elements.
Some right triangles are isosceles.
Example 2.2 Here are some additional statements. They are all false.
All right triangles are isosceles.
5=2
2∉R
Z⊆N {0, 1, 2} ∩ N = Example 2.3 Here we pair sentences or expressions that are not statements with similar expressions that are statements.
NOT Statements: Statements: Add 5 to both sides. Adding 5 to both sides of x − 5 = 37 gives x = 42. Z 42 ∈ Z 42 42 is not a number. What is the solution of 2 x = 84? The solution of 2 x = 84 is 42. Logic 30 Example 2.4 We will often use the letters P , Q , R and S to stand for
speciﬁc statements. When more letters are needed we can use subscripts.
Here are more statements, designated with letters. You decide which of
them are true and which are false.
P : For every integer n > 1, the number 2n − 1 is prime.
Q : Every polynomial of degree n has at most n roots.
R : The function f ( x) = x2 is continuous.
S1 : Z ⊆
S 2 : {0, −1, −2} ∩ N = Designating statements with letters (as was done above) is a very useful
shorthand. In discussing a particular statement, such as “The function
f ( x) = x2 is continuous,” it is convenient to just refer to it as R to avoid
having to write or say it many times.
Statements can contain variables. Here is an example.
P : If an integer x is a multiple of 6, then x is even. This is a sentence that is true. (All multiples of 6 are even, so no matter
which multiple of 6 the integer x happens to be, it is even.) Since the
sentence P is deﬁnitely true, it is a statement. When a sentence or
statement P contains a variable such as x, we sometimes denote it as P ( x)
to indicate that it is saying something about x. Thus the above statement
can be expressed as
P ( x) : If an integer x is a multiple of 6, then x is even. A statement or sentence involving two variables might be denoted
P ( x, y), and so on. It is quite possible for a sentence containing variables to not be a
statement. Consider the following example.
Q ( x) : The integer x is even. Is this a statement? Whether it is true or false depends on just which
integer x is. It is true if x = 4 and false if x = 7, etc. But without any
stipulations on the value of x it is impossible to say whether Q ( x) is true or
false. Since it is neither deﬁnitely true nor deﬁnitely false, Q ( x) is cannot
be a statement. A sentence such as this, whose truth depends on the value
of some variable, is called an open sentence. The variables in an open
sentence (or statement) can represent any type of entity, not just numbers.
Here is an open sentence where the variables are functions. Statements 31 R ( f , g) : The function f is the derivative of the function g. This open sentence is true if f ( x) = 2 x and g( x) = x2 . It is false if f ( x) = x3
and g( x) = x2 , etc. We point out that a sentence such as R ( f , g) (that
involves variables) can be denoted either as R ( f , g) or just R . We use the
expression R ( f , g) when we want to emphasize that the sentence involves
variables.
We will have more to say about open sentences later, but for now let’s
return to statements.
Statements are everywhere in mathematics. Any result or theorem
that has been proved true is a statement. The quadratic formula and the
Pythagorean theorem are both statements:
P : The solutions of the equation ax2 + bx + c = 0 are x =
Q: − b ± b2 − 4ac
.
2a If a right triangle has legs of lengths a and b and hypotenuse of
length c, then a2 + b2 = c2 . Here is a very famous statement, so famous, in fact, that it has a name.
It is called Fermat’s Last Theorem after Pierre Fermat, a seventeenth
century French mathematician who scribbled it in the margin of a notebook.
R : For all numbers a, b, c, n ∈ N with n > 2, it is the case that a n + b n = c n . Fermat believed this statement was true. He noted that he could prove
it was true, except his notebook’s margin was too narrow to contain his
proof. It is doubtful that he really had a correct proof in mind, for after his
death generations of brilliant mathematicians tried unsuccessfully to prove
that his statement was true (or false). Finally, in 1993, Andrew Wiles of
Princeton University announced that he had devised a proof. Wiles had
worked on the problem for over seven years, and his proof runs through
hundreds of pages. The moral of this story is that some true statements
are not obviously true.
Here is another statement famous enough to be named. It was ﬁrst
posed in the eighteenth century by the German mathematician Christian
Goldbach, and thus is called the Goldbach Conjecture:
S : Every even integer greater than 2 is a sum of two prime numbers. You must agree that S is either true or false. It appears to be true, because
when you examine even numbers that are bigger than 2, they seem to
be sums of two primes: 4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, 10 = 5 + 5, 12 = 5 + 7, 32 Logic 100 = 17 + 83, and so on. But that’s not to say there isn’t some large even
number that’s not the sum of two primes. If such a number exists, then S
is false. The thing is, in the over 260 years since Goldbach ﬁrst posed this problem, no one has been able to determine whether it’s true or false. But
since it is clearly either true or false, S is a statement.
This book is about the methods that can be used to prove that S (or
any other statement) is true or false. To prove that a statement is true,
we start with obvious statements (or other statements that have been
proven true) and use logic to deduce more and more complex statements
until ﬁnally we obtain a statement such as S . Of course some statements
are more diﬃcult to prove than others, and S appears to be notoriously
diﬃcult; we will concentrate on statements that are easier to prove.
But the point is this: In proving that statements are true, we use logic
to help us understand statements and to combine pieces of information
to produce new pieces of information. In the next several sections we
explore some standard ways that statements can be combined to form new
statements.
Exercises for Section 2.1
Decide whether or not the following are statements. In the case of a statement,
say if it is true or false, if possible.
1. Every real number is an even integer.
2. Every even integer is a real number.
3. If x and y are real numbers and 5 x = 5 y, then x = y.
4. Sets Z and N.
5. Sets Z and N are inﬁnite.
6. Some sets are ﬁnite.
7. The derivative of any polynomial of degree 5 is a polynomial of degree 6.
8. N ∉ P (N).
9. cos( x) = −1
10. (R × N) ∩ (N × R) = N × N
11. The integer x is a multiple of seven.
12. If the integer x is a multiple of seven, then it is divisible by seven.
13. Either x is a multiple of seven, or it is not.
14. Call me Ishmael.
15. In the beginning, God created the heaven and the earth. And, Or, Not 33 2.2 And, Or, Not
The word “and” can be used to combine two statements to form a new
statement. Consider for example the following sentence.
R 1 : The number 2 is even and the number 3 is odd. We recognize this as a true statement, based on our common-sense understanding of the meaning of the word “and.” Notice that R1 is made up of
two simpler statements:
P : The number 2 is even.
Q : The number 3 is odd. These are joined together by the word “and” to form the more complex
statement R1 . The statement R1 asserts that P and Q are both true. Since
both P and Q are in fact true, the statement R1 is also true.
Had one or both of P and Q been false, then R1 would be false. For
instance, each of the following statements is false.
R 2 : The number 1 is even and the number 3 is odd.
R 3 : The number 2 is even and the number 4 is odd.
R 4 : The number 3 is even and the number 2 is odd. From these examples we see that any two statements P and Q can
be combined to form a new statement “P and Q .” In the spirit of using
letters to denote statements, we now introduce the special symbol ∧ to
stand for the word “and.” Thus if P and Q are statements, P ∧ Q stands
for the statement “P and Q .” The statement P ∧ Q is true if both P and Q
are true; otherwise it is false. This is summarized in the following table,
called a truth table.
P Q P ∧Q T T T T F F F T F F F F In this table, T stands for “True,” and F stands for “False.” (T and F are
called truth values.) Each line lists one of the four possible combinations
or truth values for P and Q , and the column headed by P ∧ Q tells whether
the statement P ∧ Q is true or false in each case. Logic 34 Statements can also be combined using the word “or.” Consider for
example the following four statements.
S1 :
S2 :
S3 :
S4 : The
The
The
The number
number
number
number 2
1
2
3 is
is
is
is even
even
even
even or
or
or
or the
the
the
the number
number
number
number 3
3
4
2 is
is
is
is odd.
odd.
odd.
odd. In mathematics, the assertion “P or Q ” is always understood to mean that
one or both of P and Q is true. Thus statements S1 , S2 , S3 are all true,
while S4 is false. The symbol ∨ is used to stand for the word “or.” So if P
and Q are statements, P ∨ Q represents the statement “P or Q .” Here is
the truth table.
P Q P ∨Q T T T T F T F T T F F F It is important to be aware that the meaning of “or” expressed in
the above table diﬀers from the way it is sometimes used in everyday
conversation. For example, suppose a university oﬃcial makes the following
threat:
You pay your tuition or you will be withdrawn from school.
You understand that this means that either you pay your tuition or you
will be withdrawn from school, but not both. In mathematics we never use
the word “or” in such a sense. For us “or” means exactly what is stated
in the table for ∨. Thus P ∨ Q being true means one or both of P and Q
is true. If we ever need to express the fact that exactly one of P and Q is
true, we use one of the following constructions.
P or Q , but not both. Either P or Q .
If the university oﬃcial were a mathematician, he might have qualiﬁed
his statement in one of the following ways.
Pay your tuition or you will be withdrawn from school, but not both.
Either you pay your tuition or you will be withdrawn from school. And, Or, Not 35 To conclude this section, we mention another way of obtaining new
statements from old ones. Given any statement P , we can form the new
statement “It is not true that P .” For example consider the following
statement.
The number 2 is even.
This statement is true. Now change it by inserting the words “It is not
true that” at the beginning:
It is not true that the number 2 is even.
This new statement is false.
For another example, starting with the false statement “2 ∈ ,” we get
the true statement “It is not true that 2 ∈ .”
We use the symbol ∼ to stand for the words “It’s not true that,” so
∼ P means “It’s not true that P .” We often read ∼ P simply as “not P .”
Unlike ∧ and ∨, which combine two statements, the symbol ∼ just alters
a single statement. Thus its truth table has just two lines, one for each
possible truth value of P .
P ∼P T F F T The statement ∼ P is called the negation of P . The negation of a
speciﬁc statement can be expressed in numerous ways. Consider
P : The number 2 is even. Here are several ways of expressing its negation.
∼ P : It’s not true that the number 2 is even.
∼ P : It is false that the number 2 is even.
∼ P : The number 2 is not even. In this section we’ve learned how to combine or modify statements with
the operations ∧, ∨ and ∼. Of course we can also apply these operations
to open sentences or a mixture of open sentences and statements. For
example, ( x is an even integer) ∧ (3 is an odd integer) is an open sentence
that is a combination of an open sentence and a statement. Logic 36 Exercises for Section 2.2
Express each statement or open sentence in one of the forms P ∧ Q , P ∨ Q , or ∼ P .
Be sure to also state exactly what statements P and Q stand for.
1. The number 8 is both even and a power of 2.
2. The matrix A is not invertible.
3. x = y 4. x < y 5. y ≥ x 6. There is a quiz scheduled for Wednesday or Friday.
7. The number x equals zero, but the number y does not.
8. At least one of the numbers x and y equals 0.
9. x ∈ A − B 10. x ∈ A ∪ B 11. A ∈ X ∈ P (N) : | X | < ∞ 12. Happy families are all alike, but each unhappy family is unhappy in its own
way. (Leo Tolstoy, Anna Karenina)
13. Human beings want to be good, but not too good, and not quite all the time.
(George Orwell)
14. A man should look for what is, and not for what he thinks should be.
(Albert Einstein) 2.3 Conditional Statements
There is yet another way to combine two statements. Suppose we have in
mind a speciﬁc integer a. Consider the following statement about a.
R : If integer a is a multiple of 6, then a is divisible by 2. We immediately spot this as a true statement based on our knowledge of
integers and the meanings of the words “if” and “then.” If integer a is a
multiple of 6, then a is even, so therefore a is divisible by 2. Notice that R
is built up from two simpler statements:
P : Integer a is a multiple of 6.
Q : Integer a is divisible by 2.
R : If P , then Q . In general, given any two statements P and Q whatsoever, we can form
the new statement “If P , then Q .” This is written symbolically as P ⇒ Q
which we read as “If P , then Q ,” or “P implies Q .” Like ∧ and ∨, the symbol
⇒ has a very speciﬁc meaning. When we assert that the statement P ⇒ Q
is true, we mean that if P is true then Q must also be true. (In other words
we mean that the condition P being true forces Q to be true.) A statement
of form P ⇒ Q is called a conditional statement because it means Q will
be true under the condition that P is true. Conditional Statements 37 You can think of P ⇒ Q as being a promise that whenever P is true, Q
will be true also. There is only one way this promise can be broken (i.e.
be false) and that is if P is true but Q is false. Thus the truth table for
the promise P ⇒ Q is as follows.
P Q P ⇒Q T T T T F F F T T F F T Perhaps you are bothered by the fact that P ⇒ Q is true in the last two
lines of this table. Here’s an example to convince you that the table is
correct. Suppose your professor makes the following promise:
If you pass the ﬁnal exam, then you will pass the course.
Your professor is making the promise
(You pass the exam) ⇒ (You pass the course).
Under what circumstances did she lie? There are four possible scenarios,
depending on whether or not you passed the exam and whether or not you
passed the course. These scenarios are tallied in the following table.
You pass exam You pass course
T T (You pass exam) ⇒ (You pass course)
T T F F F T T F F T The ﬁrst line describes the scenario where you pass the exam and you
pass the course. Clearly the professor kept her promise, so we put a T in
the third column to indicate that she told the truth. In the second line,
you passed the exam but your professor gave you a failing grade in the
course. In this case she broke her promise, and the F in the third column
indicates that what she said was untrue.
Now consider the third row. In this scenario you failed the exam but
still passed the course. How could that happen? Maybe your professor felt
sorry for you. But that doesn’t make her a liar. Her only promise was that
if you passed the exam then you would pass the course. She did not say Logic 38 passing the exam was the only way to pass the course. Since she didn’t
lie, then she told the truth, so there is a T in the third column.
Finally look at the fourth row. In that scenario you failed the exam
and you failed the course. Your professor did not lie; she did exactly what
she said she would do. Hence the T in the third column.
In mathematics, whenever we encounter the construction “If P , then
Q ” it means exactly what the truth table for ⇒ expresses. But of course
there are other grammatical constructions that also mean P ⇒ Q . Here is
a summary of the main ones.
If P , then Q .
Q if P .
Q whenever P .
Q , provided that P .
Whenever P , then also Q .
P is a suﬃcient condition for Q .
For Q , it is suﬃcient that P .
Q is a necessary condition for P .
For P , it is necessary that Q .
P only if Q . P ⇒Q These can all be used in the place of (and mean exactly the same thing as)
“If P , then Q .” You should analyze the meaning of each one and convince
yourself that it captures the meaning of P ⇒ Q . For example, P ⇒ Q means
the condition of P being true is enough (i.e. suﬃcient) to make Q true;
hence “P is a suﬃcient condition for Q .”
The wording can be tricky. Often an everyday situation involving a
conditional statement can help clarify it. For example, consider your
professor’s promise:
(You pass the exam) ⇒ (You pass the course)
This means that your passing the exam is a suﬃcient (though perhaps
not necessary) condition for your passing the course. Thus your professor
might just as well have phrased her promise in one of the following ways.
Passing the exam is a suﬃcient condition for passing the course.
For you to pass the course, it is suﬃcient that you pass the exam.
However when we want to say “If P , then Q ” in everyday conversation,
we do not normally express this as “Q is a necessary condition for P ” or
“P only if Q .” But such constructions are not uncommon in mathematics.
To understand why they make sense, notice that P ⇒ Q being true means Biconditional Statements 39 that it’s impossible that P is true but Q is false, so in order for P to be
true it is necessary that Q is true; hence “Q is a necessary condition for
P .” And this means that P can only be true if Q is true, i.e. “P only if Q .”
Exercises for Section 2.3
Without changing their meanings, convert each of the following sentences into a
sentence having the form “If P , then Q .”
1. A matrix is invertible provided that its determinant is not zero.
2. For a function to be continuous, it is suﬃcient that it is diﬀerentiable.
3. For a function to be integrable, it is necessary that it is continuous.
4. A function is rational if it is a polynomial.
5. An integer is divisible by 8 only if it is divisible by 4.
6. Whenever a surface has only one side, it is non-orientable.
7. A series converges whenever it converges absolutely.
8. A geometric series with radius r converges if | r | < 1.
9. A function is integrable provided the function is continuous.
10. The discriminant is negative only if the quadratic equation has no real solutions.
11. You fail only if you stop writing. (Ray Bradbury)
12. People will generally accept facts as truth only if the facts agree with what
they already believe. (Andy Rooney)
13. Whenever people agree with me I feel I must be wrong. (Oscar Wilde) 2.4 Biconditional Statements
It is important to understand that P ⇒ Q is not the same as Q ⇒ P . To see
why, suppose that a is some integer and consider the statements
(a is a multiple of 6)
(a is divisible by 2) ⇒ (a is divisible by 2)
⇒ (a is a multiple of 6). The ﬁrst statement asserts that if a is a multiple of 6 then a is divisible by
2. This is clearly true, for any multiple of 6 is even and therefore divisible
by 2. The second statement asserts that if a is divisible by 2 then it is
a multiple of 6. This is not necessarily true, for a = 4 (for instance) is
divisible by 2, yet not a multiple of 6. Therefore the meanings of P ⇒ Q and
Q ⇒ P are in general quite diﬀerent. The conditional statement Q ⇒ P is
called the converse of P ⇒ Q , so a conditional statement and its converse
express entirely diﬀerent things. Logic 40 But sometimes, if P and Q are just the right statements, it can happen
that P ⇒ Q and Q ⇒ P are both necessarily true. For example, consider
the statements
(a is even) ⇒ (a is divisible by 2) (a is divisible by 2) ⇒ (a is even). No matter what value a has, both of these statements are true. Since both
P ⇒ Q and Q ⇒ P are true, it follows that (P ⇒ Q ) ∧ (Q ⇒ P ) is true.
We now introduce a new symbol ⇔ to express the meaning of the
statement (P ⇒ Q ) ∧ (Q ⇒ P ). The expression P ⇔ Q is understood to have
exactly the same meaning as (P ⇒ Q ) ∧ (Q ⇒ P ). According to the previous
section, Q ⇒ P is read as “P if Q ,” and P ⇒ Q can be read as “P only if Q .”
Therefore we pronounce P ⇔ Q as “P if and only if Q .” For example, given
an integer a, we have the true statement
(a is even) ⇔ (a is divisible by 2)
which we can read as “Integer a is even if and only if a is divisible by 2.”
The truth table for ⇔ is shown below. Notice that in the ﬁrst and last
rows, both P ⇒ Q and Q ⇒ P are true (according to the truth table for
⇒), so (P ⇒ Q ) ∧ (Q ⇒ P ) is true, and hence P ⇔ Q is true. However in the
middle two rows one of P ⇒ Q or Q ⇒ P is false, so (P ⇒ Q ) ∧ (Q ⇒ P ) is false,
making P ⇔ Q false.
P Q P ⇔Q T T T T F F F T F F F T Compare the statement R : (a is even) ⇔ (a is divisible by 2) with this
truth table. If a is even then the two statements on either side of ⇔
are true, so according to the table R is true. If a is odd then the two
statements on either side of ⇔ are false, and again according to the table
R is true. Thus R is true no matter what value a has. In general, P ⇔ Q
being true means P and Q are both true or both false.
Not surprisingly, there are many ways of saying P ⇔ Q in English. The
following constructions all mean P ⇔ Q . Truth Tables for Statements 41 P if and only if Q . P is a necessary and suﬃcient condition for Q . P ⇔Q For P it is necessary and suﬃcient that Q . If P , then Q , and conversely. The ﬁrst three of these just combine constructions from the previous
section to express that P ⇒ Q and Q ⇒ P . In the last one, the words “...and
conversely” mean that in addition to “If P , then Q ” being true, the converse
statement “If Q , then P ” is also true.
Exercises for Section 2.4
Without changing their meanings, convert each of the following sentences into a
sentence having the form “P if and only if Q .”
1. For matrix A to be invertible, it is necessary and suﬃcient that det( A ) = 0.
2. If a function has a constant derivative then it is linear, and conversely.
3. If x y = 0 then x = 0 or y = 0, and conversely.
4. If a ∈ Q then 5a ∈ Q, and if 5a ∈ Q then a ∈ Q.
5. For an occurrence to become an adventure, it is necessary and suﬃcient for
one to recount it. (Jean-Paul Sartre) 2.5 Truth Tables for Statements
You should now know the truth tables for ∧, ∨, ∼, ⇒ and ⇔. They should
be internalized as well as memorized. You must understand the symbols
thoroughly, for we now combine them to form more complex statements.
For example, suppose we want to convey that one or the other of P and
Q is true but they are not both true. No single symbol expresses this, but
we could combine them as
(P ∨ Q ) ∧ ∼ (P ∧ Q ) which literally means:
P or Q is true, and it is not the case that both P and Q are true. This statement will be true or false depending on the truth values of P
and Q . In fact we can make a truth table for the entire statement. Begin
as usual by listing the possible true/false combinations of P and Q on four
lines. The statement (P ∨ Q )∧ ∼ (P ∧ Q ) contains the individual statements
(P ∨ Q ) and (P ∧ Q ), so we next tally their truth values in the third and
fourth columns. The ﬁfth column lists values for ∼ (P ∧ Q ), and these Logic 42 are just the opposites of the corresponding entries in the fourth column.
Finally, combining the third and ﬁfth columns with ∧, we get the values
for (P ∨ Q )∧ ∼ (P ∧ Q ) in the sixth column.
P Q (P ∨ Q ) (P ∧ Q ) ∼ (P ∧ Q ) (P ∨ Q )∧ ∼ (P ∧ Q ) T T T T F F T F T F T T F T T F T T F F F F T F This truth table tells us that (P ∨ Q )∧ ∼ (P ∧ Q ) is true precisely when
one but not both of P and Q are true, so it has the meaning we intended.
(Notice that the middle three columns of our truth table are just “helper
columns” and are not necessary parts of the table. In writing truth tables,
you may choose to omit such columns if you are conﬁdent about your work.)
For another example, consider the following familiar statement concerning two real numbers x and y:
The product x y equals zero if and only if x = 0 or y = 0.
This can be modeled as ( x y = 0) ⇔ ( x = 0 ∨ y = 0). If we introduce letters
P , Q and R for the statements x y = 0, x = 0 and y = 0, it becomes P ⇔ (Q ∨ R ).
Notice that the parentheses are necessary here, for without them we
wouldn’t know whether to read the statement as P ⇔ (Q ∨ R ) or (P ⇔ Q ) ∨ R .
Making a truth table for P ⇔ (Q ∨ R ) entails a line for each T /F combination for the three statements P , Q and R . The eight possible combinations
are tallied in the ﬁrst three columns of the following table.
P Q R Q∨R P ⇔ (Q ∨ R ) T T T T T T T F T T T F T T T T F F F F F T T T F F T F T F F F T T F F F F F T We ﬁll in the fourth column using our knowledge of the truth table for
∨. Finally the ﬁfth column is ﬁlled in by combining the ﬁrst and fourth
columns with our understanding of the truth table for ⇔. The resulting
table gives the true/false values of P ⇔ (Q ∨ R ) for all values of P , Q and R . Truth Tables for Statements 43 Notice that when we plug in various values for x and y, the statements
P : x y = 0, Q : x = 0 and R : y = 0 have various truth values, but the statement
P ⇔ (Q ∨ R ) is always true. For example, if x = 2 and y = 3, then P , Q and R
are all false. This scenario is described in the last row of the table, and
there we see that P ⇔ (Q ∨ R ) is true. Likewise if x = 0 and y = 7, then P
and Q are true and R is false, a scenario described in the second line of
the table, where again P ⇔ (Q ∨ R ) is true. There is a simple reason why
P ⇔ (Q ∨ R ) is true for any values of x and y. It is that P ⇔ (Q ∨ R ) represents
( x y = 0) ⇔ ( x = 0 ∨ y = 0), which is a true mathematical statement. It is
absolutely impossible for it to be false.
This may make you wonder about the lines in the table where P ⇔ (Q ∨ R )
is false. Why are they there? The reason is that P ⇔ (Q ∨ R ) can also
represent a false statement. To see how, imagine that at the end of the
semester your professor makes the following promise.
You pass the class if and only if you get an “A” on the ﬁnal or you get
a “B” on the ﬁnal.
This promise has the form P ⇔ (Q ∨ R ), so its truth values are tabulated in
the above table. Imagine it turned out that you got an “A” on the exam
but failed the course. Then surely your professor lied to you. In fact, P is
false, Q is true and R is false. This scenario is reﬂected in the sixth line
of the table, and indeed P ⇔ (Q ∨ R ) is false (i.e. it is a lie).
The moral of this example is that people can lie, but true mathematical
statements never lie.
We close this section with a word about the use of parentheses. The
symbol ∼ is analogous to the minus sign in algebra. It negates the
expression it precedes. Thus ∼ P ∨ Q means (∼ P ) ∨ Q , not ∼ (P ∨ Q ). In
∼ (P ∨ Q ), the value of the entire expression P ∨ Q is negated.
Exercises for Section 2.5
Write a truth table for the logical statements in problems 1 − 9:
1. P ∨ (Q ⇒ R ) 4. ∼ (P ∨ Q ) ∨ (∼ P ) 7. (P ∧ ∼ P ) ⇒ Q 2. (Q ∨ R ) ⇔ (R ∧ Q ) 5. (P ∧ ∼ P ) ∨ Q 8. P ∨ (Q ∧ ∼ R ) 3. ∼ (P ⇒ Q ) 6. (P ∧ ∼ P ) ∧ Q 9. ∼ (∼ P ∨ ∼ Q ) 10. Suppose the statement ((P ∧ Q ) ∨ R ) ⇒ (R ∨ S ) is false. Find the truth values of
P , Q , R and S . (This can be done without a truth table.)
11. Suppose P is false and that the statement (R ⇒ S ) ⇔ (P ∧ Q ) is true. Find the
truth values of R and S . (This can be done without a truth table.) Logic 44
2.6 Logical Equivalence In contemplating the truth table for P ⇔ Q , you probably noticed that
P ⇔ Q is true exactly when P and Q are both true or both false. In other
words, P ⇔ Q is true precisely when at least one of the statements P ∧ Q
or ∼ P ∧ ∼ Q is true. This may tempt us to say that P ⇔ Q means the same
thing as (P ∧ Q ) ∨ (∼ P ∧ ∼ Q ).
To see if this is really so, we can write truth tables for P ⇔ Q and
(P ∧ Q ) ∨ (∼ P ∧ ∼ Q ). In doing this, it is more eﬃcient to put these two
statements into the same table, as follows. (This table has helper columns
for the intermediate expressions ∼ P , ∼ Q , (P ∧ Q ) and (∼ P ∧ ∼ Q ).)
P Q ∼P ∼Q (P ∧ Q ) (∼ P ∧ ∼ Q ) (P ∧ Q ) ∨ (∼ P ∧ ∼ Q ) P ⇔Q T T T F F F T F T T F T F F F F F T F F T F F F F F T T F T T T The table shows that P ⇔ Q and (P ∧ Q ) ∨ (∼ P ∧ ∼ Q ) have the same truth
value, no matter what values P and Q may have. It is as if P ⇔ Q and
(P ∧ Q ) ∨ (∼ P ∧ ∼ Q ) are algebraic expressions that are equal no matter what
is “plugged into” variables P and Q . We express this state of aﬀairs by
writing
P ⇔ Q = ( P ∧ Q ) ∨ (∼ P ∧ ∼ Q ) and saying that P ⇔ Q and (P ∧ Q ) ∨ (∼ P ∧ ∼ Q ) are logically equivalent.
In general, two statements are logically equivalent if their truth
values match up line-for-line in a truth table.
Logical equivalence is important because it can give us diﬀerent (and
potentially useful) ways of looking at the same thing. As an example, the
following table shows that P ⇒ Q is logically equivalent to (∼ Q ) ⇒ (∼ P ).
P Q ∼P ∼Q (∼ Q ) ⇒ (∼ P ) P ⇒Q T T F F T T T F F T F F F T T F T T F F T T T T The fact that P ⇒ Q is logically equivalent to (∼ Q ) ⇒ (∼ P ) is especially
important because so many theorems are statements having the form
P ⇒ Q . As we will see in Chapter 5, proving such a theorem may be easier
if we express it in the logically equivalent form (∼ Q ) ⇒ (∼ P ). Logical Equivalence 45 There are two other pairs of logically equivalent statements that we
will encounter again and again throughout this book and beyond. They
come up so often that they have a special name: DeMorgan’s Laws.
Fact 2.1 (DeMorgan’s Laws) 1. ∼ (P ∧ Q ) = (∼ P ) ∨ (∼ Q ) 2. ∼ (P ∨ Q ) = (∼ P ) ∧ (∼ Q ) The ﬁrst of DeMorgan’s laws is veriﬁed by the following table. You are
asked to verify the second in one of the exercises.
P Q ∼P ∼Q P ∧Q ∼ (P ∧ Q ) (∼ P ) ∨ (∼ Q ) T T T F F F T F F F T F T T F T F F T F F T T T T F T T DeMorgan’s laws are actually very natural and intuitive. Consider the
statement ∼ (P ∧ Q ), which we can interpret as meaning that it is not the
case that both P and Q are true. If it is not the case that both P and Q
are true, then at least one of P or Q is false, in which case (∼ P ) ∨ (∼ Q ) is
true. Thus ∼ (P ∧ Q ) means the same thing as (∼ P ) ∨ (∼ Q ).
DeMorgan’s laws can be very useful. Suppose we happen to know that
some statement having form ∼ (P ∨ Q ) is true. The second of DeMorgan’s
laws tells us that (∼ Q ) ∧ (∼ P ) is also true, hence ∼ P and ∼ Q are both true
as well. Being able to quickly obtain such additional pieces of information
can be extremely useful.
Exercises for Section 2.6
A. Use truth tables to show that the following statements are logically equivalent.
1. P ∧ (Q ∨ R ) = (P ∧ Q ) ∨ (P ∧ R ) 5. ∼ (P ∨ Q ∨ R ) = (∼ P ) ∧ (∼ Q ) ∧ (∼ R ) 2. P ∨ (Q ∧ R ) = (P ∨ Q ) ∧ (P ∨ R ) 6. ∼ (P ∧ Q ∧ R ) = (∼ P ) ∨ (∼ Q ) ∨ (∼ R ) 3. P ⇒ Q = (∼ P ) ∨ Q 7. P ⇒ Q = (P ∧ ∼ Q ) ⇒ (Q ∧ ∼ Q ) 4. ∼ (P ∨ Q ) = (∼ P ) ∧ (∼ Q ) 8. ∼ P ⇔ Q = (P ⇒∼ Q ) ∧ (∼ Q ⇒ P ) B. Decide whether or not the following pairs of statements are logically equivalent.
9. P ∧ Q and ∼ (∼ P ∨ ∼ Q ) 11. (∼ P ) ∧ (P ⇒ Q ) and ∼ (Q ⇒ P ) 10. (P ⇒ Q ) ∨ R and ∼ ((P ∧ ∼ Q )∧ ∼ R ) 12. ∼ (P ⇒ Q ) and P ∧ ∼ Q Logic 46
2.7 Quantiﬁers Using symbols ∧, ∨, ∼, ⇒ and ⇔, we can deconstruct many English
sentences into a symbolic form. As we have seen, this symbolic form can
help us understand the logical structure of sentences and how diﬀerent
sentences may actually have the same meaning (as in logical equivalence).
But these ﬁve symbols alone are not powerful enough to capture the
full meaning of every mathematical statement. To begin to overcome this
defect, we now introduce two new symbols that correspond to English
phrases that occur often in mathematics. The symbol “∀” stands for the
phrase “For all” or “For every,” and the symbol “∃” stands for the phrase
“There exists a” or “There is a.” Thus the statement
For every n ∈ Z, 2n is even.
can be expressed as
∀ n ∈ Z, 2 n is even. If we let E ( x) stand for “ x is even,” then the above becomes
∀ n ∈ Z, E (2 n). Likewise, a statement such as
There exists a subset X of N for which | X | = 5.
can be translated as
∃ X , ( X ⊆ N) ∧ (| X | = 5) or ∃ X ⊆ N, | X | = 5 or ∃ X ∈ P (N), | X | = 5. The symbols ∀ and ∃ are called quantiﬁers because they refer in some
sense to the quantity (i.e. all or some) of the variable that follows them.
Symbol ∀ is called the universal quantiﬁer and ∃ is called the existential quantiﬁer. Statements which contain them are called quantiﬁed
statements. A statement beginning with ∀ is called a universally quantiﬁed statement, and one beginning with ∃ is called an existentially
quantiﬁed statement.
Example 2.5 The following English statements are paired with their
translations into symbolic form.
Every integer that is not odd is even.
∀ n ∈ Z, ∼ ( n is odd ) ⇒ ( n is even) or ∀ n ∈ Z, ∼ O ( n) ⇒ E ( n)
There is an integer that is not even.
∃ n ∈ Z, ∼ E ( n ) Quantiﬁers 47 For every real number x, there is a real number y for which y3 = x.
∀ x ∈ R, ∃ y ∈ R, y3 = x Given any two rational numbers a and b, it follows that ab is rational.
∀ a, b ∈ Q, ab ∈ Q.
Given a set S (such as, but not limited to, N, Z, Q etc.), a quantiﬁed
statement of form ∀ x ∈ S , P ( x) is understood to be true if P ( x) is true
for every x ∈ S . If there is at least one x ∈ S for which P ( x) is false, then
∀ x ∈ S , P ( x) is a false statement. Similarly, ∃ x ∈ S , P ( x) is true provided that
P ( x) is true for at least one element x ∈ S ; otherwise it is false. Thus each
statement in Example 2.5 is true. Here are some examples of quantiﬁed
statements that are false.
Example 2.6 The following false quantiﬁed statements are paired with
their translations.
Every integer is even.
∀ n ∈ Z, E ( n) There is an integer n for which n2 = 2.
∃ n ∈ Z, n2 = 2 For every real number x, there is a real number y for which y2 = x.
∀ x ∈ R, ∃ y ∈ R, y2 = x Given any two rational numbers a and b, it follows that
∀ a, b ∈ Q, a b ∈ Q. a b is rational. Example 2.7 When a statement contains two quantiﬁers you must be
very alert to their order, for reversing the order can change the meaning.
Consider the following statement from Example 2.5.
∀ x ∈ R, ∃ y ∈ R, y3 = x This statement is true, for no matter what number x is there exist a
number y = 3 x for which y3 = x. Now reverse the order of the quantiﬁers
to get the new statement
∃ y ∈ R, ∀ x ∈ R, y3 = x. This new statement says that there exists a particular number y with
the property that y3 = x for every real number x. Since no number y can
have this property, the statement is false. The two statements above have
entirely diﬀerent meanings. Logic 48 Quantiﬁed statements are often misused in casual conversation. Perhaps you’ve heard someone say something like “All students do not pay
full tuition.” when they mean “Not all students pay full tuition.” While the
mistake is perhaps marginally forgivable in casual conversation, it must
never be made in a mathematical context. Do not say “All integers are not
even.” because that means there are no even integers. Instead, say “Not
all integers are even.”
Exercises for Section 2.7
Write the following as English sentences. Say whether the statements are true
or false.
6. ∃ n ∈ N, ∀ X ∈ P (N), | X | < n 1. ∀ x ∈ R, x2 > 0
n 2. ∀ x ∈ R, ∃ n ∈ N, x ≥ 0 7. ∀ X ⊆ N, ∃ n ∈ Z, | X | = n 3. ∃ a ∈ R, ∀ x ∈ R, ax = x 8. ∀ n ∈ Z, ∃ X ⊆ N, | X | = n 4. ∀ X ∈ P (N), X ⊆ R 9. ∀ n ∈ Z, ∃ m ∈ Z, m = n + 5 5. ∀ n ∈ N, ∃ X ∈ P (N), | X | < n 10. ∃ m ∈ Z, ∀ n ∈ Z, m = n + 5 2.8 More on Conditional Statements
It is time to address a very important point about conditional statements
that contain variables. To motivate this, let’s return to the following
example concerning integers x.
( x is a multiple of 6) ⇒ ( x is even) As noted earlier, since every multiple of 6 is even, this is a true statement
no matter what integer x is. We could even underscore this fact by writing
this true statement as
∀ x ∈ Z, ( x is a multiple of 6) ⇒ ( x is even). But now switch things around to get the diﬀerent statement
( x is even) ⇒ ( x is a multiple of 6). This is true for some values of x such as −6, 12, 18, etc., but false for
others (such as 2, 4, etc.). Thus we do not have a statement, but rather an
open sentence. (Recall from Section 2.1 that an open sentence is a sentence
whose truth value depends on the value of a certain variable or variables.) More on Conditional Statements 49 However, by putting a universal quantiﬁer in front we get
∀ x ∈ Z, ( x is even) ⇒ ( x is a multiple of 6), which is deﬁnitely false, so this new expression is a statement, not an open
sentence. In general, given any two open sentences P ( x) and Q ( x) about
integers x, the expression ∀ x ∈ Z, P ( x) ⇒ Q ( x) is either true or false, so it is
a statement, not an open sentence.
Now we come to the very important point. In mathematics, whenever
P ( x) and Q ( x) are open sentences concerning elements x in some set S
(depending on context), an expression of form P ( x) ⇒ Q ( x) is understood
to be the statement ∀ x ∈ S , P ( x) ⇒ Q ( x). In other words, if a conditional
statement is not explicitly quantiﬁed then there is an implied universal
quantiﬁer in front of it. This is done because statements of the form
∀ x ∈ S , P ( x) ⇒ Q ( x) are so common in mathematics that we would get tired
of putting the ∀ x ∈ S in front of them.
Thus the following sentence is a true statement (as it is true for all x).
If x is a multiple of 6, then x is even.
Likewise, the next sentence is a false statement (as it is not true for all x).
If x is even, then x is a multiple of 6.
This leads to the following signiﬁcant interpretation of a conditional
statement, which is more general than (but consistent with) the interpretation from Section 2.3.
Deﬁnition 2.1
sentence If P and Q are statements or open sentences, then the “If P , then Q ,”
is a statement. This statement is true if it’s impossible for P to be true
while Q is false. It is false if there is at least one instance in which P is
true but Q is false.
Thus the following are true statements.
If x ∈ R, then x2 + 1 > 0.
If a function f is diﬀerentiable on R, then f is continuous on R.
Likewise, the following are false statements.
If p is a prime number, then p is odd.
If f is a rational function, then f has an asymptote. (2 is prime)
( x2 is rational) Logic 50
2.9 Translating English to Symbolic Logic In writing (and reading) proofs of theorems, we must always be alert to the
logical structure and meanings of the sentences. Sometimes it is necessary
or helpful to parse them into expressions involving logic symbols. This may
be done mentally or on scratch paper, or occasionally even explicitly within
the body of a proof. The purpose of this section is to give you suﬃcient
practice in translating English sentences into symbolic form so that you
can better understand their logical structure. Here are some examples.
Example 2.8 Consider the Mean Value Theorem from Calculus:
If f is continuous on the interval [a, b] and diﬀerentiable on (a, b), then
)− f
there is a number c ∈ (a, b) for which f ( c) = f (bb−a(a) .
Here is a translation to symbolic form.
f cont. on [a, b] ∧ f is diﬀ. on (a, b) ⇒ ∃ c ∈ (a, b), f ( c) = f ( b )− f ( a )
b−a . Example 2.9 Consider Goldbach’s Conjecture, from Section 2.1:
Every even integer greater than 2 is the sum of two primes.
This can be translated in the following ways, where P denotes the set of
prime numbers and S = {4, 6, 8, 10, . . .} is the set of even integers greater
than 2.
n ∈ S ⇒ ∃ p, q ∈ P , n = p + q
∀ n ∈ S , ∃ p, q ∈ P , n = p + q These translations of Goldbach’s conjecture illustrate an important
point. The ﬁrst has the basic structure ( n ∈ S ) ⇒ Q ( n) and the second has
structure ∀ n ∈ S , Q (n), yet they have exactly the same meaning. This is
signiﬁcant. Every universally quantiﬁed statement can be expressed as a
conditional statement.
Fact 2.2 Suppose S is a set and Q ( x) is a statement about x for each
x ∈ S . The following statements mean the same thing.
∀ x ∈ S , Q ( x) ( x ∈ S ) ⇒ Q ( x) Translating English to Symbolic Logic 51 This fact is signiﬁcant because so many theorems have the form of
a conditional statement. (The Mean Value Theorem is an example!) In
proving a theorem we have to think carefully about what it says. Sometimes
a theorem will be expressed as a universally quantiﬁed statement but it will
be more convenient to think of it as a conditional statement. Understanding
the above fact allows us to switch between the two forms.
We close this section with one ﬁnal point. In translating a statement, be attentive to its intended meaning. Don’t jump into, for example,
automatically replacing every “and” with ∧ and “or” with ∨. An example:
At least one of the integers x and y is even.
Don’t be led astray by the presence of the word “and.” The meaning of
the statement is that one or both of the numbers is even, so it should be
translated with “or,” not “and.”
( x is even) ∨ ( y is even)
Exercises for Section 2.9
Translate each of the following sentences into symbolic logic.
1. If f is a polynomial and its degree is greater than 2, then f is not constant.
2. The number x is positive but the number y is not positive.
3. If x is prime then x is not a rational number. 4. For every prime number p there is another prime number q with q > p.
5. For every positive number ε, there is a positive number δ for which | x − a| < δ
implies | f ( x) − f (a)| < ε.
6. For every positive number ε there is a positive number M for which | f ( x) − b| < ε,
whenever x > M .
7. There exists a real number a for which a + x = x for every real number x.
8. I don’t eat anything that has a face.
9. If x is a rational number and x = 0, then tan( x) is not a rational number.
10. If sin( x) < 0, then it is not the case that 0 ≤ x ≤ π.
11. There is a Providence that protects idiots, drunkards, children and the United
States of America. (Otto von Bismarck)
12. You can fool some of the people all of the time, and you can fool all of the people
some of the time, but you can’t fool all of the people all of the time. (Abraham
Lincoln)
13. Everything is funny as long as it is happening to somebody else. (Will Rogers) Logic 52
2.10 Negating Statements Given a statement R , the statement ∼ R is called the negation of R . If R
is a complex statement, then it is often the case that its negation ∼ R can
be written in a simpler or more useful form. The process of ﬁnding this
form is called negating R . In proving theorems it is often necessary to
negate certain statements. We now investigate how to do this.
We have already examined part of this topic. DeMorgan’s Laws
∼ (P ∧ Q ) = (∼ P ) ∨ (∼ Q ) (2.1) ∼ (P ∨ Q ) = (∼ P ) ∧ (∼ Q ) (2.2) (from Section 2.6) can be viewed as rules that tell us how to negate the
statements P ∧ Q and P ∨ Q . Here are some examples that illustrate how
DeMorgan’s laws are used to negate statements involving “and” or “or.”
Example 2.10 Consider negating the following statement.
R : You can solve it by factoring or with the quadratic formula. Now, R means (You can solve it by factoring) ∨ (You can solve it with Q.F.),
which we will denote as P ∨ Q . The negation of this is
∼ (P ∨ Q ) = (∼ P ) ∧ (∼ Q ) Therefore, in words, the negation of R is
∼ R : You can’t solve it by factoring and you can’t solve it with the quadratic formula.
Maybe you can ﬁnd ∼ R without invoking DeMorgan’s laws. That is good;
you have internalized DeMorgan’s laws and are using them unconsciously.
Example 2.11 We will negate the following sentence.
R : The numbers x and y are both odd. This statement means ( x is odd) ∧ ( y is odd), so its negation is
∼ ( x is odd) ∧ ( y is odd) = ∼ ( x is odd) ∨ ∼ ( y is odd)
= ( x is even) ∨ ( y is even) Therefore the negation of R can be expressed in the following ways
∼ R : The number x is even or the number y is even.
∼ R : At least one of x and y is even. Negating Statements 53 Now let’s move on to a slightly diﬀerent kind of problem. It’s often
necessary to ﬁnd the negations of quantiﬁed statements. For example,
consider ∼ (∀ x ∈ N, P ( x)). Reading this in words, we have the following:
It is not the case that P ( x) is true for all natural numbers x.
This means P ( x) is false for at least one x. In symbols, this is ∃ x ∈ N, ∼ P ( x).
Thus ∼ (∀ x ∈ N, P ( x)) = ∃ x ∈ N, ∼ P ( x). Similarly, you can reason out that
∼ (∃ x ∈ N, P ( x)) = ∀ x ∈ N, ∼ P ( x). In general we have the following.
∼ (∀ x ∈ S , P ( x)) = ∃ x ∈ S , ∼ P ( x)
∼ (∃ x ∈ S , P ( x)) = ∀ x ∈ S , ∼ P ( x) Example 2.12 (2.3)
(2.4) Consider negating the following statement. R : The square of every real number is non-negative. Symbolically, R can be expressed as ∀ x ∈ R, x2 ≥ 0, and thus its negation is
∼ (∀ x ∈ R, x2 ≥ 0) = ∃ x ∈ R, ∼ ( x2 ≥ 0) = ∃ x ∈ R, x2 < 0. In words, this is
∼ R : There exists a real number whose square is negative. Observe that R is true and ∼ R is false. You may be able to get ∼ R
immediately, without using Equation (2.3) as we did above. If so, that is
good; if not, you will probably be there soon.
If a statement has multiple quantiﬁers, negating it will involve several
iterations of Equations (2.3) and (2.4). Consider the following.
S : For every real number x there is a real number y for which y3 = x. This statement asserts any real number x has a cube root y, so it’s true.
Symbolically S can be expressed as
∀ x ∈ R, ∃ y ∈ R, y3 = x. Let’s work out the negation of this statement.
∼ (∀ x ∈ R, ∃ y ∈ R, y3 = x) = ∃ x ∈ R, ∼ (∃ y ∈ R, y3 = x)
= ∃ x ∈ R, ∀ y ∈ R, ∼ ( y3 = x)
= ∃ x ∈ R, ∀ y ∈ R, y3 = x Therefore the negation is the following (false) statement.
∼ S : There is a real number x for which y3 = x for all real numbers y. Logic 54 In writing proofs you will sometimes have to negate a conditional
statement P ⇒ Q . The remainder of this section describes how to do this.
To begin, look at the expression ∼ (P ⇒ Q ), which literally says “P ⇒ Q is
false.” You know from the truth table for ⇒ that the only way that P ⇒ Q
can be false is if P is true and Q is false. Therefore ∼ (P ⇒ Q ) = P ∧ ∼ Q .
∼ (P ⇒ Q ) = P ∧ ∼ Q (2.5) (In fact, in Exercise 5 of Section 2.6, you used a truth table to verify that
these two statements are logically equivalent.)
Example 2.13 Consider negating the following statement about a particular (constant) number a.
R : If a is odd then a2 is odd. Using Equation (2.5), we get the following negation.
∼ R : a is odd and a2 is not odd. Example 2.14 This example is like the previous one, but the constant a
is replaced by a variable x. We will negate the following statement.
R : If x is odd then x2 is odd. As discussed in Section 2.8, we interpret this as the universally quantiﬁed
statement
R : ∀ x ∈ Z, ( x odd) ⇒ ( x2 odd). By Equations (2.3) and (2.5), we get the following negation for R .
∼ ∀ x ∈ Z, ( x odd) ⇒ ( x2 odd) = ∃ x ∈ Z, ∼ ( x odd) ⇒ ( x2 odd)
= ∃ x ∈ Z, ( x odd)∧ ∼ ( x2 odd) Translating back into words, we have
∼ R : There is an odd integer x whose square is not odd. Notice that R is true and ∼ R is false.
The above Example 2.14 showed how to negate a conditional statement
P ( x) ⇒ Q ( x). This type of problem can sometimes be embedded in more complex negation. See Exercise 5 below (and its solution). Logical Inference 55 Exercises for Section 2.10
Negate the following sentences.
1. The number x is positive but the number y is not positive.
2. If x is prime, then x is not a rational number. 3. For every prime number p, there is another prime number q with q > p.
4. For every positive number ε, there is a positive number δ such that | x − a| < δ
implies | f ( x) − f (a)| < ε.
5. For every positive number ε, there is a positive number M for which | f ( x) − b| < ε
whenever x > M .
6. There exists a real number a for which a + x = x for every real number x.
7. I don’t eat anything that has a face.
8. If x is a rational number and x = 0, then tan( x) is not a rational number.
9. If sin( x) < 0, then it is not the case that 0 ≤ x ≤ π.
10. If f is a polynomial and its degree is greater than 2, then f is not constant.
11. You can fool all of the people all of the time.
12. Whenever I have to choose between two evils, I choose the one I haven’t tried
yet. (Mae West) 2.11 Logical Inference
Suppose we know that a statement of form P ⇒ Q is true. This tells us
that whenever P is true, Q will also be true. By itself, P ⇒ Q being true
does not tell us that either P or Q is true (they could both be false, or P
could be false and Q true). However if in addition we happen to know
that P is true then it must be that Q is true. This is called a logical
inference: Given two true statements we can infer that a third statement
is true. In this instance true statements P ⇒ Q and P are “added together”
to get Q . This is described below with P ⇒ Q and P stacked one atop the
other with a line separating them from Q . The intended meaning is that
P ⇒ Q combined with P produces Q .
P ⇒Q
P P ⇒Q
∼Q Q ∼P P ∨Q
∼P
Q Two other logical inferences are listed above. In each case you should
convince yourself (based on your knowledge of the relevant truth tables) Logic 56 that the truth of the statements above the line forces the statement below
the line to be true.
Following are some additional useful logical inferences. The ﬁrst
expresses the obvious fact that if P and Q are both true then the statement
P ∧ Q will be true. On the other hand, P ∧ Q being true forces P (also Q )
to be true. Finally, if P is true, then P ∨ Q must be true, no matter what
statement Q is.
P
Q P ∧Q P P ∧Q P P ∨Q These inferences are so intuitively obvious that they scarcely need to
be mentioned. However, they represent certain patterns of reasoning that
we will frequently apply to sentences in proofs, so we should be cognizant
of the fact that we are using them.
2.12 An Important Note
It is important to be aware of the reasons that we study logic. There
are three very signiﬁcant reasons. First, the truth tables we studied tell
us the exact meanings of the words such as “and,” “or”, “not” and so on.
For instance, whenever we use or read the “If..., then” construction in
a mathematical context, logic tells us exactly what is meant. Second,
the rules of inference provide a system in which we can produce new
information (statements) from known information. Finally, logical rules
such as DeMorgan’s laws help us correctly change certain statements into
(potentially more useful) statements with the same meaning. Thus logic
helps us understand the meanings of statements and it also produces new
meaningful statements.
Logic is the glue that holds strings of statements together and pins down
the exact meaning of certain key phrases such as the “If..., then” or “For
all” constructions. Logic is the common language that all mathematicians
use, so we must have a ﬁrm grip on it in order to write and understand
mathematics.
But despite its fundamental role, logic’s place is in the background of
what we do, not the forefront. From here on, the beautiful symbols ∧, ∨,
⇒, ⇔, ∼, ∀ and ∃ are rarely written. But we are aware of their meanings
constantly. When reading or writing a sentence involving mathematics we
parse it with these symbols, either mentally or on scratch paper, so as to
understand the true and unambiguous meaning. CHAPTER 3 Counting t may seem peculiar that a college-level text has a chapter on counting.
At its most basic level, counting is a process of pointing to each object
in a collection and calling oﬀ “one, two, three,...” until the quantity of
objects is determined. How complex could that be? Actually, counting
can become quite subtle, and in this chapter we explore some of its more
sophisticated aspects. Our goal is still to answer the question “How many?”
but we introduce mathematical techniques that bypass the actual process
of counting individual objects.
Almost every branch of mathematics uses some form of this “sophisticated counting.” Many such counting problems can be modeled with the
idea of a list, so we start there. I 3.1 Counting Lists
A list is an ordered sequence of objects. A list is denoted by an opening
parenthesis, followed by the objects, separated by commas, followed by a
closing parenthesis. For example (a, b, c, d , e) is a list consisting of the ﬁrst
ﬁve letters of the English alphabet, in order. The objects a, b, c, d , e are
called the entries of the list; the ﬁrst entry is a, the second is b, and so
on. If the entries are rearranged we get a diﬀerent list, so, for instance,
(a, b, c, d , e) = ( b, a, c, d , e). A list is somewhat like a set, but instead of being a mere collection of
objects, the entries of a list have a deﬁnite order. Note that for sets we
have
a, b , c , d , e = b , a, c , d , e , but—as noted above—the analogous equality for lists does not hold.
Unlike sets, lists are allowed to have repeated entries. For example
(5, 3, 5, 4, 3, 3) is a perfectly acceptable list, as is (S , O , S ). The number of
entries in a list is called its length. Thus (5, 3, 5, 4, 3, 3) has length six, and
(S , O , S ) has length three. Counting 58 Occasionally we may get sloppy and write lists without parentheses
and commas; for instance we may express (S , O , S ) as SOS if there is no
danger of confusion. But be alert that doing this can lead to ambiguity.
Is it reasonable that (9, 10, 11) should be the same as 91011? If so, then
(9, 10, 11) = 91011 = (9, 1, 0, 1, 1), which makes no sense. We will thus almost
always adhere to the parenthesis/comma notation for lists.
Lists are important because many real-world phenomena can be described and understood in terms of them. For example, your phone number
(with area code) can be identiﬁed as a list of ten digits. Order is essential,
for rearranging the digits can produce a diﬀerent phone number. A byte is
another important example of a list. A byte is simply a length-eight list of
0’s and 1’s. The world of information technology revolves around bytes.
To continue our examples of lists, (a, 15) is a list of length two. Likewise
(0, (0, 1, 1)) is a list of length two whose second entry is a list of length three.
The list (N, Z, R) has length three, and each of its entries is a set. We
emphasize that for two lists to be equal, they must have exactly the same
entries in exactly the same order. Consequently if two lists are equal, then
they must have the same length. Said diﬀerently, if two lists have diﬀerent
lengths, then they are not equal. For example, (0, 0, 0, 0, 0, 0) = (0, 0, 0, 0, 0).
For another example note that
( g , r , o , c , e , r , y, l , i , s , t ) = bread
milk
eggs
mustard
coﬀee because the list on the left has length eleven but the list on the right has
just one entry (a piece of paper with some words on it).
There is one very special list which has no entries at all. It is called
the empty list, and is denoted (). It is the only list whose length is zero.
One often needs to count up the number of possible lists which satisfy
some condition or property. For example suppose we need to make a list of
length three having the property that the ﬁrst entry must be an element
of the set a, b, c , the second entry must be in 5, 7 and the third entry
must be in a, x . Thus (a, 5, a) and ( b, 5, a) are two such lists. How many
such lists are there all together? To answer this question, imagine making
the list by selecting the ﬁrst element, then the second and ﬁnally the third.
This is described in Figure 3.1. The choices for the ﬁrst list entry are a, b
or c, and the left of the diagram branches out in three directions, one for
each choice. Once this choice is made there are two choices (5 or 7) for
the second entry, and this is described graphically by two branches from
each of the three choices for the ﬁrst entry. This pattern continues for the Counting Lists 59 choice for the third entry which is either a or x. Thus, in the diagram there
are 3 · 2 · 2 = 12 paths from left to right, each corresponding to a particular
choice for each entry in the list. The corresponding lists are tallied at the
far-right end of each path. So, to answer our original question, there are
12 possible lists with the stated properties.
Resulting list
ﬁrst choice second choice 5 a
7
5 b
7
5 c
7 third choice a
x
a
x
a
x
a
x
a
x
a
x ( a , 5, a )
( a , 5, x )
( a , 7, a )
( a , 7, x )
( b, 5, a)
( b, 5, x)
( b, 7, a)
( b, 7, x)
( c , 5, a )
( c , 5, x )
( c , 7, a )
( c , 7, x ) Figure 3.1. Constructing lists of length 3 We summarize the type of reasoning used above in the following important fact, which is called the Multiplication Principle.
Fact 3.1 (Multiplication Principle) Suppose in making a list of length
n there are a 1 possible choices for the ﬁrst entry, a 2 possible choices for
the second entry, a 3 possible choices for the third entry, and so on. Then
the total number of diﬀerent lists that can be made this way is the product
a1 · a2 · a3 · · · · · a n .
So, for instance, in the above example we had a 1 = 3, a 2 = 2 and a 3 = 2,
so the total number of lists was a 1 · a 2 · a 3 = 3 · 2 · 2 = 12. Now let’s look at
some additional examples of how the Multiplication Principle can be used.
Example 3.1 A standard license plate consists of three letters followed
by four numbers. For example, JRB-4412 and MMX-8901 are two standard
license plates. (Vanity plates such as LV2COUNT are not included among
the standard plates.) How many diﬀerent standard license plates are
possible? Counting 60 To answer this question, note that any standard license plate such as
JRB-4412 corresponds to a length-7 list (J,R,B,4,4,1,2), so the question
can be answered by counting how many such lists are possible. We use the
Multiplication Principle. There are a 1 = 26 possibilities (one for each letter
of the alphabet) for the ﬁrst entry of the list. Similarly, there are a 2 = 26
possibilities for the second entry and a 3 = 26 possibilities for the third
entry. There are a 4 = 10 possibilities for the fourth entry, and likewise
a 5 = a 6 = a 7 = 10. Therefore there are a total of a 1 · a 2 · a 3 · a 4 · a 5 · a 6 · a 7 =
26 · 26 · 26 · 10 · 10 · 10 · 10 = 17,576,000 possible standard license plates.
There are two types of list-counting problems. On one hand, there are
situations in which the same symbol or symbols may appear multiple times
in diﬀerent entries of the list. For example, license plates or telephone
numbers can have repeated symbols. The sequence CCX-4144 is a perfectly
valid license plate in which the symbols C and 4 appear more than once.
On the other hand, for some lists repeated symbols do not make sense or
are not allowed. For instance, imagine drawing ﬁve cards from a standard
52-card deck and laying them in a row. Since no two cards in the deck
are identical, this list has no repeated entries. We say that repetition is
allowed in the ﬁrst type of list and repetition is not allowed in the second
kind of list. (Often we call a list in which repetition is not allowed a
non-repetitive list.) The following example illustrates the diﬀerence.
Example 3.2 Consider making lists from symbols A, B, C, D, E, F, G.
(a) How many length-4 lists are possible if repetition is allowed?
(b) How many length-4 lists are possible if repetition is not allowed?
(c) How many length-4 lists are possible if repetition is not allowed and
the list must contain an E?
(d) How many length-4 lists are possible if repetition is allowed and the
list must contain an E?
Solutions:
(a) Imagine the list as containing four boxes that we ﬁll with selections
from the letters A,B,C,D,E,F and G, as illustrated below. ( , , , ) 7 choices
7 choices
7 choices
7 choices There are seven possibilities for the contents of each box, so the total
number of lists that can be made this way is 7 · 7 · 7 · 7 = 2401. Counting Lists 61 (b) This problem is the same as the previous one except that repetition is
not allowed. We have seven choices for the ﬁrst box, but once it is ﬁlled
we can no longer use the symbol that was placed in it. Hence there are
only six possibilities for the second box. Once the second box has been
ﬁlled we have used up two of our letters, and there are only ﬁve left to
choose from in ﬁlling the third box. Finally, when the third box is ﬁlled
we have only four possible letters for the last box. ( , , , ) 7 choices
6 choices
5 choices
4 choices Thus the answer to our question is that there are 7 · 6 · 5 · 4 = 840 lists in
which repetition does not occur.
(c) We are asked to count the length-4 lists in which repetition is not
allowed and the symbol E must appear somewhere in the list. Thus E
occurs once and only once in each such list. Let us divide these lists into
four categories depending on whether the E occurs as the ﬁrst, second,
third or fourth entry. These four types of lists are illustrated below.
Type 1 (E, , Type 2 , )( , E, Type 3 , )( , , E, Type 4 )( , , , E) 6 choices
6 choices
6 choices
6 choices
5 choices
5 choices
5 choices
5 choices
4 choices
4 choices
4 choices
4 choices Consider lists of the ﬁrst type, in which the E appears in the ﬁrst entry.
We have six remaining choices (A,B,C,D,F or G) for the second entry, ﬁve
choices for the third entry, and four choices for the fourth entry. Hence
there are 6 · 5 · 4 = 120 lists having an E in the ﬁrst entry. As indicated
in the above diagram, there are also 6 · 5 · 4 = 120 lists having an E in the
second, third or fourth entry. Thus there are 120 + 120 + 120 + 120 = 480
such lists all together.
(d) Now we must ﬁnd the number of length-four lists where repetition
is allowed and the list must contain an E. Our strategy is as follows.
By Part (a) of this exercise there are 7 · 7 · 7 · 7 = 74 = 2401 lists where
repetition is allowed. Obviously this is not the answer to our current
question, for many of these lists contain no E. We will subtract from 2401
the number of lists which do not contain an E. In making a list that
does not contain an E we have six choices for each list entry (because Counting 62 we can choose any one of the six letters A,B,C,D,F or G). Thus there
are 6 · 6 · 6 · 6 = 64 = 1296 lists that do not have an E. Therefore the ﬁnal
answer to our question is that there are 2401 − 1296 = 1105 lists with
repetition allowed that contain at least one E.
Perhaps you wondered if Part (d) of Example 3.2 could be solved with
a set-up similar to that of Part (c). Let’s try doing it that way. We want
to count the length-4 lists (with repetition allowed) that contain at least
one E. The following diagram is adapted from Part (c), the only diﬀerence
being that there are now seven choices in each slot because we are allowed
to repeat any of the seven letters.
Type 1 (E, , Type 2 , )( , E, Type 3 , )( , , E, Type 4 )( , , , E) 7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices
7 choices This gives a total of 73 + 73 + 73 + 73 = 1373 lists, an answer that is
substantially larger than the (correct) value of 1105 that we got in our
solution to Part (d) above. It is not hard to see what went wrong. The
list (E , E , A , B) is of type 1 and type 2, so it got counted twice. Similarly
(E , E , C , E ) is of type 1, 3, and 4, so it got counted three times. In fact, you
can ﬁnd many similar lists that were counted multiple times.
In solving counting problems, we must always be careful to avoid this
kind of double-counting or triple-counting, or worse.
Exercises for Section 3.1
Note: A calculator may be helpful for some of the exercises in this chapter. This
is the only chapter for which a calculator may be helpful. (As for the exercises in
the other chapters, a calculator makes them harder.)
1. Consider lists made from the letters T,H,E,O,R,Y, with repetition allowed.
(a) How many length-4 lists are there?
(b) How many length-4 lists are there that begin with T?
(c) How many length-4 lists are there that do not begin with T?
2. Airports are identiﬁed with 3-letter codes. For example, the Richmond, Virginia
airport has the code RIC, and Portland, Oregon has PDX. How many diﬀerent
3-letter codes are possible?
3. How many lists of length 3 can be made from the symbols A,B,C,D,E,F if... Counting Lists
(a)
( b)
(c)
(d) ...
...
...
... 63 repetition is allowed.
repetition is not allowed.
repetition is not allowed and the list must contain the letter A.
repetition is allowed and the list must contain the letter A. 4. Five cards are dealt oﬀ of a standard 52-card deck and lined up in a row. How
many such line-ups are there in which all ﬁve cards are of the same suit?
5. Five cards are dealt oﬀ of a standard 52-card deck and lined up in a row. How
many such line-ups are there in which all ﬁve cards are of the same color (i.e.
all black or all red)?
6. Five cards are dealt oﬀ of a standard 52-card deck and lined up in a row. How
many such line-ups are there in which exactly one of the ﬁve cards is a queen?
7. This problems involves 8-digit binary strings such as 10011011 or 00001010
(i.e. 8-digit numbers composed of 0’s and 1’s).
(a) How many such strings are there?
(b) How many such strings end in 0?
(c) How many such strings have the property that their second and fourth
digits are 1’s?
(d) How many such strings have the property that their second or fourth digits
are 1’s?
8. This problem concerns lists made from the symbols A,B,C,D,E.
(a) How many such length-5 lists have at least one letter repeated?
(b) How many such length-6 lists have at least one letter repeated?
9. This problem concerns 4-letter codes made from the letters A,B,C,D,...,Z.
(a) How many such codes can be made?
(b) How many such codes have no two consecutive letters the same?
10. This problem concerns lists made from the letters A,B,C,D,E,F,G,H,I,J.
(a) How many length-5 lists can be made from these letters if repetition is not
allowed and the list must begin with a vowel?
(b) How many length-5 lists can be made from these letters if repetition is not
allowed and the list must begin and end with a vowel?
(c) How many length-5 lists can be made from these letters if repetition is not
allowed and the list must contain exactly one A?
11. This problem concerns lists of length 6 made from the letters A,B,C,D,E,F,G,H.
How many such lists are possible if repetition is not allowed and the list
contains two consecutive vowels?
12. Consider the lists of length six made with the symbols P, R, O, F, S, where
repetition is allowed. (For example, the following is such a list: (P,R,O,O,F,S).)
How many such lists can be made if the list must end in an S, and the symbol
O is used more than once? Counting 64
3.2 Factorials In working the examples from Section 3.1 you may have noticed that often
we need to count the number of non-repetitive lists of length n that are
made from n symbols. In fact, this particular problem occurs with such
frequency that a special idea, called a factorial, is introduced to handle it.
n
0 Non-repetitive lists of length n made from the symbols n! () Symbols 1 1 A ( A) 1 2 A, B ( A , B), (B, A ) 2 3 A , B, C ( A , B, C ), ( A , C , B), (B, C , A ), (B, A , C ), (C , A , B), (C , B, A ) 6 4 A , B, C , D ( A ,B,C ,D ), ( A ,B,D ,C ), ( A ,C ,B,D ), ( A ,C ,D ,B), ( A ,D ,B,C ), ( A ,D ,C ,B)
(B, A ,C ,D ), (B, A ,D ,C ), (B,C , A ,D ), (B,C ,D , A ), (B,D, A ,C ), (B,D ,C , A )
( C , A , B , D ), ( C , A , D , B ), ( C , B , A , D ), ( C , B , D , A ), ( C , D , A , B ), ( C , D , B , A )
( D , A , B , C ), ( D , A , C , B ), ( D , B , A , C ), ( D , B , C , A ), ( D , C , A , B ), ( D , C , B , A ) .
.
. .
.
. .
.
. 24
.
.
. The above table motivates this idea. The ﬁrst column contains successive integer values n (beginning with 0) and the second column contains
a set A , B, · · · of n symbols. The third column contains all the possible
non-repetitive lists of length n which can be made from these symbols.
Finally, the last column tallies up how many lists there are of that type.
Notice that when n = 0 there is only one list of length 0 that can be made
from 0 symbols, namely the empty list (). Thus the value 1 is entered in
the last column of that row.
For n > 0, the number that appears in the last column can be computed
using the Multiplication Principle. The number of non-repetitive lists of
length n that can be made from n symbols is n(n − 1)( n − 2) · · · 3 · 2 · 1. Thus, for
instance, the number in the last column of the row for n = 4 is 4 · 3 · 2 · 1 = 24.
The number that appears in the last column of Row n is called the
factorial of n. It is denoted as n! (read “ n factorial”). Here is the deﬁnition.
Deﬁnition 3.1 If n is a non-negative integer, then the factorial of n,
denoted n!, is the number of non-repetitive lists of length n that can
be made from n symbols. Thus 0! = 1, and 1! = 1. If n > 1, then n! =
n( n − 1)( n − 2) · · · 3 · 2 · 1. Factorials
It follows that 65
0!
1!
2!
3!
4!
5!
6! =
=
=
=
=
=
= 1
1
2·1 = 2
3·2·1 = 6
4 · 3 · 2 · 1 = 24
5 · 4 · 3 · 2 · 1 = 120
6 · 5 · 4 · 3 · 2 · 1 = 720, and so on. Students are often tempted to say 0! = 0, but this is wrong. The correct
value is 0! = 1, as the above deﬁnition and table tell us. Here is another
way to see that 0! must equal 1. Notice that 5! = 5 · 4 · 3 · 2 · 1 = 5 · (4 · 3 · 2 · 1) =
5 · 4!. Also 4! = 4 · 3 · 2 · 1 = 4 · (3 · 2 · 1) = 4 · 3!. Generalizing this reasoning we
have the following formula.
n! = n · ( n − 1)! (3.1) Plugging in n = 1 gives 1! = 1 · (1 − 1)! = 1 · 0!. If we mistakenly thought 0!
were 0, this would give the incorrect result 1! = 0.
We round out our discussion of factorials with an example.
Example 3.3 This problem involves making lists of length seven from
the symbols 0, 1, 2, 3, 4, 5, and 6.
(a) How many such lists are there if repetition is not allowed?
(b) How many such lists are there if repetition is not allowed and the
ﬁrst three entries must be odd?
(c) How many such lists are there in which repetition is allowed, and
the list must contain at least one repeated number?
To answer the ﬁrst question, note that there are seven symbols, so the
number of lists is 7! = 5040. To answer the second question, notice that
the set 0, 1, 2, 3, 4, 5, 6 contains three odd numbers and four even numbers.
Thus in making the list the ﬁrst three entries must be ﬁlled by odd numbers
and the ﬁnal four must be ﬁlled with even numbers. By the Multiplication
Principle, the number of such lists is 3 · 2 · 1 · 4 · 3 · 2 · 1 = 3!4! = 144.
To answer the third question, notice that there are 77 = 823, 543 lists
in which repetition is allowed. The set of all such lists includes lists
that are non-repetitive (e.g. (0, 6, 1, 2, 4, 3, 5)) as well as lists that have
some repetition (e.g. (6, 3, 6, 2, 0, 0, 0)). We want to compute the number of
lists that have at least one repeated number. To ﬁnd the answer we can
subtract the number of non-repetitive lists of length seven from the total
number of possible lists of length seven. Therefore the answer is 77 − 7! =
823, 543 − 5040 = 818, 503. Counting 66 We close this section with a formula that combines the ideas of the ﬁrst
and second sections of the present chapter. One of the main problems of
Section 3.1 was as follows: Given n symbols, how many non-repetitive lists
of length k can be made from the n symbols? We learned how to apply the
Multiplication Principle to obtain the answer
n( n − 1)( n − 2) · · · ( n − k + 1). Notice that by cancellation this value can also be written as
n( n − 1)( n − 2) · · · ( n − k + 1)( n − k)( n − k − 1) · · · 3 · 2 · 1
( n − k)( n − k − 1) · · · 3 · 2 · 1 = n!
.
( n − k)! We summarize this as follows.
Fact 3.2 The number of non-repetitive lists of length k whose entries
n
are chosen from a set of n possible entries is (n−!k)! .
For example, consider ﬁnding the number of non-repetitive lists of
length ﬁve that can be made from the symbols 1, 2, 3, 4, 5, 6, 7, 8. We will do
this two ways. By the Multiplication Principle, the answer is 8 · 7 · 6 · 5 · 4 =
40,320
8!
6720. Using the formula from Fact 3.2, the answer is (8−5)! = 8! = 6 =
3!
6720.
The new formula isn’t really necessary, but it is a nice repackaging of
an old idea and will prove convenient in the next section.
Exercises for Section 3.2
1. What is the smallest n for which n! has more than 10 digits?
2. For which values of n does n! have n or fewer digits?
3. How many 5-digit positive integers are there in which there are no repeated
digits and all digits are odd?
4. Using only pencil and paper, ﬁnd the value of
5. Using only pencil and paper, ﬁnd the value of 100!
95! .
120!
118! . 6. There are two 0’s at the end of 10! = 3, 628, 800. Using only pencil and paper,
determine how many 0’s are at the end of the number 100!.
7. Compute how many 9-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9
if repetition is not allowed and all the odd digits occur ﬁrst (on the left) followed
by all the even digits (i.e. as in 1375980264, but not 0123456789).
8. Compute how many 7-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6, 7 if
there is no repetition and the odd digits must appear in an unbroken sequence.
(Examples: 3571264 or 2413576 or 2467531, etc., but not 7234615.) Counting Subsets 67 9. There is a very interesting function Γ : [0, ∞) → R called the Gamma Function.
∞
It is deﬁned as Γ( x) = 0 t x−1 e− t dt. It has the remarkable property that if x ∈ N,
then Γ( x) = ( x − 1)!. Check that this is true for x = 1, 2, 3, 4.
Notice that this function provides a way of extending factorials to numbers other
than integers. Since Γ( n) = (n − 1)! for all n ∈ N, we have the formula n! = Γ(n + 1).
But Γ can be evaluated at any number in [0, ∞), not just at integers, so we
have a formula for n! for any n ∈ [0, ∞). Extra credit: Compute π!.
10. There is another signiﬁcant function called Stirling’s Formula that provides
n
an approximation to factorials. It states that n! ≈ 2π n n . It is an approxie
mation to n! in the sense that 2πn! n n approaches 1 as n approaches ∞. Use
n
e Stirling’s Formula to ﬁnd approximations to 5!, 10!, 20! and 50!. 3.3 Counting Subsets
The previous two sections were concerned with counting the number of
lists that can be made by selecting k entries from a set of n possible entries.
We turn now to a related question: How many subsets can be made by
selecting k elements from a set with n elements?
To highlight the diﬀerences between these two problems, look at the set
A = a, b, c, d , e . First, think of the non-repetitive lists that can be made
from selecting two entries from A . By Fact 3.2 (on the previous page),
5!
there are (5−2)! = 5! = 120 = 20 such lists. They are as follows.
3!
6
(a, b), (a, c), (a, d ), (a, e), ( b, c), ( b, d ), ( b, e), ( c, d ), ( c, e) ( d , e)
( b, a), ( c, a), ( d , a), ( e, a), ( c, b), ( d , b), ( e, b), ( d , c), ( e, c) ( e, d ) Next consider the subsets of A that can made from selecting two elements from A . There are only 10 such subsets, as follows.
a, b , a, c , a, d , a, e , b , c , b , d , b , e , c , d , c , e , d , e . The reason that there are more lists than subsets is that changing the
order of the entries of a list produces a diﬀerent list, but changing the
order of the elements of a set does not change the set. Using elements
a, b ∈ A , we can make two lists (a, b) and ( b, a), but only one subset a, b .
In this section we are concerned not with counting lists, but with
counting subsets. As was noted above, the basic question is this: How
many subsets can be made by choosing k elements from an n-element
set? We begin with some notation that gives a name to the answer to this
question. Counting 68 Deﬁnition 3.2 If n and k are integers, then n denotes the number
k
of subsets that can be made by choosing k elements from a set with n
elements. The symbol n is read “ n choose k.” (Some textbooks write
k
C ( n, k) instead of n .)
k
To illustrate this deﬁnition, the following table computes the values of
for various values of k by actually listing all the subsets of the 4-element
set A = a, b, c, d that have cardinality k. The values of k appear in the
far-left column. To the right of each k are all of the subsets (if any) of A of
size k. For example, when k = 1, set A has four subsets of size k, namely
a , b , c and d . Therefore 4 = 4. Similarly, when k = 2 there are six
1
subsets of size k so 4 = 6.
2
4
k k 4
k k-element subsets of a, b, c, d −1 4
−1 0 4
0 =1 =0 1 a,b, c,d 4
1 =4 2 a, b , a, c , a, d , b , c , b , d , c , d 4
2 =6 3 a, b , c , a, b , d , a, c , d , b , c , d 4
3 =4 4 a, b , c , d 4
4 =1 5 4
5 =0 6 4
6 =0 When k = 0, there is only one subset of A that has cardinality k, namely
the empty set, . Therefore 4 = 1.
0
Notice that if k is negative or greater than | A |, then A has no subsets
of cardinality k, so 4 = 0 in these cases. In general n = 0 whenever k < 0
k
k
or k > n. In particular this means n = 0 if n is negative.
k
Although it was not hard to work out the values of 4 by writing out
k
subsets in the above table, this method of actually listing sets would not
be practical for computing n when n and k are large. We need a formula.
k
To ﬁnd one, we will now carefully work out the value of 5 in such a way
3
that a pattern will emerge that points the way to a formula for any n .
k Counting Subsets 69 To begin, note that 5 is the number of 3-element subsets of a, b, c, d , e .
3
These are listed in the following table. We see that in fact 5 = 10.
3
5
3 3! a,b,c a,b,d a,b,e a,c,d a,c,e a,d,e b,c,d b,c,e b,d,e c,d,e The formula will emerge when we expand this table as follows. Taking
any one of the ten 3-element sets above, we can make 3! diﬀerent nonrepetitive lists from its elements. For example, consider the ﬁrst set a, b, c .
The ﬁrst column of the following table tallies the 3! = 6 diﬀerent lists that
can be the letters a, b, c . The second column tallies the lists that can be
made from a, b, d , and so on.
5
3 abc abe acd ace ade bcd bce bde cde acb adb aeb adc aec aed bdc bec bed ced bac bad bae cad cae dae cbd cbe dbe dce bca bda bea cda cea dea cdb ceb deb dec cba dba eba dca eca eda dcb ecb edb edc cab 3! abd dab eab dac eac ead dbc ebc ebd ecd The ﬁnal table has 5 columns and 3! rows, so it has a total of 3! 5 lists.
3
3
But notice also that the table consists of every non-repetitive length-3 list
that can be made from the symbols a, b, c, d , e . We know from Fact 3.2
5!
that there are (5−3)! such lists. Thus the total number of lists in the table
5!
is 3! 5 = (5−3)! . Dividing both sides of this equation by 3!, we get
3
5
5!
=
.
3
3!(5 − 3)! Working this out, you will ﬁnd that it does give the correct value of 10.
But there was nothing special about the values 5 and 3. We could
do the above analysis for any n instead of 5 . The table would have n
k
3
k
columns and k! rows. We would get
n
n!
=
.
k
k!( n − k)! We summarize this as follows: Counting 70 n
n!
n
=
. Otherwise
= 0.
k
k!( n − k)!
k Fact 3.3 If n, k ∈ Z and 0 ≤ k ≤ n, then Let’s now use our new knowledge to work some exercises.
Example 3.4 How many 4-element subsets does 1, 2, 3, 4, 5, 6, 7, 8, 9 have?
9!
· 7· 6
9!
The answer is 9 = 4!(9−4)! = 4!5! = 9·84!5! ·5! = 9·8·7·6 = 9·8·7·6 = 126.
4!
24
4
Example 3.5 A single 5-card hand is dealt oﬀ of a standard 52-card deck.
How many diﬀerent 5-card hands are possible?
To answer this, think of the deck as being a set D of 52 cards. Then a
5-card hand is just a 5-element subset of D . For example, here is one of
many diﬀerent 5-card hands that might be dealt from the deck.
7
♣ 2 , ♣ , 3
♥ , A
♠ , 5
♦ The total number of possible hands equals the number of 5-element
subsets of D , that is
52
52 · 51 · 50 · 49 · 48 · 47! 52 · 51 · 50 · 49 · 48
52!
=
=
= 2, 598, 960.
=
5! · 47!
5! · 47!
5!
5 Thus the answer to our question is that there are 2,598,960 diﬀerent
ﬁve-card hands that can be dealt from a deck of 52 cards.
Example 3.6 This problem concerns 5-card hands that can be dealt oﬀ
of a 52-card deck. How many such hands are there in which two of the
cards are clubs and three are hearts?
Solution: Think of such a hand as being described by a list of length
two of the form
∗ ♣ , ∗ ♣ , ∗ ♥ , ∗ ♥ , ∗ ♥ , where the ﬁrst entry is a 2-element subset of the set of 13 club cards, and
the second entry is a 3-element subset of the set of 13 heart cards. There
are 123 choices for the ﬁrst entry and 133 choices for the second entry, so
13! 13!
by the Multiplication Principle there are 123 133 = 2!11! 3!10! = 66, 924 such
lists. Answer: There are 66, 924 possible 5-card hands with two clubs
and three hearts.
Example 3.7 Imagine a lottery that works as follows. A bucket contains
36 balls numbered 1, 2, 3, 4, ..., 36. Six of these balls will be drawn randomly.
For $1 you buy a ticket that has six blanks:
. You ﬁll in the
blanks with six diﬀerent numbers between 1 and 36. You win $1, 000, 000 Counting Subsets 71 if you chose the same numbers that are drawn, regardless of order. What
are your chances of winning?
Solution: In ﬁlling out the ticket you are choosing six numbers from
36!
a set of 36 numbers. Thus there are 366 = 6!(36−6)! = 1, 947, 792 diﬀerent
combinations of numbers you might write. Only one of these will be a
winner. Your chances of winning are one in 1, 947, 792.
Exercises for Section 3.3
1. Suppose a set A has 37 elements. How many subsets of A have 10 elements?
How many subsets have 30 elements? How many have 0 elements?
2. Suppose A is a set for which | A | = 100. How many subsets of A have 5 elements?
How many subsets have 10 elements? How many have 99 elements?
3. A set X has exactly 56 subsets with 3 elements. What is the cardinality of X ?
4. Suppose a set B has the property that X : X ∈ P (B), | X | = 6 = 28. Find |B|. 5. How many 16-digit binary strings contain exactly seven 1’s? (Examples of such
strings include 0111000011110000 and 0011001100110010, etc.)
6. X ∈ P ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ) : | X | = 4 = 7. X ∈ P ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ) : | X | < 4 = 8. This problem concerns lists made from the symbols A,B,C,D,E,F,G,H,I.
(a) How many length-5 lists can be made if repetition is not allowed and the
list is in alphabetical order? (Example: BDEFI or ABCGH, but not BACGH.)
(b) How many length-5 lists can be made if repetition is not allowed and the
list is not in alphabetical order?
9. This problem concerns lists of length six made from the letters A,B,C,D,E,F,
without repetition. How many such lists have the property that the D occurs
before the A?
10. A department consists of ﬁve men and seven women. From this department
you select a committee with three men and two women. In how many ways
can you do this?
11. How many 10-digit integers contain no 0’s and exactly three 6’s?
12. Twenty one people are to be divided into two teams, the Red Team and the
Blue Team. There will be 10 people on Red Team and 11 people on Blue Team.
In how many ways can this be done?
13. Suppose n and k are integers for which 0 ≤ k ≤ n. Use the formula
n
to show that n = n−k .
k n
k = n!
k!( n− k)! 14. Suppose n, k ∈ Z, and 0 ≤ k ≤ n. Use Deﬁnition 3.2 alone (without using Fact 3.2)
n
to show that n = n−k .
k Counting 72
3.4 Pascal’s Triangle and the Binomial Theorem There are some beautiful and signiﬁcant patterns among the numbers n .
k
This section investigates a pattern based on one equation in particular. It
happens that
n+1
n
n
=
+
k
k−1
k (3.2) for any integers n and k with 0 < k ≤ n.
To see why this is true, recall that n+1 equals the number of k-element
k
subsets of a set with n + 1 elements. Now, the set A = 0, 1, 2, 3, . . . , n has
n + 1 elements, so n+1 equals the number of k-element subsets of A . Such
k
subsets can be divided into two types: those that contain 0 and those that
do not contain 0. To make a k-element subset that contains 0 we can start
with 0 and then append to this set an additional k − 1 numbers selected
n
from 1, 2, 3, . . . , n . There are k−1 ways to make this selection, so there
n
are k−1 k-element subsets of A that contain 0. Concerning the k-element
subsets of A that do not contain 0, there are n of these sets, for we can
k
form them by selecting k elements from the n-element set 1, 2, 3, . . . , n . In
light of all this, Equation (3.2) just expresses the obvious fact that the
number of k-element subsets of A equals the number of k-element subsets
that contain 0 plus the number of k-element subsets that do not contain 0. .
.. 7
0 6
0 5
0
7
1 4
0
6
1 3
0
5
1
7
2 2
0
4
1
6
2 1
0
3
1
5
2
7
3 0
0
2
1
4
2
6
3 .
.
. 1
1
3
2
5
3
7
4 1
2
2
4
3
6
4 1
3
3
5
4
7
5 1
4
4
6
5 1
5
5
7
6 1 6
6 .. . 3
4 1
3 6 1 1
10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 . 35 21 7 1 .
.
.
..
..
.
1 7
7 1
2 4 5 Figure 3.2. Pascal’s Triangle
Now that we have seen why Equation (3.2) is true, we are going to
arrange the numbers n in a triangular pattern that highlights various
k
relationships among them. The left-hand side of Figure 3.2 shows numbers
n
0
k arranged in a pyramid with 0 at the apex, just above a row containing
1
2
k with k = 0 and k = 1. Below this is a row listing the values of k for
k = 0, 1, 2. In general, each row listing the numbers n is just above a row
k
listing the numbers n+1 .
k Pascal’s Triangle and the Binomial Theorem 73 Any number n+1 in this pyramid falls immediately below and between
k
n
the positions of the two numbers k−1 and n in the previous row. But
k
n
Equation 3.2 says n+1 = k−1 + n , and therefore any number (other than 1)
k
k
in the pyramid is the sum of the two numbers immediately above it.
This pattern is especially evident on the right of Figure 3.2, where the
values of each n have been worked out. Notice how 21 is the sum of the
k
two numbers 6 and 15 above it. Similarly, 5 is the sum of the numbers 1
and 4 above it and so on.
The arrangement of numbers on the right of Figure 3.2 is called
Pascal’s Triangle. (It is named after Blaise Pascal, 1623–1662, a french
mathematician and philosopher who discovered many of its properties.)
Although we have written only the ﬁrst eight rows (beginning with Row 0
at the apex) of Pascal’s triangle, it obviously could be extended downward
indeﬁnitely. We could add an additional row at the bottom by placing a
1 at each end and obtaining each remaining number by adding the two
numbers above its position. Doing this would give the following row.
1 8 28 56 70 56 28 8 1 This row consists of the numbers 8 for 0 ≤ k ≤ 8, and we have computed
k
8!
them without using the formula 8 = k!(8−k)! . Any number n can be
k
k
computed in this fashion.
We call the very top row (containing only 1) Row 0. Row 1 is the next
row down, followed by Row 2, then Row 3, etc. With this labeling, Row n
consists of the numbers n for 0 ≤ k ≤ n.
k
1
1
1
1
1
1
. .. 3 3 .
.
. 1 x3 + 3 x2 y + 3 x y2 + 1 y3 1
4 10 + 1y 1 x2 + 2 x y + 1 y2 1 6
10 1x 1
2 4
5 1 1 x4 + 4 x3 y2 + 6 x2 y2 + 4 x y3 + 1 y4 1
5 1
.. . 1 x5 + 5 x4 y + 10 x3 y2 + 10 x2 y3 + 5 x y4 + 1 y5
.
..
.
.
.
. .. Figure 3.3. The n th row of Pascal’s Triangle lists the coeﬃcients of ( x + y)n
Notice that the numbers in Row n of Pascal’s triangle appear to be the
same as the coeﬃcients of the expansion of ( x + y)n . For example, for n = 2
we have ( x + y)2 = 1 x2 + 2 x y + 1 y2 and Row 2 of Pascal’s Triangle lists the Counting 74 coeﬃcients 1 2 1. Similarly ( x + y)3 = 1 x3 + 3 x2 y + 3 x y2 + 1 y3 , and Row 3 lists
the coeﬃcients 1 3 3 1. Pascal’s Triangle is shown on the left of Figure
3.3 and immediately to the right of each Row n is listed the expansion of
( x + y)n . In every case (at least as far as you care to check) the numbers in
Row n match up with the coeﬃcients of ( x + y)n .
In fact this turns out to be true for every n. This result is known as
the Binomial Theorem, and it is worth mentioning here. It tells how to
raise a binomial x + y to a non-negative integer power n.
Theorem 3.1
( x + y) n = n
0 xn + (Binomial Theorem) If n is a non-negative integer, then
n n−1
n
y + n x n−2 y2 + n x n−3 y3 + · · · + n−1 x yn−1 + n yn .
1x
2
3
n For now we will be content to accept the Binomial Theorem without
proof. (You will be asked to prove it in an exercise in Chapter 10.) But
until then you may ﬁnd it useful from time to time. For instance you
can apply it if you ever need to expand an expression such as ( x + y)7 . To
do this, look at Row 7 of Pascal’s Triangle in Figure 3.2 and apply the
Binomial Theorem to get
( x + y)7 = x7 + 7 x6 y + 21 x5 y2 + 35 x4 y3 + 35 x3 y4 + 21 x2 y5 + 7 x y6 + y7 . Exercises for Section 3.4
1. Write out Row 11 of Pascal’s triangle.
2. Use the Binomial Theorem to ﬁnd the coeﬃcient of x8 y5 in ( x + y)13 .
3. Use the Binomial Theorem to ﬁnd the coeﬃcient of x8 in ( x + 2)13 .
4. Use the Binomial Theorem to ﬁnd the coeﬃcient of x6 y3 in (3 x − 2 y)9 .
n
n
= 2n .
k=0 k
n
Deﬁnition 3.2 and Fact 1.3 to show k=0 n = 2n .
k
the Binomial Theorem to show n=0 3k n = 4n .
k
k 5. Use the Binomial Theorem to show
6. Use
7. Use 8. Use Fact 3.3 to derive Equation 3.2.
n
0 9. Use the Binomial Theorem to show
10. Show that the formula k n
k =n n−1
k−1 − n
1 13. Show that n
k
n
3 k
m = n n− m
m k− m
2
3
4
2+2+2 = 5
2 − n
3 n
(−1)k n
k
k=0 .
+ n
2 + n
4 − · · · + (−1)n n
n = 0. is true for all integers n, k with 0 ≤ k ≤ n. 11. Use the Binomial Theorem to show 9n =
12. Show that + +···+ n−1
2 . 10n−k . Inclusion-Exclusion 75 3.5 Inclusion-Exclusion
Many counting problems involve computing the cardinality of a union A ∪ B
of two ﬁnite sets A and B. We examine this kind of problem now.
First we develop a formula for | A ∪ B|. It is tempting to say that | A ∪ B|
must equal | A | + |B|, but that is not quite right. If we count the elements
of A and then count the elements of B and add the two ﬁgures together,
we get | A | + |B|. But if A and B have some elements in common then we
have counted each element in A ∩ B twice. B A Therefore | A | + |B| exceeds | A ∪ B| by | A ∩ B|, and consequently | A ∪ B| =
| A | + |B| − | A ∩ B|. This can be a useful equation.
| A ∪ B | = | A | + |B | − | A ∩ B | (3.3) Notice that the sets A , B and A ∩ B are all generally smaller than A ∪ B, so
Equation (3.3) has the potential of reducing the problem of determining
| A ∪ B| to three simpler counting problems. It is sometimes called an
inclusion-exclusion formula because elements in A ∩ B are included (twice)
in | A |+|B|, then excluded when | A ∩ B| is subtracted. Notice that if A ∩ B = ,
then we do in fact get | A ∪ B| = | A | + |B|; conversely if | A ∪ B| = | A | + |B|, then
it must be that A ∩ B = .
Example 3.8 A three-card hand is dealt oﬀ of a standard 52-card deck.
How many diﬀerent such hands are there for which all three cards are red
or all three cards are face cards?
Solution: Let A be the set of 3-card hands where all three cards are
red (i.e. either ♥ or ♦). Let B be the set of 3-card hands in which all three
cards are face cards (i.e. J,K or Q of any suit). These sets are illustrated
below.
A = B = 5
♥ K
♠ , , K
♦ K
♦ , , 2
♥ J
♣ K , ♥ , K
♥ J , ♥ , J
♥ Q , ♥ , Q
♥ A , ♦ , Q
♦ 6 , ♦ , Q
♣ 6 , ,... ♥ , Q
♥ ,... (Red cards) (Face cards) Counting 76 We seek the number of 3-card hands that are all red or all face cards,
and this number is | A ∪ B|. By Formula (3.3), | A ∪ B| = | A | + |B| − | A ∩ B|.
Let’s examine | A |, |B| and | A ∩ B| separately. Any hand in A is formed
by selecting three cards from the 26 red cards in the deck, so | A | = 236 .
Similarly, any hand in B is formed by selecting three cards from the 12
face cards in the deck, so |B| = 132 . Now think about A ∩ B. It contains all
the 3-card hands made up of cards that are red face cards.
A∩B = K
♥ , K
♦ , J
♥ , K
♥ , J
♥ , Q
♥ ,, Q
♦ , J
♦ , Q
♥ ,... (Red face
cards) The deck has only six red face cards, so | A ∩ B| = 6 .
3
Now we can answer our question. The number of 3-card hands that
are all red or all face cards is | A ∪ B| = | A | + |B| − | A ∩ B| = 236 + 132 − 6 =
3
2600 + 220 − 20 = 2800.
There is an analogue to Equation (3.3) that involves three sets. Consider
three sets A , B and C , as represented in the following Venn Diagram.
C A B Using the same kind of reasoning that resulted in Equation (3.3), you can
convince yourself that
| A ∪ B ∪ C | = | A | + | B | + | C | − | A ∩ B | − | A ∩ C | − | B ∩ C | + | A ∩ B ∩ C |. (3.4) There’s probably not much harm in ignoring this one for now, but if you
ﬁnd this kind of thing intriguing you should deﬁnitely take a course in
combinatorics. (Ask your instructor!)
As we’ve noted, Equation (3.3) becomes | A ∪ B| = | A | + |B| if it happens
that A ∩ B = . Also, in Equation (3.4), note that if A ∩ B = , A ∩ C = , and
B ∩ C = we get the simple formula | A ∪ B ∪ C | = | A | + |B| + |C |. In general
we have the following formula for n sets, none of which overlap. It is
sometimes called the Addition Principle.
Fact 3.4
Ai ∩ A j = (The Addition Principle) If A 1 , A 2 , . . . , A n are sets for which
whenever i = j , then | A 1 ∪ A 2 ∪ · · · ∪ A n | = | A 1 | + | A 2 | + · · · + | A n |. Inclusion-Exclusion 77 Example 3.9 How many 7-digit binary strings (0010100, 1101011, etc.)
have an odd number of 1’s?
Solution: Let A be the set of all 7-digit binary strings with an odd
number of 1’s, so the answer to the question will be | A |. To compute | A |,
we break A up into smaller parts. Notice any string in A will have either
one, three, ﬁve or seven 1’s. Let A 1 be the set of 7-digit binary strings
with only one 1. Let A 3 be the set of 7-digit binary strings with three 1’s.
Let A 5 be the set of 7-digit binary strings with ﬁve 1’s, and let A 7 be the
set of 7-digit binary strings with seven 1’s. Therefore A = A 1 ∪ A 3 ∪ A 5 ∪ A 7 .
Notice that any two of the sets A i have empty intersection, so Fact 3.4
gives | A | = | A 1 | + | A 3 | + | A 5 | + | A 7 |.
Now the problem is to ﬁnd the values of the individual terms of this
sum. For instance take A 3 , the set of 7-digit binary strings with three 1’s.
Such a string can be formed by selecting three out of seven positions for
the 1’s and putting 0’s in the other spaces. Therefore | A 3 | = 7 . Similarly
3
| A 1 | = 7 , | A 5 | = 7 , and | A 7 | = 7 . Finally the answer to our question is
1
5
7
| A | = | A 1 | + | A 3 | + | A 5 | + | A 7 | = 7 + 7 + 7 + 7 = 7 + 35 + 21 + 1 = 64. There
1
3
5
7
are 64 seven-digit binary strings with an odd number of 1’s.
You may already have been using the Addition Principle intuitively,
without thinking of it as a free-standing result. For instance, we used it
in Example 3.2(c) when we divided lists into four types and computed the
number of lists of each type.
Exercises for Section 3.5
1. At a certain university 523 of the seniors are history majors or math majors
(or both). There are 100 senior math majors, and 33 seniors are majoring in
both history and math. How many seniors are majoring in history?
2. How many 4-digit positive integers are there for which there are no repeated
digits, or all digits are odd?
3. How many 4-digit positive integers are there that are even or contain no 0’s?
4. This problem involves lists made from the letters T,H,E,O,R,Y, with repetition
allowed.
(a) How many 4-letter lists are there that don’t begin with T, or don’t end in
Y?
(b) How many 4-letter lists are there in which the sequence of letters T,H,E
appears consecutively?
(c) How many 5-letter lists are there in which the sequence of letters T,H,E
appears consecutively? 78 Counting 5. How many 7-digit binary strings begin in 1 or end in 1 or have exactly four 1’s?
6. Is the following statement true or false? Explain. If A 1 ∩ A 2 ∩ A 3 = , then
| A 1 ∪ A 2 ∪ A 3 | = | A 1 | + | A 2 | + | A 3 |.
7. This problem concerns 4-card hands dealt oﬀ of a standard 52-card deck. How
many 4-card hands are there for which all four cards are of the same suit or
all four cards are red?
8. This problem concerns 4-card hands dealt oﬀ of a standard 52-card deck. How
many 4-card hands are there for which all four cards are of diﬀerent suits or
all four cards are red?
9. A 4-letter list is made from the letters L,I,S,T,E,D according to the following
rule: Repetition is allowed, and the ﬁrst two letters on the list are vowels or
the list ends in D. How many such lists are possible?
10. A 5-card poker hand is called a ﬂush if all cards are the same suit. How many
diﬀerent ﬂushes are there? Part II
How to Prove Conditional
Statements CHAPTER 4 Direct Proof t is time to prove some theorems. There are various methods of doing
this; we now examine the most straightforward approach, a technique
called direct proof. As we begin, it is important to keep in mind the
meanings of three key terms: Theorem, proof and deﬁnition.
A theorem is a mathematical statement that is true, and can be (and
has been) veriﬁed as true. A proof of a theorem is a written veriﬁcation
that shows that the theorem is deﬁnitely and unequivocally true. A proof
should be understandable and convincing to anyone who has the requisite
background and knowledge. This knowledge includes an understanding of
the meanings of the mathematical words, phrases and symbols that occur
in the theorem and its proof. It is crucial that both the writer of the proof
and the readers of the proof agree on the exact meanings of all the words,
for otherwise there is an intolerable level of ambiguity. A deﬁnition is an
exact, unambiguous explanation of the meaning of a mathematical word or
phrase. We will elaborate on the terms theorem and deﬁnition in the next
two sections, and then ﬁnally we will be ready to begin writing proofs. I 4.1 Theorems
A theorem is a statement that is true and has been proved to be true.
You have encountered many theorems in your mathematical education.
Here are some theorems taken from an undergraduate calculus text. They
will be familiar to you, though you may not have read all the proofs.
Theorem: Let f be continuous on an open interval I and let c ∈ I . If
f ( c) is the maximum or minimum value of f on I and if f ( c) exists,
then f ( c) = 0.
Theorem: If ∞
a
k=1 k converges, then limk→∞ a k = 0. Theorem: Suppose f is continuous on the interval [a, b]. Then f is
integrable on [a, b]. 82 Direct Proof Theorem: Every absolutely convergent series converges.
Observe that each of these theorems either has the conditional form “If
P , then Q ,” or can be put into that form. The ﬁrst theorem has an initial
sentence “Let f be continuous on an open interval I , and let c ∈ I ,” which
sets up some notation, but a conditional statement follows it. The third
theorem has form “Suppose P . Then Q ,” but this means the same thing
as “If P , then Q .” The last theorem can be re-expressed as “If a series is
absolutely convergent, then it is convergent.”
A theorem of form “If P , then Q ,” can be regarded as a device that
produces new information from P . Whenever we are dealing with a
situation where P is true, then the theorem guarantees that, in addition,
Q is true. Since this kind of expansion of information is useful, theorems
of form “If P , then Q ,” are very common.
But not every theorem is a conditional statement. Some have the form
of the biconditional P ⇔ Q , but, as we know, that can be expressed as two
conditional statements. Other theorems simply state facts about speciﬁc
things. For example, here is another theorem from your study of calculus.
1
Theorem: The series 1 + 1 + 3 + 1 + 1 + · · · diverges.
2
4
5 It would be diﬃcult (or at least awkward) to restate this as a conditional
statement. Still, it is true that most theorems are conditional statements,
so much of this book will concentrate on that type of theorem.
It is important to be aware that there are a number of words that mean
essentially the same thing as the word “theorem,” but which are used in
slightly diﬀerent ways. In general the word “theorem” is reserved for a
statement that is considered important or signiﬁcant (the Pythagorean
Theorem, for example). A statement that is true but not as signiﬁcant
is sometimes called a proposition. A lemma is a theorem whose main
purpose is to help prove another theorem. A corollary is a result that is
an immediate consequence of a theorem or proposition. It is not important
that you remember all these words now, for their meanings will become
clear with usage.
Our main task is to learn how to prove theorems. As the above examples
suggest, proving theorems requires a clear understanding of the meaning
of the conditional statement, and that is the primary reason we studied it
so extensively in Chapter 2. In addition, it is also crucial to understand
the role of deﬁnitions. Deﬁnitions 83 4.2 Deﬁnitions
A proof of a theorem should be absolutely convincing. Ambiguity must
be avoided. Thus everyone must agree on the exact meaning of each
mathematical term. In Chapter 1 we deﬁned the meanings of the sets N,
Z, R, Q and , as well as the meanings of the symbols ∈ and ⊆, and we
shall make frequent use of these things. Here is another deﬁnition that
we use often.
Deﬁnition 4.1 An integer n is even if n = 2a for some integer a ∈ Z. Thus, for example, 10 is even because 10 = 2 · 5. Also, according to the
deﬁnition, 7 is not even because there is no integer a for which 7 = 2a.
While there would be nothing wrong with deﬁning an integer to be odd if
it’s not even, the following deﬁnition is more concrete.
Deﬁnition 4.2 An integer n is odd if n = 2a + 1 for some integer a ∈ Z. Thus 7 is odd because 7 = 2 · 3 + 1. We will use these deﬁnitions whenever
the concept of even or odd numbers arises. If in a proof a certain number
turns out to be even, the deﬁnition allows us to write it as 2a for an
appropriate integer a. If some quantity has form 2b + 1 where b is an
integer, then the deﬁnition tells us the quantity is odd.
Deﬁnition 4.3 Two integers have the same parity if they are both even
or they are both odd. Otherwise they have opposite parity.
Thus 5 and −17 have the same parity, as do 8 and 0; but 3 and 4 have
opposite parity.
Two points about deﬁnitions are in order. First, in this book the word
or term being deﬁned appears in boldface type. Second, it is common to
express deﬁnitions as conditional statements even though the biconditional
would more appropriately convey the meaning. Consider the deﬁnition
of an even integer. You understand full well that if n is even then n = 2a
(for a ∈ Z), and if n = 2a, then n is even. Thus, technically the deﬁnition
should read “An integer n is even if and only if n = 2a for some a ∈ Z.”
However, it is an almost-universal convention that deﬁnitions are phrased
in the conditional form, even though they are interpreted as being in
the biconditional form. There is really no good reason for this, other
than economy of words. It has just become the standard way of writing
deﬁnitions, and we have to get used to it.
Here is another deﬁnition that we will use often. Direct Proof 84 Deﬁnition 4.4 Suppose a and b are integers. We say that a divides b,
written a | b, if b = ac for some c ∈ Z. In this case we also say that a is a
divisor of b, and that b is a multiple of a.
For example, 5 divides 15 because 15 = 5 · 3. We write this as 5 | 15.
Similarly 8 | 32 since 32 = 8 · 4, and −6 | 6 since 6 = −6 · −1. However, 6 does
not divide 9 because there is no integer for which 9 = 6 · c. We express this
as 6 9, which we read as “6 does not divide 9.”
Be careful of your interpretation of the symbols. There is a big diﬀerence
between the expressions a | b and a/ b. The expression a | b is a statement,
while a/b is a fraction. For example, 8 | 16 is true and 8 | 20 is false. By
contrast, 8/16 = 0.5 and 8/20 = 0.4 are numbers, not statements. Be careful
to not write one when you mean the other.
Every integer has a set of integers which divide it. For example the set
of divisors of 6 is a ∈ Z : a | 6 = − 6, −3, −2, −1, 1, 2, 3, 6 . The set of divisors
of 5 is − 5, −1, 1, 5 . The set of divisors of 0 is Z. This brings us to the
following deﬁnition, with which you are already familiar.
Deﬁnition 4.5 A natural number p is prime if it has exactly two positive
divisors, 1 and p.
Of course not all terms can be deﬁned. If we deﬁned every word that
appeared in a deﬁnition, we would need separate deﬁnitions for the words
that appeared in those deﬁnitions, and so on until the chain of deﬁned
terms became circular. Thus we accept some ideas as being so intuitively
clear that they do not require deﬁnitions. For example, we will not ﬁnd
it necessary to deﬁne exactly what an integer is, nor will we deﬁne what
addition, multiplication, subtraction and division are. We will freely use
such things as the distributive and commutative properties of addition
and multiplication, along with all the usual properties of arithmetic and
algebra. In addition, we accept the following statements as being so
obviously true that they do not need to be proved.
Fact 4.1 Suppose a and b are integers. Then:
• a+b ∈Z
• a−b ∈Z
• ab ∈ Z These three statements can be combined. For example, we see that if
a, b and c are integers, then a2 b − ca + b is also an integer. Another fact that we will accept without proof (at least for now) is
that every natural number greater than 1 has a unique factorization into Direct Proof 85 primes. For example, the number 1176 can be factored into primes as
1176 = 2 · 2 · 2 · 3 · 7 · 7 = 23 · 3 · 72 . By unique we mean that any factorization
of 1176 into primes will have exactly the same factors (i.e. three 2’s, one 3,
and two 7’s). Thus, for example, there is no valid factorization of 1176 that
has a factor of 5. You may be so used to factoring numbers into primes that you think it is obvious that there cannot be diﬀerent prime factorizations
of the same number, but in fact this is a fundamental result whose proof
is not transparent. Nonetheless, we will be content to assume that every
natural number greater than 1 has a unique factorization into primes.
(Though you may wish to revisit the issue of a proof once you become
ﬂuent at proving theorems.)
We will introduce other accepted facts, as well as other deﬁnitions, as
they are needed.
4.3 Direct Proof
This section explains a simple technique for proving theorems or propositions which have the form of conditional statements. The technique is
called direct proof. To simplify the discussion, our ﬁrst examples will
involve proving statements that are almost obviously true. Thus we will
call the statements propositions rather than theorems. (Remember, a
proposition is a statement that, although true, is not as signiﬁcant as a
theorem.)
To understand how the technique of direct proof works, suppose we
have some proposition of the following form.
Proposition If P , then Q . This proposition is a conditional statement of form P ⇒ Q . Our goal
is to show that this conditional statement is true. To see how to proceed,
look at the truth table.
P Q P ⇒Q T T T T F F F T T F F T The table shows that if P is false, the statement P ⇒ Q is automatically
true. This means that if we are concerned with showing P ⇒ Q is true, we
don’t have to worry about the situations where P is false (as in the last
two lines of the table) because the statement P ⇒ Q will be automatically Direct Proof 86 true in those cases. But we must be very careful about the situations
where P is true (as in the ﬁrst two lines of the table). We must show that
the condition of P being true forces Q to be true also, for that means the
second line of the table cannot happen.
This gives a fundamental outline for proving statements of the form
P ⇒ Q . Begin by assuming that P is true (remember, we don’t need to worry
about P being false) and show this forces Q to be true. We summarize this
as follows.
Outline for Direct Proof.
Proposition If P , then Q . Proof. Suppose P .
.
.
. Therefore Q .
So the setup for direct proof is remarkably simple. The ﬁrst line of the
proof is the sentence “Suppose P .” The last line is the sentence “Therefore
Q .” Between the ﬁrst and last line we use logic, deﬁnitions and standard
math facts to transform the statement P to the statement Q . It is common
to use the word “Proof” to indicate the beginning of a proof, and the symbol
to indicate the end.
As our ﬁrst example, let’s prove that if x is odd then x2 is also odd.
(Granted, this is not a terribly impressive result, but we will move on to
more signiﬁcant things in due time.) The ﬁrst step in the proof is to ﬁll
in the outline for direct proof. This is a lot like painting a picture, where
the basic structure is sketched in ﬁrst. We leave some space between the
ﬁrst and last line of the proof. The following series of frames indicates the
steps you might take to ﬁll in this space with a logical chain of reasoning.
Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Therefore x2 is odd.
Now that we have written the ﬁrst and last lines, we need to ﬁll in the
space with a chain of reasoning that shows that x being odd forces x2 to be
odd. In doing this it’s always advisable to use any deﬁnitions that apply. Direct Proof 87 The ﬁrst line says x is odd, and by Deﬁnition 4.2 it must be that x = 2a + 1
for some a ∈ Z, so we write this in as our second line.
Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd.
Then x = 2a + 1 for some a ∈ Z, by deﬁnition of an odd number.
Therefore x2 is odd.
Now jump down to the last line, which says x2 is odd. Think about what
the line immediately above it would have to be in order for us to conclude
that x2 is odd. By the deﬁnition of an odd number, we would have to have
x2 = 2a + 1 for some a ∈ Z. However, the symbol a now appears earlier in
the proof in a diﬀerent context, so we should use a diﬀerent symbol, say b.
Proposition If x is odd, then x2 is odd.
Proof. Suppose x is odd.
Then x = 2a + 1 for some a ∈ Z, by deﬁnition of an odd number.
Thus x2 = 2b + 1 for an integer b.
Therefore x2 is odd, by deﬁnition of an odd number. We are almost there. We can bridge the gap as follows.
Proposition If x is odd, then x2 is odd.
Proof. Suppose x is odd.
Then x = 2a + 1 for some a ∈ Z, by deﬁnition of an odd number.
Thus x2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 + 2a) + 1.
So x2 = 2b + 1 where b is the integer b = 2a2 + 2a.
Thus x2 = 2b + 1 for an integer b.
Therefore x2 is odd, by deﬁnition of an odd number. Finally, we may wish to clean up our work and write the proof in paragraph
form. Here is our ﬁnal version. Direct Proof 88
Proposition If x is odd, then x2 is odd. Proof. Suppose x is odd. Then x = 2a + 1 for some a ∈ Z, by deﬁnition
of an odd number. Thus x2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 + 2a) + 1,
so x2 = 2b + 1 where b = 2a2 + 2a ∈ Z. Thus x2 = 2b + 1 for an integer
b. Therefore x2 is odd, by deﬁnition of an odd number.
It’s generally a good idea to write the ﬁrst and last line of your proof
ﬁrst, and then ﬁll in the gap, sometimes jumping alternately between
top and bottom until you meet in the middle, as we did above. This way
you are constantly reminded that you are aiming for the statement at
the bottom. Sometimes you will leave too much space, sometimes not
enough. Sometimes you will get stuck before ﬁguring out what to do. This
is normal. Mathematicians do scratch work just as artists do sketches for
their paintings.
Here is another example. Consider writing a proof of the following
proposition.
Proposition Let a, b and c be integers. If a | b and b | c, then a | c. Let’s apply the basic outline for direct proof. To clarify the procedure
we will write the proof in stages again.
Proposition Let a, b and c be integers. If a | b and b | c, then a | c. Proof. Suppose a | b and b | c. Therefore a | c.
Our ﬁrst step is to apply Deﬁnition 4.4 to the ﬁrst line. The deﬁnition
says a | b means b = ac for some integer c, but since c already appears in
a diﬀerent context on the ﬁrst line, we must use a diﬀerent letter, say d .
Similarly let’s use a new letter e in the deﬁnition of b | c.
Proposition Let a, b and c be integers. If a | b and b | c, then a | c. Proof. Suppose a | b and b | c.
By Deﬁnition 4.4, we know a | b means there is an integer d with b = ad .
Likewise, b | c means there is an integer e for which c = be.
Therefore a | c. Direct Proof 89 We have almost bridged the gap. The line immediately above the last line
should show that a | c. According to Deﬁnition 4.4, this line should say
that c = ax for some integer x. We can get this equation from the lines at
the top, as follows.
Proposition Let a, b and c be integers. If a | b and b | c, then a | c. Proof. Suppose a | b and b | c.
By Deﬁnition 4.4, we know a | b means there is an integer d with b = ad .
Likewise, b | c means there is an integer e for which c = be.
Thus c = be = (ad ) e = a( de), so c = ax for the integer x = de.
Therefore a | c.
Here is another example, though this time the proof is presented all at
once rather than in stages.
Proposition If x is an even integer, then x2 − 6 x + 5 is odd. Proof. Suppose x is an even integer.
Then x = 2a for some a ∈ Z, by deﬁnition of an even integer.
So x2 −6 x+5 = (2a)2 −6(2a)+5 = 4a2 −12a +5 = 4a2 −12a +4+1 = 2(2a2 −6a +2)+1.
Therefore we have x2 − 6 x + 5 = 2b + 1, where b = 2a2 − 6a + 2 ∈ Z.
Consequently x2 − 6 x + 5 is odd, by deﬁnition of an odd number.
In writing a proof you do not have to put each sentence on a separate
line. But we will do this for clarity in the ﬁrst few chapters of this book.
The examples we’ve looked at so far have all been proofs of statements
about integers. In our next example, we are going to prove that if x and y
are positive real numbers for which x ≤ y, then x ≤ y. You may feel that
the proof is not as “automatic” as the proofs we have done so far. Finding
the right steps in a proof can be challenging, and that is part of the fun.
Proposition Let x and y be positive numbers. If x ≤ y, then Proof. Suppose x ≤ y.
Subtracting y from both sides gives x − y ≤ 0.
2
This can be written as x − y2 ≤ 0.
Factor this to get ( x − y)( x + y) ≤ 0.
Dividing both sides by the positive number x + y produces
Adding y to both sides gives x ≤ y. x≤ x− y. y ≤ 0. Direct Proof 90 This proposition tells us that whenever x ≤ y, we can take the square
root of both sides and be assured that x ≤ y. This can be useful, as we
will see in our next proposition.
That proposition will concern the expression 2 x y ≤ x + y. Notice when
you plug in random positive values for the variables, the expression is true.
For example, in plugging in x = 6 and y = 4, the left side is 2 6 · 4 = 4 6
≈ 9.79, which is less than the right side 6 + 4 = 10. Is it true that 2 x y ≤ x + y
for any positive x and y? How could we prove it?
To see how, let’s ﬁrst cast this into the form of a conditional statement:
If x and y are positive real numbers, then 2 x y ≤ x + y. The proof begins
with the assumption that x and y are positive, and ends with 2 x y ≤ x + y.
In mapping out a strategy, it can be helpful to work backwards, working
from 2 x y ≤ x + y to something that is obviously true. Then the steps can
be reversed in the proof. In this case, squaring both sides of 2 x y ≤ x + y
gives us
4 x y ≤ x 2 + 2 x y + y2 . Now subtract 4 x y from both sides and factor.
0≤ x2 − 2 x y + y2 0 ≤ ( x − y)2 But this last line is clearly true, since the square of x − y cannot be negative!
This gives us a strategy for the proof, which follows.
Proposition If x and y are positive real numbers, then 2 x y ≤ x + y. Proof. Suppose x and y are positive real numbers.
Then 0 ≤ ( x − y)2 , so 0 ≤ x2 − 2 x y + y2 .
Adding 4 x y to both sides gives 4 x y ≤ x2 + 2 x y + y2 , which gives 4 x y ≤ ( x + y)2 .
Taking the square root of both sides produces 2 x y ≤ x + y.
Notice that in going from the next-to-last line to the last line of the
proof, we took the square root of both sides of 4 x y ≤ ( x + y)2 and got 4 x y ≤
( x + y)2 . The fact that taking roots of both sides does not alter the ≤ follows
from our previous proposition. This is an important point. Sometimes the
proof of a proposition or theorem uses another proposition or theorem.
4.4 Using Cases
In proving a statement is true, we sometimes have to examine multiple
cases before showing the statement is true in all possible scenarios. This
section illustrates a few examples. Using Cases 91 Our examples will concern the expression 1 + (−1)n (2n − 1). Here is a
table that shows what we get when plugging in various integers for n.
Notice that 1 + (−1)n (2n − 1) is a multiple of 4 in every line.
n 1 + (−1)n (2 n − 1) 1
2
3
4
5
6 0
4
−4
8
−8
12 Is 1+(−1)n (2 n−1) always a multiple of 4? We will prove the answer is “yes”
in our next example. Notice, however, that the expression 1 + (−1)n (2n − 1)
behaves diﬀerently depending on whether n is even or odd, for in the ﬁrst
case (−1)n = 1, and in the second (−1)n = −1. Thus the proof must examine
these two possibilities separately.
Proposition If n ∈ N, then 1 + (−1)n (2 n − 1) is a multiple of 4. Proof. Suppose n ∈ N.
Then n is either even or odd. Let’s consider these two cases separately.
Case 1. Suppose n is even. Then n = 2 k for some k ∈ Z, and (−1)n = 1.
Thus 1 + (−1)n (2 n − 1) = 1 + (1)(2 · 2k − 1) = 4k, which is a multiple of 4.
Case 2. Suppose n is odd. Then n = 2k + 1 for some k ∈ Z, and (−1)n = −1.
Thus 1 + (−1)n (2 n − 1) = 1 − (2(2 k + 1) − 1) = −4k, which is a multiple of 4.
These cases show that 1 + (−1)n (2n − 1) is always a multiple of 4.
Now let’s examine the ﬂip side of the question. We just proved that
1 + (−1)n (2 n − 1) is always a multiple of 4, but can we get every multiple of 4
this way? The following proposition and proof give an aﬃrmative answer.
Proposition Every multiple of 4 has form 1 + (−1)n (2n − 1) for some n ∈ N. Proof. In conditional form, the proposition is as follows:
If k is a multiple of 4, then there is an n ∈ N for which 1 + (−1)n (2n − 1) = k.
What follows is a proof of this conditional statement.
Suppose k is a multiple of 4.
This means k = 4a for some integer a.
We must produce an n ∈ N for which 1 + (−1)n (2n − 1) = k.
This is done by cases, depending on whether a is zero, positive or negative. Direct Proof 92 Case 1. Suppose a = 0. Let n = 1. Then 1 + (−1)n (2 n − 1) = 1 + (−1)1 (2 − 1) = 0
= 4 · 0 = 4a = k.
Case 2. Suppose a > 0. Let n = 2a, which is in N because a is positive. Also
n is even, so (−1)n = 1. Thus 1 + (−1)n (2 n − 1) = 1 + (2 n − 1) = 2 n = 2(2a) = 4a = k.
Case 3. Suppose a < 0. Let n = 1 − 2a, which is an element of N because
a is negative, making 1 − 2a positive. Also n is odd, so (−1)n = −1. Thus
1 + (−1)n (2 n − 1) = 1 − (2 n − 1) = 1 − (2(1 − 2a) − 1) = 4a = k.
The above cases show that no matter whether a multiple k = 4a of 4 is
zero, positive or negative, it always equals 1 + (−1)n (2n − 1) for some natural
number n.
4.5 Treating Similar Cases
Occasionally two or more cases in a proof will be so similar that writing
them separately seems tedious or unnecessary. Here is an example.
Proposition If two integers have opposite parity, then their sum is odd. Proof. Suppose m and n are two integers with opposite parity.
We need to show that m + n is odd. This is done in two cases, as follows.
Case 1. Suppose m is even and n is odd. Thus m = 2a and n = 2b + 1 for
some integers a and b. Therefore m + n = 2a + 2b + 1 = 2(a + b) + 1, which is
odd (by Deﬁnition 4.2).
Case 2. Suppose m is odd and n is even. Thus m = 2a + 1 and n = 2b for
some integers a and b. Therefore m + n = 2a + 1 + 2b = 2(a + b) + 1, which is
odd (by Deﬁnition 4.2).
In either case, m + n is odd.
The two cases in this proof are entirely alike except for the order in
which the even and odd terms occur. It is entirely appropriate to just do
one case and indicate that the other case is nearly identical. The phrase
“Without loss of generality...” is a common way of signaling that the proof
is treating just one of several nearly identical cases. Here is a second
version of the above example.
Proposition If two integers have opposite parity, then their sum is odd. Proof. Suppose m and n are two integers with opposite parity.
We need to show that m + n is odd.
Without loss of generality, suppose m is even and n is odd.
Thus m = 2a and n = 2b + 1 for some integers a and b.
Therefore m + n = 2a + 2b + 1 = 2(a + b) + 1, which is odd (by Deﬁnition 4.2). Treating Similar Cases 93 In reading proofs in other texts, you may sometimes see the phrase
“Without loss of generality” abbreviated as “WLOG.” However, in the
interest of transparency we will avoid writing it this way. In a similar
spirit, it is advisable—at least until you become more experienced in proof
writing—that you write out all cases, no matter how similar they appear
to be.
Please check your understanding by doing the following exercises.
Exercises for Chapter 4
Use the method of direct proof to prove the following statements.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12. If x is an even integer, then x2 is even.
If x is an odd integer, then x3 is odd.
If a is an odd integer, then a2 + 3a + 5 is odd.
Suppose x, y ∈ Z. If x and y are odd, then x y is odd.
Suppose x, y ∈ Z. If x is even, then x y is even.
Suppose a, b, c ∈ Z. If a | b and a | c, then a | ( b + c).
Suppose a, b ∈ Z. If a | b, then a2 | b2 .
Suppose a is an integer. If 5 | 2a, then 5 | a.
Suppose a is an integer. If 7 | 4a, then 7 | a.
Suppose a and b are integers. If a | b, then a | (3 b3 − b2 + 5b).
Suppose a, b, c, d ∈ Z. If a | b and c | d , then ac | bd .
If x ∈ R and 0 < x < 4, then x(44 x) ≥ 1.
− 13.
14.
15.
16.
17.
18.
19.
20.
21.
22. Suppose x, y ∈ R. If x2 + 5 y = y2 + 5 x, then x = y or x + y = 5.
If n ∈ Z, then 5n2 + 3n + 7 is odd. (Try cases.)
If n ∈ Z, then n2 + 3n + 4 is even. (Try cases.)
If two integers have the same parity, then their sum is even. (Try cases.)
If two integers have opposite parity, then their product is even.
Suppose x and y are positive real numbers. If x < y, then x2 < y2 .
Suppose a, b and c are integers. If a2 | b and b3 | c, then a6 | c.
If a is an integer and a2 | a, then a ∈ − 1, 0, 1 .
p
If p is prime and k is an integer for which 0 < k < p, then p divides k .
If n ∈ N, then n2 = 2 n + n . (You may need a separate case for n = 1.)
2
1 n
23. If n ∈ N, then 2n is even.
24. If n ∈ N and n ≥ 2, then the numbers n! + 2, n! + 3, n! + 4, n! + 5, . . . n! + n are
all composite. (This means that for any n, no matter how big, there exist n
consecutive composite numbers. In other words there are arbitrarily large
“gaps” between prime numbers.)
a
25. If a, b, c ∈ N and c ≤ b ≤ a, then a b = b− c a−b+ c .
bc
c
26. Every odd integer is a diﬀerence of two squares. (Example 7 = 42 − 32 , etc.) CHAPTER 5 Contrapositive Proof his chapter explains an alternative to direct proof called contrapositive proof. Like direct proof, the technique of contrapositive proof is
used to prove conditional statements of the form “If P , then Q .” Although
it is possible to use direct proof exclusively, there are occasions where
contrapositive proof is much easier. T 5.1 Contrapositive Proof
To understand how contrapositive proof works, imagine that you need to
prove a proposition of the following form.
Proposition If P , then Q . This is a conditional statement of form P ⇒ Q . Our goal is to show
that this conditional statement is true. To see how to proceed, recall that
in Section 2.6 we observed that P ⇒ Q is logically equivalent to ∼ Q ⇒∼ P .
For convenience, we duplicate the truth table that veriﬁes this fact.
P Q ∼Q ∼P P ⇒Q ∼ Q ⇒∼ P T T F F T T T F T F F F F T F T T T F F T T T T According to the table, statements P ⇒ Q and ∼ Q ⇒∼ P are diﬀerent
ways of expressing exactly the same thing. The expression ∼ Q ⇒∼ P is
called the contrapositive form of P ⇒ Q . 1
1 Do not confuse the words contrapositive and converse. Recall from Section 2.4 that the converse of P ⇒ Q is the statement Q ⇒ P , which is not logically equivalent to P ⇒ Q . Contrapositive Proof 95 Since P ⇒ Q is logically equivalent to ∼ Q ⇒∼ P , it follows that to prove
P ⇒ Q is true, it suﬃces to instead prove that ∼ Q ⇒∼ P is true. If we were
to use direct proof to show ∼ Q ⇒∼ P is true, we would assume ∼ Q is true
use this to deduce that ∼ P is true. This in fact is the basic approach of
contrapositive proof, summarized as follows.
Outline for Contrapositive Proof.
Proposition If P , then Q . Proof. Suppose ∼ Q .
.
.
. Therefore ∼ P .
So the setup for contrapositive proof is very simple. The ﬁrst line of the
proof is the sentence “Suppose Q is not true.” (Or something to that eﬀect.)
The last line is the sentence “Therefore P is not true.” Between the ﬁrst
and last line we use logic and deﬁnitions to transform the statement ∼ Q
to the statement ∼ P .
To illustrate this new technique, and to contrast it with direct proof,
we now prove a proposition in two ways: ﬁrst with direct proof and then
with contrapositive proof.
Proposition Suppose x ∈ Z. If 7 x + 9 is even, then x is odd. Proof. (Direct) Suppose 7 x + 9 is even.
Thus 7 x + 9 = 2a for some integer a.
Subtracting 6 x + 9 from both sides, we get x = 2a − 6 x − 9.
Thus x = 2a − 6 x − 9 = 2a − 6 x − 10 + 1 = 2(a − 3 x − 5) + 1.
Consequently x = 2b + 1, where b = a − 3 x − 5 ∈ Z.
Therefore x is odd.
Here is a contrapositive proof of the same statement.
Proposition Suppose x ∈ Z. If 7 x + 9 is even, then x is odd. Proof. (Contrapositive) Suppose x is not odd.
Thus x is even, so x = 2a for some integer a.
Then 7 x + 9 = 7(2a) + 9 = 14a + 8 + 1 = 2(7a + 4) + 1.
Therefore 7 x + 9 = 2b + 1, where b is the integer 7a + 4.
Consequently 7 x + 9 is odd.
Therefore 7 x + 9 is not even. Contrapositive Proof 96 Though the proofs are of equal length, you may feel that the contrapositive proof ﬂowed more smoothly. This is because it is easier to
transform information about x into information about 7 x + 9 than the other
way around. For our next example, consider the following proposition
concerning an integer x.
Proposition If x2 − 6 x + 5 is even, then x is odd. A direct proof would be problematic. We would begin by assuming that
x − 6 x + 5 is even, so x2 − 6 x + 5 = 2a. Then we would need to transform this
into x = 2b + 1 for b ∈ Z. But it is not quite clear how that could be done,
for it would involve isolating an x from the quadratic expression. However
the proof becomes very simple if we use contrapositive proof.
2 Proposition Suppose x ∈ Z. If x2 − 6 x + 5 is even, then x is odd. Proof. (Contrapositive) Suppose x is not odd.
Thus x is even, so x = 2a for some integer a.
So x2 −6 x+5 = (2a)2 −6(2a)+5 = 4a2 −12a +5 = 4a2 −12a +4+1 = 2(2a2 −6a +2)+1.
Therefore x2 − 6 x + 5 = 2b + 1, where b is the integer 2a2 − 6a + 2.
Consequently x2 − 6 x + 5 is odd.
Therefore x2 − 6 x + 5 is not even.
In summary, since x being not odd (∼ Q ) resulted in x2 − 6 x + 5 being not
even (∼ P ), then x2 − 6 x + 5 being even (P ) means that x is odd (Q ). Thus
we have proved P ⇒ Q by proving ∼ Q ⇒∼ P . Here is another example.
Proposition Suppose x, y ∈ R. If y3 + yx2 ≤ x3 + x y2 , then y ≤ x. Proof. (Contrapositive) Suppose it is not true that y ≤ x, so y > x.
Then y − x > 0. Multiply both sides of y − x > 0 by the positive value x2 + y2 .
( y − x)( x2 + y2 ) > 0 yx2 + y3 − x3 − x y2
3 y + yx 2 >0
> x 3 + x y2 Therefore y3 + yx2 > x3 + x y2 , so it is not true that y3 + yx2 ≤ x3 + x y2 .
Proving “If P , then Q ,” with the contrapositive approach necessarily
involves the negated statements ∼ P and ∼ Q . In working with these we
may have to use the techniques for negating statements (e.g. DeMorgan’s
Laws) discussed in Section 2.10. We consider such an example next. Congruence of Integers
Proposition 97 Suppose x, y ∈ Z. If 5 x y, then 5 x and 5 y. Proof. (Contrapositive) Suppose it is not true that 5 x and 5 y.
By DeMorgan’s Law, it is not true that 5 x or it is not true that 5 y.
Therefore 5 | x or 5 | y. We consider these possibilities separately.
Case 1. Suppose 5 | x. Then x = 5a for some a ∈ Z.
From this we get x y = 5(a y), and that means 5 | x y.
Case 2. Suppose 5 | y. Then y = 5a for some a ∈ Z.
From this we get x y = 5(ax), and that means 5 | x y.
The above cases show that 5 | x y, so it is not true that 5 x y.
5.2 Congruence of Integers
This is a good time to introduce a new deﬁnition. It is not necessarily
related to contrapositive proof, but introducing it now ensures that we
have a suﬃcient variety of exercises to practice all our proof techniques on.
This new deﬁnition occurs in many branches of mathematics, and it will
surely play a role in some of your later courses. But our primary reason
for introducing it is that it will give us more practice in writing proofs.
Deﬁnition 5.1 Given integers a and b and an n ∈ N, we say that a and b
are congruent modulo n if n | (a − b). We express this as a ≡ b (mod n).
If a and b are not congruent modulo n, we write this as a ≡ b (mod n).
Example 5.1 Here are some examples 1. 9 ≡ 1 (mod 4) because 4 | (9 − 1).
2. 6 ≡ 10 (mod 4) because 4 | (6 − 10).
3. 14 ≡ 8 (mod 4) because 4 (14 − 8).
4. 20 ≡ 4 (mod 8) because 8 | (20 − 4).
5. 17 ≡ −4 (mod 3) because 3 | (17 − (−4)).
In practical terms, a ≡ b (mod n) means that a and b have the same
remainder when divided by n. For example, we saw above that 6 ≡ 10
(mod 4) and indeed 6 and 10 both have remainder 2 when divided by 4.
Also we saw 14 ≡ 8 (mod 4), and sure enough 14 has remainder 2 when
divided by 4, while 8 has remainder 0.
To see that this is true in general, note that if a and b both have the
same remainder r when divided by n, then it follows that a = kn + r and
b = n + r for some k, ∈ Z. Then a − b = ( kn + r ) − ( n + r ) = n( k − ). But
a − b = n( k − ) means n | (a − b), so a ≡ b (mod n). Conversely, one of the
exercises for this chapter asks you to show that if a ≡ b (mod n), then a
and b have the same remainder when divided by n. Contrapositive Proof 98 We conclude this section with a couple of proofs involving congruence of
integers, but you will also test your skills with other proofs in the exercises.
Proposition
(mod n). Suppose a, b ∈ Z and n ∈ N. If a ≡ b (mod n), then a2 ≡ b2 Proof. We will use direct proof. Suppose a ≡ b (mod n).
By deﬁnition of congruence of integers, this means n | (a − b).
Then by deﬁnition of divisibility, there is an integer c for which a − b = nc.
Now multiply both sides of this equation by a + b.
a−b = (a − b)(a + b) =
2 a −b 2 = nc
nc(a + b)
nc(a + b) Since c(a + b) ∈ Z, the above equation tells us n | (a2 − b2 ).
According to Deﬁnition 5.1, this gives a2 ≡ b2 (mod n).
Let’s stop and think about what this proposition means. It says a ≡ b
(mod n) implies a2 ≡ b2 (mod n). In other words, it says that if integers a
and b have the same remainder when divided by n, then a2 and b2 also
have the same remainder when divided by n. As an example of this, 6 and
10 have the same remainder (2) when divided by n = 4, and their squares
36 and 100 also have the same remainder (0) when divided by n = 4. The
proposition promises this will happen for all a, b and n. In our examples
we tend to concentrate more on how to prove propositions than on what
the propositions mean. This is reasonable since our main goal is to learn
how to prove statements. But it is helpful to sometimes also think about
the meaning of what we prove.
Proposition
(mod n). Suppose a, b, c ∈ Z and n ∈ N. If a ≡ b (mod n), then ac ≡ bc Proof. We employ direct proof. Suppose a ≡ b (mod n). By Deﬁnition 5.1,
it follows that n | (a − b). Therefore, by deﬁnition of divisibility, there exists
an integer k for which a − b = nk. Multiply both sides of this equation
by c to get ac − bc = nkc. Thus ac − bc = n( kc) where kc ∈ Z, which means
n | (ac − bc). By Deﬁnition 5.1, we have ac ≡ bc (mod n).
Contrapositive proof seems to be the best approach in the next example,
since it will eliminate the symbols | and ≡. Mathematical Writing
Proposition 99 Suppose a, b ∈ Z and n ∈ N. If 12a ≡ 12 b (mod n), then n 12. Proof. (Contrapositive) Suppose n | 12, so there is an integer c for which
12 = nc. Now reason as follows.
12 =
12(a − b) =
12a − 12 b = nc
nc(a − b)
n( ca − cb) Since ca − cb ∈ Z, the equation 12a − 12 b = n( ca − cb) implies n | (12a − 12b).
This in turn means 12a ≡ 12b (mod n).
5.3 Mathematical Writing
Now that you have begun writing proofs, it is the right time to address
issues concerning writing. Unlike logic and mathematics, where there
is a clear-cut distinction between what is right or wrong, the diﬀerence
between good and bad writing is sometimes a matter of opinion. But there
are some standard guidelines that will make your writing clearer. Some
of these are listed below.
1. Never begin a sentence with a mathematical symbol. The reason
is that sentences begin with capital letters, but mathematical symbols
are case sensitive. Since x and X can have entirely diﬀerent meanings,
putting such symbols at the beginning of a sentence can lead to ambiguity.
Following are some examples of bad usage (marked with ×) and good
usage (marked with ).
A is a subset of B.
The set A is a subset of B. × x is an integer, so 2 x + 5 is an integer.
Since x is an integer, 2 x + 5 is an integer. × x2 − x + 2 = 0 has two solutions.
X 2 − x + 2 = 0 has two solutions.
The equation x2 − x + 2 = 0 has two solutions. ×
× (and silly too) 2. End each sentence with a period. Do this even when the sentence
ends with a mathematical symbol or expression.
∞ Euler proved that 1
1
=
s
1
p∈P 1 − p s
k=1 k
∞ Euler proved that 1
1
=
.
s
1
p∈P 1 − p s
k=1 k × Contrapositive Proof 100 Mathematical statements (equations, inequalities, etc.) are like English phrases that happen to contain special symbols, so use normal
punctuation.
3. Separate mathematical symbols and expressions with words.
Failure to do this can cause confusion by making distinct expressions
appear to merge into one. Compare the clarity of the following examples.
Since x2 − 1 = 0, x = 1 or x = −1.
Since x2 − 1 = 0, it follows that x = 1 or x = −1. × Unlike A ∪ B, A ∩ B equals .
Unlike A ∪ B, the set A ∩ B equals × . 4. Avoid misuse of symbols. Symbols such as =, ≤, ⊆, ∈, etc. are not
words. While it is appropriate to use them in mathematical expressions,
they are out of place in other contexts.
Since the two sets are =, one is a subset of the other.
Since the two sets are equal, one is a subset of the other. × The empty set is a ⊆ of every set.
The empty set is a subset of every set. × Since a is odd and x odd ⇒ x2 odd, a2 is odd.
×
2
Since a is odd and any odd number squared is odd, then a is odd.
5. Avoid using unnecessary symbols. Mathematics is confusing enough
without them. Don’t muddy the water even more.
No set X has negative cardinality.
No set has negative cardinality. × 6. Use the ﬁrst person plural. In mathematical writing, it is common
to use the words “we” and “us” rather than “I,” “you” or “me.” It is as if
the reader and writer are having a conversation, with the writer guiding
the reader through the details of the proof.
7. Use the active voice. This is just a suggestion, but the active voice
makes your writing more lively.
The value x = 3 is obtained through the division of both sides by 5.×
Dividing both sides by 5, we get the value x = 3.
8. Explain each new symbol. In writing a proof, you must explain the
meaning of every new symbol you introduce. Failure to do this can lead
to ambiguity, misunderstanding and mistakes. For example, consider
the following two possibilities for a sentence in a proof, where a and b
have been introduced on a previous line. Mathematical Writing 101 Since a | b, it follows that b = ac.
Since a | b, it follows that b = ac for some integer c. × If you use the ﬁrst form, then a reader who has been carefully following
your proof may momentarily scan backwards looking for where the c
entered into the picture, not realizing at ﬁrst that it comes from the
deﬁnition of divides.
9. Watch out for “It.” The pronoun “it” can cause confusion when it is
unclear what it refers to. If there is any possibility of confusion, you
should avoid the word “it.” Here is an example.
Since X ⊆ Y , and 0 < | X |, we see that it is not empty. × Is “it” X or Y ? Either one would make sense, but which do we mean?
Since X ⊆ Y , and 0 < | X |, we see that Y is not empty.
10. Since, because, as for, so. In proofs, it is common to use these
words as conjunctions joining two statements, and meaning that one
statement is true and as a consequence the other true. The following
statements all mean that P is true (or assumed to be true) and as a
consequence Q is true also.
Q since P
Since P , Q Q because P
Because P , Q Q , as P
as P , Q Q , for P P , so Q Notice that the meaning of these constructions is diﬀerent from that of
“If P , then Q ,” for they are asserting not only that P implies Q , but also
that P is true. Exercise care in using them. It must be the case that P
and Q are both statements and that Q really does follow from P .
x ∈ N, so Z
x ∈ N, so x ∈ Z × 11. Thus, hence, therefore consequently. These adverbs precede a
statement that follows logically from previous sentences or clauses. Be
sure that a statement follows them.
Therefore 2k + 1.
Therefore a = 2k + 1. × Your mathematical writing will get better with practice. One of the
best ways to develop a good mathematical writing style is to read other
people’s proofs. Adopt what works and avoid what doesn’t. Contrapositive Proof 102
Exercises for Chapter 5 A. Use the method of contrapositive proof to prove the following statements. (In
each case you should also think about how a direct proof would work. You will
ﬁnd in most cases that contrapositive is easier.)
1. Suppose n ∈ Z. If n2 is even, then n is even.
2. Suppose n ∈ Z. If n2 is odd, then n is odd.
3. Suppose a, b ∈ Z. If a2 (b2 − 2 b) is odd, then a and b are odd.
4. Suppose a, b, c ∈ Z. If a does not divide bc, then a does not divide b.
5. Suppose x ∈ R. If x2 + 5 x < 0 then x < 0.
6. Suppose x ∈ R. If x3 − x > 0 then x > −1.
7. Suppose a, b ∈ Z. If both ab and a + b are even, then both a and b are even.
8. Suppose x ∈ R. If x5 − 4 x4 + 3 x3 − x2 + 3 x − 4 ≥ 0, then x ≥ 0.
9. Suppose n ∈ Z. If 3 | n2 , then 3 | n.
10. Suppose x, y, z ∈ Z and x = 0. If x | yz, then x | y and x | z.
11. Suppose x, y ∈ Z. If x2 ( y + 3) is even, then x is even or y is odd.
12. Suppose a ∈ Z. If a2 is not divisible by 4, then a is odd.
13. Suppose x ∈ R. If x5 + 7 x3 + 5 x ≥ x4 + x2 + 8, then x ≥ 0.
B. Prove the following statements using either direct or contrapositive proof.
Sometimes one approach will be much easier than the other.
14. If a, b ∈ Z and a and b have the same parity, then 3a + 7 and 7 b − 4 have
opposite parity.
15. Suppose x ∈ Z. If x3 − 1 is even, then x is odd.
16. Suppose x ∈ Z. If x + y is even, then x and y have the same parity.
17. If n is odd, then 8 | (n2 − 1).
18. For any a, b ∈ Z, it follows that (a + b)3 ≡ a3 + b3 (mod 3).
19. Let a, b ∈ Z and n ∈ N. If a ≡ b (mod n) and a ≡ c (mod n), then c ≡ b (mod n).
20. If a ∈ Z and a ≡ 1 (mod 5), then a2 ≡ 1 (mod 5).
21. Let a, b ∈ Z and n ∈ N. If a b (mod n), then a3 ≡ b3 (mod n)
22. Let a ∈ Z and n ∈ N. If a has remainder r when divided by n, then a ≡ r
(mod n).
23. Let a, b, c ∈ Z and n ∈ N. If a ≡ b (mod n), then ca ≡ cb (mod n).
24. If a ≡ b (mod n) and c ≡ d (mod n), then ac ≡ bd (mod n).
25. If n ∈ N and 2n − 1 is prime, then n is prime.
26. If n = 2k − 1 for some k ∈ N, then every entry in Row n of Pascal’s Triangle
is odd.
27. If a ≡ 0 (mod 4) or a ≡ 1 (mod 4), then
28. If n ∈ Z, then 4 (n2 − 3). a
2 is even. CHAPTER 6 Proof by Contradiction e now explore a third method of proof: proof by contradiction.
This method is not limited to proving just conditional statements –
it can be used to prove any kind of statement whatsoever. The basic idea
is to assume that the statement we want to prove is false, and then show
that this assumption leads to nonsense. We are then led to conclude that
we were wrong to assume the statement was false, so the statement must
be true. As an example, consider the following proposition and its proof. W Proposition If a, b ∈ Z, then a2 − 4b = 2. Proof. Suppose this proposition is false.
This conditional statement being false means there exist numbers a and b
for which a, b ∈ Z is true but a2 − 4b = 2 is false.
Thus there exist integers a, b ∈ Z for which a2 − 4b = 2.
From this equation we get a2 = 4b + 2 = 2(2 b + 1), so a2 is even.
Since a2 is even, it follows that a is even, so a = 2 c for some integer c.
Now plug a = 2 c back into the boxed equation a2 − 4b = 2.
We get (2 c)2 − 4b = 2, so 4 c2 − 4b = 2. Dividing by 2, we get 2 c2 − 2b = 1.
Therefore 1 = 2( c2 − b), and since c2 − b ∈ Z, it follows that 1 is even.
Since we know 1 is not even, something went wrong.
But all the logic after the ﬁrst line of the proof is correct, so it must be
that the ﬁrst line was incorrect. In other words, we were wrong to assume
the proposition was false. Thus the proposition is true.
You may be a bit suspicious of this line of reasoning, but in the next
section we will see that it is logically sound. For now, notice that at
the end of the proof we deduced that 1 is even, which conﬂicts with our
knowledge that 1 is odd. In essence, we have obtained the statement
(1 is odd)∧ ∼ (1 is odd), which has the form C ∧ ∼ C . Notice that no matter
what statement C is, and whether or not it is true, the statement C ∧ ∼ C
must be false. A statement—like this one—that cannot be true is called a
contradiction. Contradictions play a key role in our new technique. Proof by Contradiction 104
6.1 Proving Statements with Contradiction Let’s now see why the proof on the previous page is logically valid. In
that proof we needed to show that a statement P : (a, b ∈ Z) ⇒ (a2 − 4b = 2)
was true. The proof began with the assumption that P was false, that is
that ∼ P was true, and from this we deduced C ∧ ∼ C . In other words we
proved that ∼ P being true forces C ∧ ∼ C to be true, and this means that
we proved that the conditional statement (∼ P ) ⇒ (C ∧ ∼ C ) is true. To see
that this is the same as proving P is true, look at the following truth table
for (∼ P ) ⇒ (C ∧ ∼ C ). Notice that the columns for P and (∼ P ) ⇒ (C ∧ ∼ C )
are exactly the same, so P is logically equivalent to (∼ P ) ⇒ (C ∧ ∼ C ).
P C ∼P C∧∼C (∼ P ) ⇒ ( C ∧ ∼ C ) T T F F T T F F F T F T T F F F F T F F Therefore to prove a statement P , it suﬃces to instead prove the conditional
statement (∼ P ) ⇒ (C ∧ ∼ C ). This can be done with direct proof: Assume
∼ P and deduce C ∧ ∼ C . Here is the outline.
Outline for Proof by Contradiction.
Proposition P. Proof. Suppose ∼ P .
.
.
. Therefore C ∧ ∼ C .
One slightly unsettling feature of this method is that we may not know
at the beginning of the proof what the statement C is going to be. In
doing the scratch work for the proof, you assume that ∼ P is true, then
deduce new statements until you have deduced some statement C and its
negation ∼ C .
If this method seems confusing, look at it this way. In the ﬁrst line of
the proof we suppose ∼ P is true, that is we assume P is false. But if P is
really true then this contradicts our assumption that P is false. But we
haven’t yet proved P to be true, so the contradiction is not obvious. We
use logic and reasoning to transform the non-obvious contradiction ∼ P to
an obvious contradiction C ∧ ∼ C . Proving Statements with Contradiction 105 The idea of proof by contradiction is quite ancient, and goes back at
least as far as the Pythagoreans, who used it to prove that certain numbers
are irrational. Our next example follows their logic to prove that 2 is
irrational. Recall that a number is rational if it equals a fraction of two
integers, and it is irrational if it cannot be expressed as a fraction of two
integers. Here is the exact deﬁnition.
Deﬁnition 6.1 A real number x is rational if x = a , for some a, b ∈ Z.
b
The number x is irrational if it is not rational, that is if x = a for every
b
a , b ∈ Z.
We are now ready to use contradiction to prove that 2 is irrational.
According to the outline, the ﬁrst line of the proof should be “Suppose that
it is not true that 2 is irrational." But in writing the proof, it is helpful
(though not mandatory) to tip our reader oﬀ to the fact that we are using
proof by contradiction. One standard way of doing this is to make the ﬁrst
line “Suppose for the sake of contradiction that it is not true that 2 is
irrational."
Proposition The number 2 is irrational. Proof. Suppose for the sake of contradiction that it is not true that 2 is
irrational. Then 2 is rational, so there are integers a and b for which
2= a
.
b (6.1) Let this fraction be fully reduced; in particular, this means that a and b are
not both even. (If they were both even, the fraction could be further reduced
by factoring 2’s from the numerator and denominator and canceling.)
2
Squaring both sides of Equation 6.1 gives 2 = a2 , and therefore
b
a2 = 2 b 2 . (6.2) From this it follows that a2 is even. But we proved earlier (Exercise 5.1)
that a2 being even implies a is even. Thus, as we know that a and b are
not both even, it follows that b is odd. Now, since a is even there is an
integer c for which a = 2 c. Plugging this value for a into Equation 6.2, we
get (2 c)2 = 2b2 , so 4 c2 = 2b2 , and hence b2 = 2 c2 . This means b2 is even, so
b is even also. But previously we deduced that b is odd. Thus we have the
contradiction b is even and b is odd.
To appreciate the power of proof by contradiction, imagine trying to
prove that 2 is irrational without it. Where would we begin? What would Proof by Contradiction 106 be our initial assumption? There are no clear answers to these questions.
Proof by contradiction gives us a starting point: assume 2 is rational,
and work from there.
In the above proof we got the contradiction (b is even) ∧ ∼(b is even)
which has the form C ∧ ∼ C . In general, your contradiction need not
necessarily be of this form. Any statement that is clearly false is suﬃcient.
For example 2 = 2 would be a ﬁne contradiction, as would be 4 | 2, provided
that you could deduce them.
Here is another ancient example, dating back at least as far as Euclid.
Proposition There are inﬁnitely many prime numbers. Proof. For the sake of contradiction, suppose there are only ﬁnitely many
prime numbers. Then we can list all the prime numbers as p 1 , p 2 , p 3 , . . . p n ,
where p 1 = 2, p 2 = 3, p 3 = 5, p 4 = 7, and so on. Thus p n is the nth and largest
prime number. Now consider the number a = ( p 1 p 2 p 3 · · · p n ) + 1, that is, a is
the product of all prime numbers, plus 1. Now a, like any natural number,
has at least one prime divisor, and that means p k | a for at least one of our
n prime numbers p k . Thus there is an integer c for which a = c p k , which
is to say
( p 1 p 2 p 3 · · · p k−1 p k p k+1 · · · p n ) + 1 = c p k . Dividing both sides of this by p k gives us
( p 1 p 2 p 3 · · · p k−1 p k+1 · · · p n ) + so 1
= c,
pk 1
= c − ( p 1 p 2 p 3 · · · p k−1 p k+1 · · · p n ).
pk The expression on the right is an integer, while the expression on the left
is not an integer. Thus we have an integer that equals a non-integer value,
a contradiction.
Proof by contradiction often works well in proving statements of the
form ∀ x, P ( x). The reason is that the proof set-up involves assuming
∼ ∀ x, P ( x), which as we know from Section 2.10 is equivalent to ∃ x, ∼ P ( x).
This gives us a speciﬁc x for which ∼ P ( x) is true, and often that is enough
to produce a contradiction. Here is an example.
Proposition For every real number x ∈ [0, π/2], we have sin x + cos x ≥ 1. Proof. Suppose for the sake of contradiction that this is not true.
Then there exists an x ∈ [0, π/2] for which sin x + cos x < 1. Proving Conditional Statements by Contradiction 107 Since x ∈ [0, π/2], neither sin x nor cos x is negative, so 0 ≤ sin x + cos x < 1.
Thus 02 ≤ (sin x + cos x)2 < 12 , which gives 02 ≤ sin2 x + 2 sin x cos x + cos2 x < 12 .
As sin2 x + cos2 x = 1, this becomes 0 ≤ 1 + 2 sin x cos x < 1, so 1 + 2 sin x cos x < 1.
Subtracting 1 from both sides gives 2 sin x cos x < 0.
But this contradicts the fact that neither sin x nor cos x is negative.
6.2 Proving Conditional Statements by Contradiction
Since the previous two chapters dealt exclusively with proving conditional
statements, we now formalize the procedure in which contradiction is used
to prove a conditional statement. Suppose we want to prove a proposition
of the following form.
Proposition If P , then Q . Thus we need to prove that P ⇒ Q is a true statement. Proof by
contradiction begins with the assumption that ∼ (P ⇒ Q ) it true, that is
that P ⇒ Q is false. But we know that P ⇒ Q being false means that P is
true and Q is false. Thus the ﬁrst step in the proof it to assume P and
∼ Q . Here is an outline.
Outline for Proving a Conditional
Statement with Contradiction.
Proposition If P , then Q . Proof. Suppose P and ∼ Q .
.
.
. Therefore C ∧ ∼ C .
To illustrate this new technique, we revisit a familiar result: If a2 is
even, then a is even. According to the outline, the ﬁrst line of the proof
should be “Suppose for the sake of contradiction that a2 is even and a is
not even."
Proposition Suppose a ∈ Z. If a2 is even, then a is even. Proof. For the sake of contradiction suppose a2 is even and a is not even.
Then a2 is even, and a is odd.
Since a is odd, there is an integer c for which a = 2 c + 1.
Then a2 = (2 c + 1)2 = 4 c2 + 4 c + 1 = 2(2 c2 + 2 c) + 1, so a2 is odd.
Thus a2 is even and a2 is not even, a contradiction. Proof by Contradiction 108
Here is another example.
Proposition If a, b ∈ Z and a ≥ 2, then a b or a ( b + 1). Proof. Suppose for the sake of contradiction there exist a, b ∈ Z with a ≥ 2,
and for which it is not true that a b or a ( b + 1).
By DeMorgan’s Law, we have a | b and a | ( b + 1).
The deﬁnition of divisibility says there are c, d ∈ Z with b = ac and b + 1 = ad .
Subtracting one equation from the other gives ad − ac = 1, so a(d − c) = 1.
Since a is positive, d − c is also positive (otherwise a( d − c) would be negative).
Then d − c is a positive integer and a( d − c) = 1, so a = 1/(d − c) < 2.
Thus we have a ≥ 2 and a < 2, a contradiction.
6.3 Combining Techniques
Often, especially in more complex proofs, several proof techniques are
combined within a single proof. For example, in proving a conditional
statement P ⇒ Q , we might begin with direct proof and thus assume P to
be true with the aim of ultimately showing Q is true. But the truth of
Q might hinge on the truth of some other statement R which—together
with P —would imply Q . We would then need to prove R , and we would
use whichever proof technique seems most appropriate. This can lead to
“proofs inside of proofs." Consider the following result. The overall approach
is direct, but inside the direct proof is a separate proof by contradiction.
Proposition Every nonzero rational number can be expressed as a
product of two irrational numbers.
Proof. This proposition can be reworded as follows: If r is a nonzero
rational number, then r is a product of two irrational numbers. In what
follows, we prove this with direct proof.
Suppose r is a nonzero rational number. Then r = a for integers a and
b
b. Also, r can be written as a product of two numbers as follows.
r= 2· r
2 . We know 2 is irrational, so to complete the proof we must show r / 2 is
also irrational.
To show this, assume for the sake of contradiction that r / 2 is rational.
This means
r 2 = c
d Some Words of Advice 109 for integers c and d , so d
2=r .
c But we know r = a/b, which combines with the above equation to give
2=r d a d ad
=
=
.
c bc
bc This means 2 is rational, which is a contradiction because we know it is
irrational. Therefore r / 2 is irrational.
Consequently r = 2 · r / 2 is a product of two irrational numbers.
For another example of a proof-within-a-proof, try Exercise 5 of this
chapter and then check its solution. That exercise asks you to prove that
3 is irrational. This turns out to be slightly trickier than proving that
2 is irrational.
6.4 Some Words of Advice
Despite the power of proof by contradiction, it’s best to use it only when the
direct and contrapositive approaches do not seem to work. The reason for
this is that a proof by contradiction can often have hidden in it a simpler
contrapositive proof, and if this is the case it’s better to go with the simpler
approach. Consider the following example.
Proposition Suppose a ∈ Z. If a2 − 2a + 7 is even, then a is odd. Proof. To the contrary, suppose a2 − 2a + 7 is even and a is not odd.
That is, suppose a2 − 2a + 7 is even and a is even.
Since a is even, there is an integer c for which a = 2 c.
Then a2 − 2a + 7 = (2 c)2 − 2(2 c) + 7 = 2(2 c2 − 2 c + 3) + 1, so a2 − 2a + 7 is odd.
Thus a2 − 2a + 7 is both even and odd, a contradiction.
Though there is nothing really wrong with this proof, notice that part
of it assumes a is not odd and deduces that a2 − 2a + 7 is not even. That is
the contrapositive approach! Thus it would be more eﬃcient to proceed as
follows, using contrapositive proof.
Proposition Suppose a ∈ Z. If a2 − 2a + 7 is even, then a is odd. Proof. (Contrapositive) Suppose a is not odd.
Then a is even, so there is an integer c for which a = 2 c.
Then a2 − 2a + 7 = (2 c)2 − 2(2 c) + 7 = 2(2 c2 − 2 c + 3) + 1, so a2 − 2a + 7 is odd.
Thus a2 − 2a + 7 is not even. Proof by Contradiction 110
Exercises for Chapter 6 A. Use the method of proof by contradiction to prove the following statements.
(In each case you should also think about how a direct or contrapositive proof
would work. You will ﬁnd in most cases that proof by contradiction is easier.)
1. Suppose n ∈ Z. If n is odd, then n2 is odd.
2. Suppose n ∈ Z. If n2 is odd, then n is odd.
3. Prove that 3 2 is irrational. 4. Prove that 6 is irrational. 5. Prove that 3 is irrational. 6. If a, b ∈ Z, then a2 − 4b − 2 = 0.
7. If a, b ∈ Z, then a2 − 4 b − 3 = 0.
8. Suppose a, b, c ∈ Z. If a2 + b2 = c2 , then a or b is even.
9. Suppose a, b ∈ R. If a is rational and ab is irrational, then b is irrational.
10. There exist no integers a and b for which 21a + 30 b = 1.
11. There exist no integers a and b for which 18a + 6b = 1.
12. For every positive rational number x, there is a positive rational number y
for which y < x.
13. For every x ∈ [π/2, π], sin x − cos x ≥ 1.
14. If A and B are sets, then A ∩ (B − A ) = .
15. If b ∈ Z and b k for every k ∈ N, then b = 0.
16. If a and b are positive real numbers, then a + b ≥ 2 a b.
17. For every n ∈ Z, 4 | (n2 + 2).
18. Suppose a, b ∈ Z. If 4 |(a2 + b2 ), then a and b are not both odd.
B. Prove the following statements using any method from chapters 4, 5 or 6.
19. The product of any ﬁve consecutive integers is divisible by 120. (For
example, the product of 3,4,5,6 and 7 is 2520, and 2520 = 120 · 21.)
20. We say that a point P = ( x, y) in R2 is rational if both x and y are rational.
More precisely, P is rational if P = ( x, y) ∈ Q2 . An equation F ( x, y) = 0 is said
to have a rational point if there exists x0 , y0 ∈ Q such that F ( x0 , y0 ) = 0. For
example, the curve x2 + y2 − 1 = 0 has rational point ( x0 , y0 ) = (1, 0). Show that
the curve x2 + y2 − 3 = 0 has no rational points.
21. Exercise 20 involved showing that there are no rational points on the curve
x2 + y2 − 3 = 0. Use this fact to show that 3 is irrational.
22. Explain why x2 + y2 − 3 = 0 not having any rational solutions (Exercise 20)
implies x2 + y2 − 3k = 0 has no rational solutions for k an odd, positive integer.
23. Use the above result to prove that
24. The number log2 3 is irrational. 3k is irrational for all odd, positive k. Part III
More on Proof CHAPTER 7 Proving Non-conditional Statements he past three chapters have introduced three major proof techniques:
direct, contrapositive and contradiction. These three techniques are
used to prove statements of the form “If P , then Q .” As we know, most theorems and propositions have this conditional form or they can be reworded
to have this form. Thus the three main techniques are quite important.
But some theorems and propositions cannot be put into conditional form.
For example, some theorems have form “P if and only if Q .” Such theorems
are biconditional statements, not conditional statements. In this chapter
we examine ways of proving theorems of this form. In addition to learning
how to prove if-and-only-if theorems, we will also look at two other types
of theorems. T 7.1 If-And-Only-If Proof
Some propositions have the form
P if and only if Q . We know from Section 2.4 that this statement asserts that both of the
following two conditional statements are true.
If P , then Q .
If Q , then P .
So to prove “P if and only if Q ,” we need to prove two conditional statements. Recall from Section 2.4 that Q ⇒ P is called the converse of P ⇒ Q .
Thus we need to prove both P ⇒ Q and its converse. Since these are both
conditional statements we may prove them with either direct, contrapositive or contradiction proof. Here is an outline.
Outline for If-And-Only-If Proof.
Proposition P if and only if Q . Proof.
[Prove P ⇒ Q using direct, contrapositive or contradiction proof.]
[Prove Q ⇒ P using direct, contrapositive or contradiction proof.] 114 Proving Non-conditional Statements Let’s start with a very simple example. You already know that an
integer n is odd if and only if n2 is odd, but let’s prove it anyway, just
to illustrate the outline. In this example we prove ( n is odd)⇒( n2 is odd)
using direct proof and (n2 is odd)⇒(n is odd) using contrapositive proof.
Proposition The integer n is odd if and only if n2 is odd. Proof. First we show that n being odd implies that n2 is odd. Suppose n
is odd. Then, by deﬁnition of an odd number, n = 2a + 1 for some integer a.
Thus n2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 + 2a) + 1. This expresses n2 as twice
an integer, plus 1, so n2 is odd.
Conversely, we need to prove that n2 being odd implies that n is odd.
We use contrapositive proof. Suppose n is not odd. Then n is even, so
n = 2a for some integer a (by deﬁnition of an even number). Thus n2 =
(2a)2 = 2(2a2 ), so n2 is even because it’s twice an integer. Thus n2 is not
odd. We’ve now proved that if n is not odd, then n2 is not odd, and this is
a contrapositive proof that if n2 is odd then n is odd.
In proving “P if and only if Q ,” you should always begin a new paragraph
when starting the proof of Q ⇒ P . Since this is the converse of P ⇒ Q , it’s
a good idea to begin the paragraph with the word “Conversely” (as we did
above) to remind the reader that you’ve ﬁnished the ﬁrst part of the proof
and are moving on to the second. Likewise, it’s a good idea to remind the
reader of exactly what statement that paragraph is proving.
The next example uses direct proof in both parts of the proof.
Proposition Suppose a and b are integers. Then a ≡ b (mod 6) if and
only if a ≡ b (mod 2) and a ≡ b (mod 3).
Proof. First we prove that if a ≡ b (mod 6), then a ≡ b (mod 2) and a ≡ b
(mod 3). Suppose a ≡ b (mod 6). This means 6|(a − b), so there is an integer
n for which
a − b = 6 n. From this we get a − b = 2(3n), which implies 2|(a − b), so a ≡ b (mod 2). But
we also get a − b = 3(2n), which implies 3|(a − b), so a ≡ b (mod 3). Therefore
a ≡ b (mod 2) and a ≡ b (mod 3).
Conversely, suppose a ≡ b (mod 2) and a ≡ b (mod 3). Since a ≡ b (mod 2)
we get 2|(a − b), so there is an integer k for which a − b = 2k. Therefore a − b
is even. Also, from a ≡ b (mod 3) we get 3|(a − b), so there is an integer
for which
a−b =3 . Equivalent Statements 115 But since we know a − b is even, it follows that must be even also, for
if it were odd then a − b = 3 would be odd. (Because a − b would be the
product of two odd integers.) Hence = 2m for some integer m. Thus
a − b = 3 = 3 · 2 m = 6 m. This means 6|(a − b), so a ≡ b (mod 6).
Since if-and-only-if proofs simply combine methods with which we are
already familiar, we will not do any further examples in this section.
However it is of utmost importance that you practice your skill on some of
this chapter’s exercises.
7.2 Equivalent Statements
In other courses you will sometimes encounter a certain kind of theorem
that is neither a conditional nor a biconditional statement. Instead, it
asserts that a list of statements is “equivalent.” You saw this (or will see
it) in your linear algebra textbook, which featured the following theorem.
Theorem Suppose A is an n × n matrix. The following statements are
equivalent.
(a)
(b)
(c)
(d)
(e)
(f) The matrix A is invertible.
The equation A x = b has a unique solution for every b ∈ Rn .
The equation A x = 0 has only the trivial solution.
The reduced row echelon form of A is I n .
det( A ) = 0.
Matrix A does not have 0 as an eigenvector. When a theorem asserts that a list of statements is “equivalent,” it is
asserting that either the statements are all true, or they are all false.
Thus the above theorem tells us that whenever we are dealing with a
particular n × n matrix A , then either the statements (a) through (f) are all
true for A , or statements (a) through (f) are all false for A . For example, if
we happen to know that det( A ) = 0, the theorem assures us that in addition
to statement (e) being true, all the statements (a) through (f) are true. On
the other hand, if it happens that det( A ) = 0, the theorem tells us that all
statements (a) through (f) are false. In this way, the theorem multiplies
our knowledge of A by a factor of six. Obviously that can be very useful.
What method would we use to prove such a theorem? In a certain
sense, the above theorem is like an if-and-only-if theorem. An if-and-only-if
theorem of form P ⇔ Q asserts that P and Q are either both true or both
false, that is that P and Q are equivalent. To prove P ⇔ Q we prove P ⇒ Q
followed by Q ⇒ P , essentially making a “cycle” of implications from P to Q Proving Non-conditional Statements 116 and back to P . Similarly, one approach to proving the theorem cited at the
beginning of this section would be to prove (a) ⇒ (b), then (b) ⇒ (c), then
(c) ⇒ (d), then (d) ⇒ (e), then (e) ⇒ (f), and ﬁnally (f) ⇒ (a). This pattern is
illustrated below.
(a) =⇒ ( b) =⇒ ( c)
⇑
⇓
( f ) ⇐= ( e) ⇐= ( d ) Notice that if these six implications have been proved, then it really does
follow that the statements (a) through (f) are either all true or all false.
If one of them is true then the circular chain of implications forces them
all to be true. On the other hand, if one of them (say (c)) is false, the fact
that (b) ⇒ (c) is true forces (b) to be false. This combined with the truth of
(a) ⇒ (b) makes (a) false, and so on counterclockwise around the circle.
Thus to prove that n statements are equivalent, it suﬃces to prove n
conditional statements showing each statement implies another, in circular
pattern. But it is not necessary that the pattern be circular. The following
schemes would also do the job.
(a) =⇒ ( b) ⇐⇒ ( c)
⇑
⇓
( f ) ⇐= ( e) ⇐⇒ ( d ) (a) ⇐⇒ ( b) ⇐⇒
( f ) ⇐⇒ ( e) ( c) ⇐⇒ ( d ) However, a circular pattern results in the fewest number of conditional
statements that must be proved. Whatever the pattern, each conditional
statement can be proved with either direct, contrapositive or contradiction
proof.
Though we shall not do any of these proofs in this text, you are sure to
encounter them in subsequent courses.
7.3 Existence Proofs
Up until this point, we have dealt with proving conditional statements
or with statements that can be expressed with two or more conditional
statements. Generally, these conditional statements have form P ( x) ⇒ Q ( x).
(Possibly with more than one variable.) We saw in Section 2.8 that this
can be interpreted as a universally quantiﬁed statement ∀ x, P ( x) ⇒ Q ( x). Existence Proofs 117 Thus, conditional statements are universally quantiﬁed statements, so
in proving a conditional statement—whether we use direct, contrapositive
or contradiction proof—we are really proving a universally quantiﬁed
statement.
But how would we prove an existentially quantiﬁed statement? What
technique would we employ to prove a theorem of the following form?
∃ x, R ( x) This statement asserts that there exists some speciﬁc object x for which
R ( x) is true. To prove ∃ x, R ( x) is true, all we would have to do is ﬁnd and
display an example of a speciﬁc x that makes R ( x) true.
Though most theorems and propositions are conditional (or if-andonly-if) statements, a few have the form ∃ x, R ( x). Such statements are
called existence statements, and theorems that have this form are called
existence theorems. To prove an existence theorem, all you have to do
is provide a particular example that shows it is true. This is often quite
simple. (But not always!) Here are some examples.
Proposition There exists an even prime number. Proof. Observe that 2 is an even prime number. Proposition There exists an integer that can be expressed as the sum
of two perfect cubes in two diﬀerent ways.
Proof. Consider the number 1729. Note that 13 + 123 = 1729 and 93 + 103 =
1729. Thus the number 1729 can be expressed as the sum of two perfect
cubes in two diﬀerent ways.
Sometimes in the proof of an existence statement, a little veriﬁcation is
needed to show that the example really does work. For example, the above
proof would be incomplete if we just asserted that 1729 can be written as
a sum of two cubes in two ways without showing how this is possible.
WARNING: Although an example suﬃces to prove an existence statement,
a mere example never proves a conditional statement. 118 Proving Non-conditional Statements Exercises for Chapter 7
Prove the following statements. These exercises are cumulative, covering all
techniques addressed in Chapters 4–7.
1. Suppose x ∈ Z. Then x is even if and only if 3 x + 5 is odd.
2. Suppose x ∈ Z. Then x is odd if and only if 3 x + 6 is odd.
3. Given an integer a, then a3 + a2 + a is even if and only if a is even.
4. Given an integer a, then a2 + 4a + 5 is odd if and only if a is even.
5. An integer a is odd if and only if a3 is odd.
6. Suppose x, y ∈ R. Then x3 + x2 y = y2 + x y if and only if y = x2 or y = − x.
7. Suppose x, y ∈ R. Then ( x + y)2 = x2 + y2 if and only if x = 0 or y = 0.
8. Suppose a, b ∈ Z. Prove that a ≡ b (mod 10) if and only if a ≡ b (mod 2) and a ≡ b
(mod 5).
9. Suppose a ∈ Z. Prove that 14|a if and only if 7|a and 2|a.
10. If a ∈ Z, then a3 ≡ a (mod 3).
11. Suppose a, b ∈ Z. Prove that (a − 3) b2 is even if and only if a is odd or b is even.
12. There exist a positive real number x for which x2 < x.
13. Suppose a, b ∈ Z. If a + b is odd, then a2 + b2 is odd.
14. Suppose a ∈ Z. Then a2 |a if and only if a ∈ − 1, 0, 1 .
15. Suppose a, b ∈ Z. Prove that a + b is even if and only if a and b have the same
parity.
16. Suppose a, b ∈ Z. If ab is odd, then a2 + b2 is even.
17. There is a prime number between 90 and 100.
18. There is a set X for which N ∈ X and N ⊆ X .
19. If n ∈ N, then 20 + 21 + 22 + 23 + 24 + · · · + 2n = 2n+1 − 1.
20. There exists an n ∈ N for which 11|(2n − 1).
21. Every real solution of x3 + x + 3 = 0 is irrational.
22. If n ∈ Z, then 4 | n2 or 4 | (n2 − 1).
23. Suppose a, b and c are integers. If a | b and a | ( b2 − c), then a | c.
24. If a ∈ Z, then 4 (a2 − 3).
25. If p > 1 is an integer and n p for each integer n for which 2 ≤ n ≤
prime. p, then p is 26. The product of any n consecutive positive integers is divisible by n!.
27. Suppose a, b ∈ Z. If a2 + b2 is a perfect square, then a and b are not both odd. CHAPTER 8 Proofs Involving Sets tudents taking their ﬁrst advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most
of the proofs they encounter are proofs about sets. Perhaps you’ve already
seen such proofs in your linear algebra course, where a vector space was
deﬁned to be a set of objects (called vectors) that obey certain properties.
Your text proved many things about vector spaces, such as the fact that
the intersection of two vector spaces is also a vector space, and the proofs
used ideas from set theory. As you go deeper into mathematics, you will
encounter more and more ideas, theorems and proofs that involve sets.
The purpose of this chapter is to give you a foundation that will prepare
you for this new outlook.
We will discuss how to show that an object is an element of a set, how
to prove one set is a subset of another, and how to prove two sets are
equal. As you read this chapter, you may need to occasionally refer back
to Chapter 1 to refresh your memory. For your convenience, the main
deﬁnitions from Chapter 1 are summarized below. If A and B are sets,
then: S A×B = ( x , y) : x ∈ A , y ∈ B A∪B = x : ( x ∈ A ) ∨ ( x ∈ B) A∩B = x : ( x ∈ A ) ∧ ( x ∈ B) A−B = x : ( x ∈ A ) ∧ ( x ∉ B) A = U−A Also, recall that A ⊆ B means that every element of A is also an element
of B.
8.1 How to Prove a ∈ A
We will begin with a review of set-builder notation, and then review how
to show that a given object a is an element of some set A . Proofs Involving Sets 120 Generally, a set A will be expressed in set-builder notation A = x : P ( x) ,
where P ( x) is some statement (or open sentence) about x. The set A is
understood to have as elements all those things x for which P ( x) is true.
For example,
x : x is an odd integer = . . . , −5, −3, −1, 1, 3, 5, . . . . A common variation of this notation is to express a set as A = x ∈ S : P ( x) .
Here it is understood that A consists of all elements x of the (predetermined)
set S for which P ( x) is true. Keep in mind that, depending on context, x
could be any kind of object (integer, ordered pair, set, function, etc.). There
is also nothing special about the particular variable x; any reasonable
symbol x, y, k, etc. would do. Some examples follow.
n ∈ Z : n is odd = . . . , −5, −3, −1, 1, 3, 5, . . . x ∈ N : 6| x = 6, 12, 18, 24, 30, . . . ( a, b ) ∈ Z × Z : b = a + 5 = . . . , (−2, 3), (−1, 4), (0, 5), (1, 6), . . . X ∈ P (Z) : | X | = 1 = ..., −1 , 0 , 1 , 2 , 3 , 4 ,... Now it should be clear how to prove that an object a belongs to a set
x : P ( x) . Since x : P ( x) consists of all things x for which P ( x) is true, to
show that a ∈ x : P ( x) we just need to show that P (a) is true. Likewise, to
show a ∈ x ∈ S : P ( x) , we need to conﬁrm that a ∈ S and that P (a) is true.
These ideas are summarized below. However, you should not memorize
these methods, you should understand them. With contemplation and
practice, using them becomes natural and intuitive.
How to show a ∈ x : P ( x)
Show that P (a) is true. How to show a ∈ x ∈ S : P ( x)
1. Verify that a ∈ S .
2. Show that P (a) is true. Example 8.1 Let’s investigate elements of A = x : x ∈ N and 7 | x . This set
has form A = x : P ( x) where P ( x) is the open sentence P ( x) : ( x ∈ N) ∧ (7 | x).
Thus 21 ∈ A because P (21) is true. Similarly, 7, 14, 28, 35, etc. are all
elements of A . But 8 ∉ A (for example) because P (8) is false. Likewise
−14 ∉ A because P (−14) is false.
Example 8.2 Consider the set A = X ∈ P (N) : | X | = 3 . We know that
4, 13, 45 ∈ A because 4, 13, 45 ∈ P (N) and 4, 13, 45 = 3. Also 1, 2, 3 ∈ A ,
10, 854, 3 ∈ A , etc. However 1, 2, 3, 4 ∉ A because 1, 2, 3, 4 = 3. Further,
− 1, 2, 3 ∉ A because − 1, 2, 3 ∉ P (N). How to Prove A ⊆ B 121 Example 8.3 Consider the set B = ( x, y) ∈ Z × Z : x ≡ y (mod 5) . Notice
(8, 23) ∈ B because (8, 23) ∈ Z × Z and 8 ≡ 23 (mod 5). Likewise, (100, 75) ∈ B,
(102, 77) ∈ B, etc., but (6, 10) ∉ B.
Now suppose n ∈ Z and consider the ordered pair (4 n + 3, 9n − 2). Does
this ordered pair belong to B? To answer this, we ﬁrst observe that
(4 n +3, 9 n −2) ∈ Z×Z. Next, we observe that (4 n +3)−(9 n −2) = −5 n +5 = 5(1− n),
so 5 |((4n + 3) − (9n − 2)), which means (4n + 3) ≡ (9n − 2) (mod 5). Therefore we
have established that (4 n + 3, 9n − 2) meets the requirements for belonging
to B, so (4n + 3, 9n − 2) ∈ B for every n ∈ Z.
Example 8.4 This illustrates another common way of deﬁning a set.
Consider the set C = 3 x3 + 2 : x ∈ Z . Elements of this set consist of all the
values 3 x3 + 2 where x is an integer. Thus −22 ∈ C because −22 = 3(−2)3 + 2.
1
Also you can conﬁrm −1 ∈ C and 5 ∈ C , etc, as well as 0 ∉ C and 2 ∉ C , etc.
8.2 How to Prove A ⊆ B
In this course (and more importantly, beyond it) you will encounter many
circumstances where it is necessary to prove that one set is a subset of another. This section explains how to do this. The methods we discuss should
improve your skills in both writing your own proofs and in comprehending
the proofs that you read.
Recall (Deﬁnition 1.3) that if A and B are sets, then A ⊆ B means that
every element of A is also an element of B. In other words, it means if
a ∈ A , then a ∈ B. Therefore to prove that A ⊆ B, we just need to prove that
the conditional statement
“If a ∈ A , then a ∈ B,”
is true. This can be proved directly, by assuming a ∈ A and deducing a ∈ B.
The contrapositive approach is another option: assume a ∉ B and deduce
a ∉ A . Each of these two approaches is outlined below.
How to Prove A ⊆ B
(Direct approach)
Proof. Suppose a ∈ A . How to Prove A ⊆ B
(Contrapositive approach)
Proof. Suppose a ∉ B. Therefore a ∈ B.
Thus a ∈ A implies a ∈ B,
so it follows that A ⊆ B. Therefore a ∉ A .
Thus a ∉ B implies a ∉ A ,
so it follows that A ⊆ B. .
.
. .
.
. Proofs Involving Sets 122 In practice, the direct approach usually results in the most straightforward and easy proof, though occasionally the contrapositive is the
most expedient. (You can even prove A ⊆ B by contradiction: assume
(a ∈ A ) ∧ (a ∉ B), and deduce a contradiction.) The remainder of this section
consists of examples with occasional commentary. Unless stated otherwise,
we will use the direct approach in all proofs; pay special attention to how
the above outline for the direct approach is used.
Example 8.5 Prove that x ∈ Z : 18 | x ⊆ x ∈ Z : 6 | x .
Proof. Suppose a ∈ x ∈ Z : 18 | x .
This means that a ∈ Z and 18 | a.
By deﬁnition of divisibility, there is an integer c for which a = 18 c.
Consequently a = 6(3 c), and from this we deduce that 6 | a.
Therefore a is one of the integers that 6 divides, so a ∈ x ∈ Z : 6 | x .
We’ve shown a ∈ x ∈ Z : 18 | x implies a ∈ n ∈ Z : 6 | x , so it follows that
x ∈ Z : 18 | x ⊆ x ∈ Z : 6 | x .
Example 8.6 Prove that x ∈ Z : 2 | x ∩ x ∈ Z : 9 | x ⊆ x ∈ Z : 6 | x . Proof. Suppose a ∈ x ∈ Z : 2 | x ∩ x ∈ Z : 9 | x .
By deﬁnition of intersection, this means a ∈ x ∈ Z : 2 | x and a ∈ x ∈ Z : 9 | x .
Since a ∈ x ∈ Z : 2 | x we know 2 | a, so a = 2 c for some c ∈ Z. Thus a is even.
Since a ∈ x ∈ Z : 9 | x we know 9 | a, so a = 9d for some d ∈ Z.
As a is even, a = 9d implies d is even. (Otherwise a = 9d would be odd.)
Then d = 2 e for some integer e, and we have a = 9d = 9(2 e) = 6(3 e).
From a = 6(3 e), we conclude 6 | a, and this means a ∈ x ∈ Z : 6 | x .
We have shown that a ∈ x ∈ Z : 2 | x ∩ x ∈ Z : 9 | x implies a ∈ x ∈ Z : 6 | x ,
so it follows that x ∈ Z : 2 | x ∩ x ∈ Z : 9 | x ⊆ x ∈ Z : 6 | x .
Example 8.7 Show ( x, y) ∈ Z×Z : x ≡ y (mod 6) ⊆ ( x, y) ∈ Z×Z : x ≡ y (mod 3) .
Proof. Suppose (a, b) ∈ ( x, y) ∈ Z × Z : x ≡ y (mod 6) .
This means (a, b) ∈ Z × Z and a ≡ b (mod 6).
Consequently 6 |(a − b), so a − b = 6 c for some integer c.
It follows that a − b = 3(2 c), and this means 3 |(a − b), so a ≡ b (mod 3).
Thus (a, b) ∈ ( x, y) ∈ Z × Z : x ≡ y (mod 3) .
We’ve now seen that (a, b) ∈ ( x, y) ∈ Z × Z : x ≡ y (mod 6) implies (a, b) ∈
( x, y) ∈ Z × Z : x ≡ y (mod 3) , so it follows that ( x, y) ∈ Z × Z : x ≡ y (mod 6) ⊆
( x, y) ∈ Z × Z : x ≡ y (mod 3) . How to Prove A ⊆ B 123 Some statements involving subsets are transparent enough that we
often accept (and use) them without proof. For example, if A and B are any
sets, then it’s very easy to conﬁrm A ∩ B ⊆ A . (Reason: Suppose x ∈ A ∩ B.
Then x ∈ A and x ∈ B by deﬁnition of intersection, so in particular x ∈ A .
Thus x ∈ A ∩ B implies x ∈ A , so A ∩ B ⊆ A .) Other statements of this nature
include A ⊆ A ∪ B and A − B ⊆ A , as well as conditional statements such as
( A ⊆ B) ∧ (B ⊆ C ) ⇒ ( A ⊆ C ) and ( X ⊆ A ) ⇒ ( X ⊆ A ∪ B). Our point of view in
this text is that we do not need to prove such obvious statements unless we
are explicitly asked to do so in an exercise. (Still, you should do some quick
mental proofs to convince yourself that the above statements are true. If
you don’t see that A ∩ B ⊆ A is true but that A ⊆ A ∩ B is not necessarily
true, then you need to spend more time on this topic.)
The next example will show that if A and B are sets, then P ( A ) ∪ P (B) ⊆
P ( A ∪ B). Before beginning our proof, let’s look at an example to see if
this statement really makes sense. Suppose A = 1, 2 and B = 2, 3 . Then
P ( A ) ∪ P (B ) =
= , 1 , 2 , 1, 2 ∪ , 2 , 3 , 2, 3 , 1 , 2 , 3 , 1, 2 , 2, 3 . Also P ( A ∪B) = P ( 1, 2, 3 ) = , 1 , 2 , 3 , 1, 2 , 2, 3 , 1, 3 , 1, 2, 3 . Thus,
even though P ( A ) ∪ P (B) = P ( A ∪ B), it is true that P ( A ) ∪ P (B) ⊆ P ( A ∪ B)
for this particular A and B. Now let’s prove P ( A ) ∪ P (B) ⊆ P ( A ∪ B) no
matter what sets A and B are.
Example 8.8 Prove that if A and B are sets, then P ( A ) ∪ P (B) ⊆ P ( A ∪ B). Proof. Suppose X ∈ P ( A ) ∪ P (B).
By deﬁnition of union, this means X ∈ P ( A ) or X ∈ P (B).
Therefore X ⊆ A or X ⊆ B (by deﬁnition of power sets). We consider cases.
Case 1. Suppose X ⊆ A . Then X ⊆ A ∪ B, and this means X ∈ P ( A ∪ B).
Case 2. Suppose X ⊆ B. Then X ⊆ A ∪ B, and this means X ∈ P ( A ∪ B).
(We do not need to consider the case where X ⊆ A and X ⊆ B because that
is taken care of by either of cases 1 or 2.) The above cases show that
X ∈ P ( A ∪ B ).
Thus we’ve shown that X ∈ P ( A ) ∪ P (B) implies X ∈ P ( A ∪ B), and this
completes the proof that P ( A ) ∪ P (B) ⊆ P ( A ∪ B). In our next example, we prove a conditional statement. Direct proof is
used, and in the process we use our new technique for showing A ⊆ B. Proofs Involving Sets 124 Example 8.9 Suppose A and B are sets. If P ( A ) ⊆ P (B), then A ⊆ B.
Proof. We use direct proof. Assume P ( A ) ⊆ P (B).
Based on this assumption, we must now show that A ⊆ B.
To show A ⊆ B, suppose that a ∈ A .
Then the one-element set a is a subset of A , so a ∈ P ( A ).
But then, since P ( A ) ⊆ P (B), it follows that a ∈ P (B).
This means that a ⊆ B, hence a ∈ B.
We’ve shown that a ∈ A implies a ∈ B, so therefore A ⊆ B.
8.3 How to Prove A = B
In proofs it is often necessary to show that two sets are equal. There is a
standard way of doing this. Suppose we want to show A = B. If we show
A ⊆ B, then every element of A is also in B, but there is still a possibility
that B could have some elements that are not in A , so we can’t conclude
A = B. But if in addition we also show B ⊆ A , then B can’t contain anything
that is not in A , so A = B. This is the standard procedure for proving A = B:
prove both A ⊆ B and B ⊆ A .
How to Prove A = B
Proof.
[Prove that A ⊆ B.]
[Prove that B ⊆ A .]
Therefore, since A ⊆ B and B ⊆ A ,
it follows that A = B.
Example 8.10 Prove that n ∈ Z : 35 | n = n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n .
Proof. First we show n ∈ Z : 35 | n ⊆ n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n . Suppose
a ∈ n ∈ Z : 35 | n . This means 35 | a, so a = 35 c for some c ∈ Z. Thus a = 5(7 c)
and a = 7(5 c). From a = 5(7 c) it follows that 5 | a, so a ∈ n ∈ Z : 5 | n . From
a = 7(5 c) it follows that 7 | a, which means a ∈ n ∈ Z : 7 | n . As a belongs
to both n ∈ Z : 5 | n and n ∈ Z : 7 | n , we get a ∈ n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n .
Thus we’ve shown that n ∈ Z : 35 | n ⊆ n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n .
Next we show n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n ⊆ n ∈ Z : 35 | n . Suppose that
a ∈ n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n . By deﬁnition of intersection, this means that
a ∈ n ∈ Z : 5 | n and a ∈ n ∈ Z : 7 | n . Therefore it follows that 5 | a and 7 | a.
By deﬁnition of divisibility, there are integers c and d with a = 5 c and a = 7d .
Then a has both 5 and 7 as prime factors, so the prime factorization of a How to Prove A = B 125 must include factors of 5 and 7. Hence 5·7 = 35 divides a, so a ∈ n ∈ Z : 35 | n .
We’ve now shown that n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n ⊆ n ∈ Z : 35 | n .
At this point we’ve shown that n ∈ Z : 35 | n ⊆ n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n
and n ∈ Z : 5 | n ∩ n ∈ Z : 7 | n ⊆ n ∈ Z : 35 | n , so we’ve proved n ∈ Z : 35 | n =
n ∈ Z : 5| n ∩ n ∈ Z : 7| n .
You know from algebra that if c = 0 and ac = bc, then a = b. The next
example shows that an analogous statement holds for sets A , B and C . The
example asks us to prove a conditional statement. We will prove it with
direct proof. In carrying out the process of direct proof, we will have to
use the new techniques from this section.
Example 8.11 Suppose A , B, and C are sets, and C = . Prove that if
A × C = B × C , then A = B.
Proof. Suppose A × C = B × C . We must now show A = B.
First we will show A ⊆ B. Suppose a ∈ A . Since C = , there exists
an element c ∈ C . Thus, since a ∈ A and c ∈ C , we have (a, c) ∈ A × C , by
deﬁnition of the Cartesian product. But then, since A × C = B × C , it follows
that (a, c) ∈ B × C . Again by deﬁnition of the Cartesian product, it follows
that a ∈ B. We have shown a ∈ A implies a ∈ B, so A ⊆ B.
Next we show B ⊆ A . We use the same argument as above, with the
roles of A and B reversed. Suppose a ∈ B. Since C = , there exists an
element c ∈ C . Thus, since a ∈ B and c ∈ C , we have (a, c) ∈ B × C . But then,
since B × C = A × C , we have (a, c) ∈ A × C . It follows that a ∈ A . We have
shown a ∈ B implies a ∈ A , so B ⊆ A .
The previous two paragraphs have shown A ⊆ B and B ⊆ A , so A = B. In
summary, we have shown that if A × C = B × C , then A = B. This completes
the proof.
Now we’ll look at another way that set operations are similar to operations on numbers. From algebra you are familiar with the distributive
property a · ( b + c) = a · b + a · c. Replace the numbers a, b, c with sets A , B, C ,
and replace · with × and + with ∩. We get A × (B ∩ C ) = ( A × B) ∩ ( A × C ).
This statement turns out to be true, as we now prove.
Example 8.12 Given sets A , B, and C , prove A × (B ∩ C ) = ( A × B) ∩ ( A × C ).
Proof. First we will show that A × (B ∩ C ) ⊆ ( A × B) ∩ ( A × C ).
Suppose (a, b) ∈ A × (B ∩ C ).
By deﬁnition of the Cartesian product, this means a ∈ A and b ∈ B ∩ C .
By deﬁnition of intersection, it follows that b ∈ B and b ∈ C . Proofs Involving Sets 126 Thus, since a ∈ A and b ∈ B, it follows that (a, b) ∈ A × B (by deﬁnition of ×).
Also, since a ∈ A and b ∈ C , it follows that (a, b) ∈ A × C (by deﬁnition of ×).
Now we have (a, b) ∈ A × B and (a, b) ∈ A × C , so (a, b) ∈ ( A × B) ∩ ( A × C ).
We’ve shown that (a, b) ∈ A × (B ∩ C ) implies (a, b) ∈ ( A × B) ∩ ( A × C ) so we
have A × (B ∩ C ) ⊆ ( A × B) ∩ ( A × C ).
Next we will show that ( A × B) ∩ ( A × C ) ⊆ A × (B ∩ C ).
Suppose (a, b) ∈ ( A × B) ∩ ( A × C ).
By deﬁnition of intersection, this means (a, b) ∈ A × B and (a, b) ∈ A × C .
By deﬁnition of the Cartesian product, (a, b) ∈ A × B means a ∈ A and b ∈ B.
By deﬁnition of the Cartesian product, (a, b) ∈ A × C means a ∈ A and b ∈ C .
We now have b ∈ B and b ∈ C , so b ∈ B ∩ C , by deﬁnition of intersection.
Thus we’ve deduced that a ∈ A and b ∈ B ∩ C , so (a, b) ∈ A × (B ∩ C ).
In summary, we’ve shown that (a, b) ∈ ( A ×B)∩( A ×C ) implies (a, b) ∈ A ×(B ∩C )
so we have ( A × B) ∩ ( A × C ) ⊆ A × (B ∩ C ).
The previous two paragraphs show that A × (B ∩ C ) ⊆ ( A × B) ∩ ( A × C ) and
( A × B) ∩ ( A × C ) ⊆ A × (B ∩ C ), so it follows that ( A × B) ∩ ( A × C ) = A × (B ∩ C ).
Occasionally you can prove two sets are equal by working out a series of
equalities leading from one set to the other. This is analogous to showing
two algebraic expressions are equal by manipulating one until you obtain
the other. We illustrate this in the following example, which gives an
alternate solution to the previous example. You are cautioned that this
approach is sometimes diﬃcult to apply, but when it works it can shorten
a proof dramatically.
Before beginning the example, a note is in order. Notice that any
statement P is logically equivalent to P ∧ P . (Write out a truth table if you
are in doubt.) At one point in the following example we will replace the
expression x ∈ A with the logically equivalent statement ( x ∈ A ) ∧ ( x ∈ A ).
Example 8.13 Given sets A , B, and C , prove A × (B ∩ C ) = ( A × B) ∩ ( A × C ).
Proof. Just observe the following sequence of equalities.
A × (B ∩ C ) =
=
=
=
=
= ( x, y) : ( x ∈ A ) ∧ ( y ∈ B ∩ C )
( x, y) : ( x ∈ A ) ∧ ( y ∈ B) ∧ ( y ∈ C )
( x, y) : ( x ∈ A ) ∧ ( x ∈ A ) ∧ ( y ∈ B) ∧ ( y ∈ C )
( x, y) : (( x ∈ A ) ∧ ( y ∈ B)) ∧ (( x ∈ A ) ∧ ( y ∈ C ))
( x, y) : ( x ∈ A ) ∧ ( y ∈ B) ∩ ( x, y) : ( x ∈ A ) ∧ ( y ∈ C )
( A × B) ∩ ( A × C ) The proof is complete. (def. of ×)
(def. of ∩)
(P = P ∧ P ) (rearrange)
(def. of ∩)
(def. of ×) Examples: Perfect Numbers 127 The equation A × (B ∩ C ) = ( A × B) ∩ ( A × C ) just obtained is a fundamental
law that you may actually use fairly often as you continue with mathematics.
Some similar equations are listed below. Each of these can be proved with
this section’s techniques, and the exercises will ask that you do so.
A∩B = A∪B
A∪B = A∩B
A ∩ (B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
A ∪ (B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
A × (B ∪ C ) = ( A × B ) ∪ ( A × C )
A × (B ∩ C ) = ( A × B ) ∩ ( A × C ) DeMorgan’s laws for sets
Distributive laws for sets
Distributive laws for sets It is very good practice to prove these equations. Depending on your
learning style, it is probably not necessary to commit them to memory.
But don’t forget them entirely. They may well be useful later in your
mathematical education. If so, you can look them up or re-derive them on
the spot. If you go on to study mathematics deeply, you will at some point
realize that you’ve internalized them without even being cognizant of it.
8.4 Examples: Perfect Numbers
Sometimes it takes a good bit of work and creativity to show that one set
is a subset of another or that they are equal. We illustrate this now with
examples from number theory involving what are called perfect numbers.
Even though this topic is quite old, dating back more than two-thousand
years, it leads to some questions that are unanswered even today.
The problem involves adding up the positive divisors of a natural
number. To begin the discussion, consider the number 12. If we add up the
positive divisors of 12 that are less than 12, we obtain 1 + 2 + 3 + 4 + 6 = 16,
which is greater than 12. Doing the same thing for 15, we get 1 + 3 + 5 = 9
which is less than 15. For the most part, given a natural number p, the
sum of its positive divisors less than itself will either be greater than p
or less than p. But occasionally the divisors add up to exactly p. If this
happens, then p is said to be a perfect number.
Deﬁnition 8.1 A number p ∈ N is perfect if it equals the sum of its
positive divisors less than itself. Some examples follow.
• The number 6 is perfect since 6 = 1 + 2 + 3.
• The number 28 is perfect since 28 = 1 + 2 + 4 + 7 + 14.
• The number 496 is perfect since 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248. Proofs Involving Sets 128 Though it would take a while to ﬁnd it by trial-and-error, the next
perfect number after 496 is 8128. You can check that 8128 is perfect. Its
divisors are 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and indeed
8128 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064. Are there other perfect numbers? How can they be found? Do they obey any
patterns? These questions fascinated the ancient Greek mathematicians.
In what follows we will develop an idea—recorded by Euclid—that partially
answers these questions. Although Euclid did not use sets1 , we will
nonetheless phrase his idea using the language of sets.
Since our goal is to understand what numbers are perfect, let’s deﬁne
the following set.
P = p ∈ N : p is perfect .
Therefore P = 6, 28, 496, 8128, . . . , but it is unclear what numbers are in
P other than the ones listed. Our goal is to gain a better understanding
of just which numbers the set P includes. To do this, we will examine
the following set A . It looks more complicated than P , but it will be very
helpful for understanding P , as you will soon see.
A = 2n−1 (2n − 1) : n ∈ N, and 2n − 1 is prime In words, A consists of every natural number of form 2n−1 (2n − 1), where
2n − 1 is prime. To get a feel for what numbers belong to A , look at the
following table. For each natural number n, it tallies the corresponding
numbers 2n−1 and 2n − 1. If 2n − 1 happens to be prime, then the product
2n−1 (2n − 1) is given; otherwise that entry is labeled with an ∗.
n 2n−1 2n − 1 2n−1 (2n − 1) 1
2
3
4
5
6
7
8
9
10
11
12
13 1
2
4
8
16
32
64
128
256
512
1024
2048
4096 1
3
7
15
31
63
127
255
511
1023
2047
4095
8191 ∗
6
28
∗
496
∗
8128
∗
∗
∗
∗
∗
33, 550, 336 1 Set theory was invented over 2000 years after Euclid died. Examples: Perfect Numbers 129 Notice that the ﬁrst four entries of A are the perfect numbers 6, 28, 496
and 8128. At this point you may want to jump to the conclusion that A = P .
But it is a shocking fact that in over 2000 years no one has ever been able
to determine whether or not A = P . But it is known that A ⊆ P , and we
will now prove it. In words, we are going to show that every element of A
is perfect. (But by itself, that leaves open the possibility that there may
be some perfect numbers in P that are not in A .)
The main ingredient for the proof will be the formula for the sum of a
geometric series. You probably saw this most recently in Calculus II. The
formula is
n k=0 rk = r n+1 − 1
.
r−1 We will need this for the case r = 2, which is
n 2k = 2n+1 − 1. (8.1) k=0 (See the solution for exercise 19 in Section 7.3 for a proof of this formula.) Now we are ready to prove our result. Let’s draw attention to its
signiﬁcance by calling it a theorem rather than a proposition.
Theorem 8.1 If A = 2n−1 (2n − 1) : n ∈ N, and 2n − 1 is prime and P =
p ∈ N : p is perfect , then A ⊆ P .
Proof. Assume A and P are as stated. To show A ⊆ P , we must show that
p ∈ A implies p ∈ P . Thus suppose p ∈ A . By deﬁnition of A , this means
p = 2n−1 (2n − 1) (8.2) for some n ∈ N for which 2n − 1 is prime. We want to show that p ∈ P , that
is we want to show p is perfect. Thus, we need to show that the sum of
the positive divisors of p that are less than p add up to p. Notice that
since 2n − 1 is prime, any divisor of p = 2n−1 (2n − 1) must have the form 2k
or 2k (2n − 1) for 0 ≤ k ≤ n − 1. Thus the positive divisors of p are as follows. 20
20 (2n − 1) 21
21 (2n − 1) 22
22 (2n − 1) . . . 2n−2
. . . 2n−2 (2n − 1) 2n−1
2n−1 (2n − 1) Notice that this list starts with 20 = 1 and ends with 2n−1 (2n − 1) = p. Proofs Involving Sets 130 If we add up all these divisors except for the last one (which equals p)
we get the following. n−1
k=0 2k + n−2
k=0 2k (2n − 1) = n−1
k=0 2k + (2n − 1) n−2 2k k=0 = (2n − 1) + (2n − 1)(2n−1 − 1) (by Equation (8.1))
= [1 + (2n−1 − 1)](2n − 1)
= 2n−1 (2n − 1)
=p
(by Equation (8.2)) This shows that the positive divisors of p that are less than p add up to p.
Therefore p is perfect, by deﬁnition of a perfect number. Thus p ∈ P , by
deﬁnition of P .
We have shown that p ∈ A implies p ∈ P , which means A ⊆ P . Combined with the chart on the previous page, this theorem gives us
a new perfect number! The element p = 213−1 (213 − 1) = 33, 550, 336 in A is
perfect.
Observe also that every element of A is a multiple of a power of 2,
and therefore even. But this does not mean every perfect number is even,
because we’ve only shown A ⊆ P , not A = P . Are there any odd perfect
numbers? No one knows.
In over 2000 years, no one has ever found an odd perfect number nor
has anyone been able to prove that every perfect number is even. But it
is known that the set A does contain every even perfect number. This
fact was ﬁrst proved by Euler, and we duplicate his reasoning in the next
theorem, which proves that A = E , where E is the set of all even perfect
numbers. It is a good example of how to prove two sets are equal.
For convenience, we are going to use a slightly diﬀerent deﬁnition of a
perfect number. A number p ∈ N is perfect if its positive divisors add up
to 2 p. For example the number 6 is perfect since the sum of its divisors
is 1 + 2 + 3 + 6 = 2 · 6. This deﬁnition is simpler than the ﬁrst one because
we do not have to stipulate that we are adding up the divisors that are
less than p. Instead we add in the last divisor p, and that has the eﬀect
of adding an additional p, thereby doubling the answer. Examples: Perfect Numbers 131 Theorem 8.2 If A = 2n−1 (2n − 1) : n ∈ N, and 2n − 1 is prime and E =
p ∈ N : p is perfect and even , then A = E .
Proof. To show that A = E , we need to show A ⊆ E and E ⊆ A .
First we will show that A ⊆ E . Suppose p ∈ A . This means p is even,
because the deﬁnition of A shows that every element of A is a multiple of
a power of 2. Also, p is a perfect number because Theorem 8.1 states that
every element of A is also an element of P , hence perfect. Thus p is an
even perfect number, so p ∈ E . Therefore A ⊆ E .
Next we show that E ⊆ A . Suppose p ∈ E . This means p is an even
perfect number. Write the prime factorization of p as p = 2k 3n1 5n2 7n2 . . .,
where the powers n1 , n2 , n3 . . . may be zero. But, as p is even, the integer
k must be greater than zero. It follows p = 2k q for some positive integer k
and an odd integer q. Now, our aim is to show that p ∈ A , which means we
must show p has form p = 2n−1 (2n − 1). To get our current p = 2k q closer to
this form, let n = k + 1, so we now have
p = 2n−1 q. (8.3) List the positive divisors of q as d1 , d2 , d3 , . . . , d m . (Where d1 = 1 and d m = q.)
Then the divisors of p are:
20 d 1
21 d 1
22 d 1
23 d 1
.
.
.
n−1
2
d1 20 d 2
21 d 2
22 d 2
23 d 2
.
.
.
n−1
2
d2 20 d 3
21 d 3
22 d 3
23 d 3
.
.
.
n−1
2
d3 ...
...
...
...
... 20 d m
21 d m
22 d m
23 d m
.
.
.
n−1
2
dm Since p is perfect, these divisors add up to 2 p. By Equation (8.3), their
sum is 2 p = 2(2n−1 q) = 2n q. Adding the divisors column-by-column, we get
n−1
k=0 2k d 1 + n−1
k=0 2k d 2 + n−1 2k d 3 + · · · + k=0 n−1 2k d m = 2n q. k=0 Applying Equation (8.1), this becomes
(2n − 1) d 1 + (2n − 1) d 2 + (2n − 1) d 3 + · · · + (2n − 1) d m = 2n q
(2n − 1)( d 1 + d 2 + d 3 + · · · + d m ) = 2n q d1 + d2 + d3 + · · · + d m = 2n q
2n − 1 Proofs Involving Sets 132
so that
d1 + d2 + d3 + · · · + d m = q
(2n − 1 + 1) q (2n − 1) q + q
=
= q+ n
.
2n − 1
2n − 1
2 −1 From this we see that 2nq 1 is an integer. It follows that both q and 2nq 1
−
−
are positive divisors of q. Since their sum equals the sum of all positive
divisors of q, it follows that q has only two positive divisors, q and 2nq 1 .
−
Since one of its divisors must be 1, it must be that 2nq 1 = 1, which means
−
q = 2n − 1. Now a number with just two positive divisors is prime, so
q = 2n − 1 is prime. Plugging this into Equation (8.3) gives p = 2n−1 (2n − 1),
where 2n − 1 is prime. This means p ∈ A , by deﬁnition of A . We have now
shown that p ∈ E implies p ∈ A , so E ⊆ A .
Since A ⊆ E and E ⊆ A , it follows that A = E .
Do not be alarmed if you feel that you wouldn’t have thought of this
proof. It took the genius of Euler to discover this approach.
We’ll conclude this chapter with some facts about perfect numbers.
• The sixth perfect number is p = 217−1 (217 − 1) = 8589869056.
• The seventh perfect number is p = 219−1 (219 − 1) = 137438691328.
• The eighth perfect number is p = 231−1 (231 − 1) = 2305843008139952128.
• The 20 th perfect number is p = 24423−1 (24423 − 1). It has 2663 digits.
• The 23rd perfect number is p = 211213−1 (211213 − 1). It has 6957 digits.
As mentioned earlier, no one knows whether or not there are any odd
perfect numbers. It is not even known whether there are ﬁnitely many or
inﬁnitely many perfect numbers. It is known that the last digit of every
even perfect number is either a 6 or an 8. Perhaps this is something you’d
enjoy proving.
We’ve seen that perfect numbers are closely related to prime numbers
that have the form 2n − 1. Prime numbers that have this form are called
Mersenne primes, after the French scholar Marin Mersenne (1588–
1648) who popularized them. The ﬁrst several Mersenne primes are
22 − 1 = 3, 23 − 1 = 7, 25 − 1 = 31, 27 − 1 = 127 and 213 − 1 = 8191. To date,
only 45 Mersenne primes are known, the largest of which is 243,112,609 − 1.
There is a substantial cash prize for anyone who ﬁnds a 46th. (See
http://www.mersenne.org/prime.htm.) You will probably have better
luck with the exercises. Examples: Perfect Numbers 133 Exercises for Chapter 8
Use the methods introduced in this chapter to prove the following statements.
1. Prove that 12 n : n ∈ Z ⊆ 2n : n ∈ Z ∩ 3n : n ∈ Z .
2. Prove that 6 n : n ∈ Z = 2n : n ∈ Z ∩ 3n : n ∈ Z .
3. If k ∈ Z, then n ∈ Z : n | k ⊆ n ∈ Z : n | k2 .
4. If m, n ∈ Z, then x ∈ Z : mn | x ⊆ x ∈ Z : m | x ∩ x ∈ Z : n | x .
5. If p and q are positive integers, then p n : n ∈ N ∩ q n : n ∈ N = .
6. Suppose A , B and C are sets. Prove that if A ⊆ B, then A − C ⊆ B − C .
7. Suppose A , B and C are sets. If B ⊆ C , then A × B ⊆ A × C .
8. If A , B and C are sets then A ∪ (B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ).
9. If A , B and C are sets then A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ).
10. If A and B are sets in a universal set U , then A ∩ B = A ∪ B.
11. If A and B are sets in a universal set U , then A ∪ B = A ∩ B.
12. If A , B and C are sets, then A − (B ∩ C ) = ( A − B) ∪ ( A − C ).
13. If A , B and C are sets, then A − (B ∪ C ) = ( A − B) ∩ ( A − C ).
14. If A , B and C are sets, then ( A ∪ B) − C = ( A − C ) ∪ (B − C ).
15. If A , B and C are sets, then ( A ∩ B) − C = ( A − C ) ∩ (B − C ).
16. If A , B and C are sets, then A × (B ∪ C ) = ( A × B) ∪ ( A × C ).
17. If A , B and C are sets, then A × (B ∩ C ) = ( A × B) ∩ ( A × C ).
18. If A , B and C are sets, then A × (B − C ) = ( A × B) − ( A × C ).
19. Prove that 9n : n ∈ Z ⊆ 3n : n ∈ Z , but 9n : n ∈ Z = 3n : n ∈ Z
20. Prove that 9n : n ∈ Q = 3n : n ∈ Q .
21. Suppose A and B are sets. Prove A ⊆ B if and only if A − B = .
22. Let A and B be sets. Prove that A ⊆ B if and only if A ∩ B = A .
23. For each a ∈ R, let A a = ( x, a( x2 − 1)) ∈ R2 : x ∈ R . Prove that
a∈R 24. Prove that A a = (−1, 0), (1, 0)) . [3 − x2 , 5 + x2 ] = [3, 5]. x∈R 25. Suppose A , B, C and D are sets. Prove that ( A × B) ∪ (C × D ) ⊆ ( A ∪ C ) × (B ∪ D ).
26. Prove 4k + 5 : k ∈ Z = 4k + 1 : k ∈ Z .
27. Prove 12a + 4b : a, b ∈ Z = 4 c : c ∈ Z .
28. Prove 12a + 25b : a, b ∈ Z = Z.
29. Suppose A = . Prove that A × B ⊆ A × C , if and only if B ⊆ C .
30. Prove that (Z × N) ∩ (N × Z) = N × N.
31. Suppose B = and A × B ⊆ B × C . Prove A ⊆ C . CHAPTER 9 Disproof ver since Chapter 4 we have dealt with one major theme: given a
statement, prove that is it true. In every example and exercise we
were handed a true statement and charged with the task of proving it.
Have you ever wondered what would happen if you were given a false
statement to prove? The answer is that no (correct) proof would be possible,
for if it were, the statement would be true, not false.
But how would you convince someone that a statement is false? The
mere fact that you could not produce a proof does not automatically mean
the statement is false, for you know (perhaps all too well) that proofs
can be diﬃcult to construct. It turns out that there is a very simple and
utterly convincing procedure that proves a statement is false. The process
of carrying out this procedure is called disproof. Thus this chapter is
concerned with disproving statements.
Before describing the new method, we will set the stage with some
relevant background information. First, we point out that mathematical
statements can be divided into three categories, described below.
One category consists of all those statements that have been proved to be
true. For the most part we regard these statements as signiﬁcant enough
to be designated with special names such as “theorem,” “proposition,”
“lemma,” and “corollary.” Some examples of statements in this category are
listed in the left-hand box in the diagram on the following page. There are
also some wholly uninteresting statements (such as 2 = 2) in this category,
and although we acknowledge their existence we certainly do not dignify
them with terms such as “theorem” or “proposition.”
At the other extreme is a category consisting of statements that are
known to be false. Examples are listed in the box on the right. Since
mathematicians are not very interested in them, these types of statements
do not get any special names, other than the blanket term “false statement.”
But there is a third (and quite interesting) category between these two
extremes. It consists of statements whose truth or falsity has not been
determined. Examples include things like “Every perfect number is even,” E 135
or “Every even integer greater than 2 is the sum of two primes.” (You may
recall the latter statement is sometimes called the Goldbach Conjecture.
See Section 2.1.) Mathematicians have a special name for the statements
in this category that they suspect (but haven’t yet proved) are true. Such
statements are called conjectures. T T S Known to be true Truth unknown (Theorems & propositions) : (Conjectures) Examples:
• Pythagorean Theorem
• Fermat’s Last Theorem (Section 2.1) Examples:
• All perfect numbers are even.
• Any even number greater Known to be false
Examples:
• All prime numbers are odd.
• Some quadratic equations have three solutions.
than 2 is the sum of two
• 0=1
primes. (Goldbach’s
number is odd.
conjecture, Section 2.1)
∞1
• There exist natural
diverges. • There are inﬁnitely many
• The series
numbers a, b and c
k
k=1
for which a3 + b3 = c3 .
prime numbers of form
n − 1, with n ∈ N.
2
• The square of an odd Professional mathematicians spend much of their time and energy
attempting to prove or disprove conjectures. (They also expend considerable
mental energy in creating new conjectures based on collected evidence or
intuition.) When a conjecture is proved (or disproved) the proof or disproof
will typically appear in a published paper (provided the conjecture is of
suﬃcient interest). If it is proved, the conjecture attains the status of
a theorem or proposition. If it is disproved, then no one is really very
interested in it anymore – mathematicians do not care much for false
statements.
Most conjectures that mathematicians are interested in are quite
diﬃcult to prove or disprove. We are not at that level yet. In this text, the
“conjectures” that you will encounter are the kinds of statements that a
professional mathematician would immediately spot as true or false, but
you may have to do some work before ﬁguring out a proof or disproof. But
in keeping with the cloud of uncertainty that surrounds conjectures at the
advanced levels of mathematics, most exercises in this chapter (and many
of them beyond it) will ask you to prove or disprove statements without
giving any hint as to whether they are true or false. Your job will be to
decide whether or not they are true and to either prove or disprove them.
The examples in this chapter will illustrate the processes one typically Disproof 136 goes through in deciding whether a statement is true or false, and then
verifying that it’s true or false.
You know the three major methods of proving a statement: direct proof,
contrapositive proof, and proof by contradiction. Now we are ready to
understand the method of disproving a statement. Suppose you want to
disprove a statement P . In other words you want to prove that P is false.
The way to do this is to prove that ∼ P is true, for if ∼ P is true, it follows
immediately that P has to be false.
How to disprove P : Prove ∼ P . Our approach is incredibly simple. To disprove P , prove ∼ P . In theory,
this proof can be carried out by direct, contrapositive or contradiction
approaches. However, in practice things can be even easier than that
if we are disproving a universally quantiﬁed statement or a conditional
statement. That is our next topic.
9.1 Disproving Universal Statements: Counterexamples
A conjecture may be described as a statement that we hope is a theorem.
As we know, many theorems (hence many conjectures) are universally
quantiﬁed statements. Thus it seems reasonable to begin our discussion
by investigating how a universally quantiﬁed statement such as
∀ x ∈ S , P ( x) can be disproved. To disprove this statement, we must prove its negation.
Its negation is
∼ (∀ x ∈ S , P ( x)) = ∃ x ∈ S , ∼ P ( x). The negation is an existence statement. To prove the negation is true,
we just need to produce an example of an x ∈ S that makes ∼ P ( x) true,
that is an x that makes P ( x) false. This leads to the following outline for
disproving a universally quantiﬁed statement.
How to disprove ∀ x ∈ S , P ( x).
Produce an example of an x ∈ S
that makes P ( x) false.
Things are even simpler if we want to disprove a conditional statement
P ( x) ⇒ Q ( x). This statement asserts that for every x that makes P ( x) true, Counterexamples 137 Q ( x) will also be true. The statement can only be false if there is an x that
makes P ( x) true and Q ( x) false. This leads to our next outline for disproof. How to disprove P ( x) ⇒ Q ( x).
Produce an example of an x that
makes P ( x) true and Q ( x) false.
In both of the above outlines, the statement is disproved simply by
exhibiting an example that shows the statement is not always true. (Think
of it as an example that proves the statement is a promise that can be
broken.) There is a special name for an example that disproves a statement:
it is called a counterexample.
Example 9.1 As our ﬁrst example, we will work through the process of
deciding whether or not the following conjecture is true.
Conjecture: For every n ∈ Z, the integer f (n) = n2 − n + 11 is prime.
In resolving the truth or falsity of a conjecture, it’s a good idea to gather
as much information about the conjecture as possible. In this case let’s
start by making a table that tallies the values of f ( n) for some integers n.
n −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 f ( n) 23 17 13 11 11 13 17 23 31 41 53 67 83 101 This looks encouraging. In every case, f ( n) is prime, so you may begin to
suspect that the conjecture is true. Before attempting a proof, let’s try one
more n. Unfortunately, f (11) = 112 − 11 + 11 = 112 is not prime. Thus the
conjecture is false because n = 11 is a counterexample. We summarize our
disproof as follows.
Disproof. The statement “For every n ∈ Z, the integer f ( n) = n2 − n + 11 is
prime,” is false. For a counterexample, note that for n = 11, the integer
f (11) = 121 = 11 · 11 is not prime. Notice that, in disproving a statement by presenting a counterexample it is
important to explain exactly how the counterexample makes the statement
false. Our work would not have been complete if we had just said “for a
counterexample, consider n = 11,” and left it at that. We need to show that
the answer f (11) is not prime. Showing that it factors as a product of two
integers suﬃces for this. Disproof 138 Example 9.2 Either prove or disprove the following conjecture.
Conjecture If A , B and C are sets, then A − (B ∩ C ) = ( A − B) ∩ ( A − C ).
Disproof. This conjecture is false because of the following counterexample.
Let A = 1, 2, 3 , B = 1, 2 and C = 2, 3 . Notice that A − (B ∩ C ) = 1, 3 and
( A − B) ∩ ( A − C ) = , so A − (B ∩ C ) = ( A − B) ∩ ( A − C ).
(To see where this counterexample came from, draw Venn diagrams for
A − (B ∩ C ) and ( A − B) ∩ ( A − C ). You will see that the diagrams are diﬀerent.
The numbers 1, 2, and 3 can then be inserted into the regions of the
diagrams in such a way as to create the above counterexample.)
9.2 Disproving Existence Statements
We have seen that we can disprove a universally quantiﬁed statement or a
conditional statement simply by ﬁnding a counterexample. Now let’s turn
to the problem of disproving an existence statement such as
∃ x ∈ S , P ( x). Proving this would involve simply ﬁnding an example of an x that makes
P ( x) true. To disprove it, we have to prove its negation ∼ (∃ x ∈ S , P ( x)) =
∀ x ∈ S , ∼ P ( x). But this negation is universally quantiﬁed. Proving it
involves showing that ∼ P ( x) is true for all x ∈ S , and for this an example
does not suﬃce. Instead we must use direct, contrapositive or contradiction
proof to prove the conditional statement “If x ∈ S , then ∼ P ( x).” As an
example, here is a conjecture to either prove or disprove.
Example 9.3 Either prove or disprove the following conjecture.
Conjecture: There is a real number x for which x4 < x < x2 .
This may not seem like an unreasonable statement at ﬁrst glance. After
all, if the statement were asserting the existence of a real number for
which x3 < x < x2 , then it would be true: just take x = −2. But it asserts
there is an x for which x4 < x < x2 . When we apply some intelligent guessing
to locate such an x we run into trouble. If x = 1 , then x4 < x, but we don’t
2
have x < x2 ; similarly if x = 2, we have x < x2 but not x4 < x. Since ﬁnding
an x with x4 < x < x2 seems problematic, we may begin to suspect that the
given statement is false.
Let’s see if we can disprove it. According to our strategy for disproof,
to disprove it we must prove its negation. Symbolically, the statement is Disproof by Contradiction 139 ∃ x ∈ R, x4 < x < x2 , so its negation is
∼ (∃ x ∈ R, x4 < x < x2 ) = ∀ x ∈ R, ∼ ( x4 < x < x2 ). Thus, in words the negation is:
For every real number x, it is not the case that x4 < x < x2 .
This can be proved with contradiction, as follows. Suppose for the
sake of contradiction that there is an x for which x4 < x < x2 . Then x must
be positive since it’s greater than the non-negative number x4 . Dividing
all parts of x4 < x < x2 by the positive number x produces x3 < 1 < x. Now
subtract 1 from all parts of x3 < 1 < x to obtain x3 − 1 < 0 < x − 1 and reason
as follows.
x3 − 1 < 0 < x−1 ( x − 1)( x2 + x + 1) < 0 < x−1 2 x + x+1 < 0 < 1 Now we have x2 + x + 1 < 0, which is a contradiction because x is positive.
We summarize our work as follows.
The statement “There is a real number x for which x4 < x < x2 ,” is false
because we have proved its negation “For every real number x, it not the
case that x4 < x < x2 .”
As you work the exercises, keep in mind that not every conjecture will be
false. If one is true, then a disproof is impossible and you must produce a
proof. Here is an example.
Example 9.4 Either prove or disprove the following conjecture.
Conjecture There exist three integers x, y, z, all greater than 1 and no
two equal, for which x y = y z .
This conjecture is true. It is an existence statement, so to prove it we
just need to give an example of three integers x, y, z, all greater than 1 and
no two equal, so that x y = y z . A proof follows.
Proof. Note that if x = 2, y = 16 and z = 4, then x y = 216 = (24 )4 = 164 = y z .
9.3 Disproof by Contradiction
Contradiction can be a very useful way to disprove a statement. To see
how this works, suppose we wish to disprove a statement P . We know Disproof 140 that to disprove P , we must prove ∼ P . To prove ∼ P with contradiction,
we assume ∼∼ P is true and deduce a contradiction. But since ∼∼ P = P ,
this boils down to assuming P is true and deducing a contradiction. Here
is an outline.
How to disprove P with contradiction:
Assume P is true, and deduce a contradiction.
To illustrate this, let’s revisit Example 9.3 but do the disproof with
contradiction. You will notice that the work duplicates much of what we
did in Example 9.3, but is it much more streamlined because here we do
not have to negate the conjecture.
Example 9.5 Disprove the following conjecture.
Conjecture: There is a real number x for which x4 < x < x2 .
Disproof. Suppose for the sake of contradiction that this conjecture is true.
Let x be a real number for which x4 < x < x2 . Then x is positive, since it is
greater than the positive number x4 . Dividing all parts of x4 < x < x2 by
the positive number x produces x3 < 1 < x. Now subtract 1 from all parts
of x3 < 1 < x to obtain x3 − 1 < 0 < x − 1 and reason as follows.
x3 − 1 < 0 < x−1 ( x − 1)( x + x + 1) < 0 < x−1 2 2 x + x+1 < 0 < 1 Now we have x2 + x + 1 < 0, which is a contradiction because x is positive.
Thus the conjecture must be false. Exercises for Chapter 9
Each of the following statements is either true or false. If a statement is true,
prove it. If a statement is false, disprove it. These exercises are cumulative,
covering all topics addressed in chapters 1–9.
1. If x, y ∈ R, then | x + y| = | x| + | y|.
2. For every natural number n, the integer 2n2 − 4n + 31 is prime.
3. If n ∈ Z and n5 − n is even, then n is even.
4. For every natural number n, the integer n2 + 17 n + 17 is prime.
5. If A , B, C and D are sets, then ( A × B) ∪ (C × D ) = ( A ∪ C ) × (B ∪ D ).
6. If A , B, C and D are sets, then ( A × B) ∩ (C × D ) = ( A ∩ C ) × (B ∩ D ). Disproof by Contradiction 141 7. If A , B and C are sets, and A × C = B × C , then A = B.
8. If A , B and C are sets, then A − (B ∪ C ) = ( A − B) ∪ ( A − C ).
9. If A and B are sets, then P ( A ) − P (B) ⊆ P ( A − B).
10. If A and B are sets and A ∩ B = , then P ( A ) − P (B) ⊆ P ( A − B).
11. If a, b ∈ N, then a + b < ab.
12. If a, b, c ∈ N and ab, bc and ac all have the same parity, then a, b and c all have
the same parity.
13. There exists a set X for which R ⊆ X and ∈ X. 14. If A and B are sets, then P ( A ) ∩ P (B) = P ( A ∩ B).
15. Every odd integer is the sum of three odd integers.
16. If A and B are ﬁnite sets, then | A ∪ B| = | A | + |B|.
17. For all sets A and B, if A − B = , then B = .
18. If a, b, c ∈ N, then at least one of a − b, a + c and b − c is even.
19. For every r , s ∈ Q with r < s, there is an irrational number u for which r < u < s.
20. There exist prime numbers p and q for which p − q = 1000.
21. There exist prime numbers p and q for which p − q = 97.
22. If p and q are prime numbers for which p < q, then 2 p + q2 is odd.
23. If x, y ∈ R and x3 < y3 , then x < y.
24. The inequality 2 x ≥ x + 1 is true for all positive real numbers x.
25. For all a, b, c ∈ Z, if a | bc, then a | b or a | c.
26. Suppose A , B and C are sets. If A = B − C , then B = A ∪ C .
27. The equation x2 = 2 x has three real solutions.
28. Suppose a, b ∈ Z. If a | b and b | a, then a = b.
29. If x, y ∈ R and | x + y| = | x − y|, then y = 0.
30. There exist integers a and b for which 42a + 7b = 1.
31. No number (other than 1) appears in Pascal’s Triangle more than four times.
32. If n, k ∈ N and n
k is a prime number, then k = 1 or k = n − 1. 33. Suppose f ( x) = a 0 + a 1 x + a 2 x2 + · · · + a n x n is a polynomial of degree 1 or greater,
and for which each coeﬃcient a i is in N. Then there is an n ∈ N for which the
integer f (n) is not prime.
34. If X ⊆ A ∪ B, then X ⊆ A or X ⊆ B. CHAPTER 10 Mathematical Induction his chapter introduces a very powerful proof technique called mathematical induction (or just induction for short). To motivate the
discussion, let’s ﬁrst examine the kinds of statements that induction is
used to prove. Consider the following statement. T Conjecture. The sum of the ﬁrst n odd natural numbers equals n2 .
The following table illustrates what this conjecture is saying. Each row
is headed by a natural number n, followed by the sum of the ﬁrst n odd
natural numbers, followed by n2 .
n sum of the ﬁrst n odd natural numbers n2 1
2
3
4
5
.
.
.
n
.
.
. 1= .....................................
1+3 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1+3+5 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1+3+5+7 = . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1+3+5+7+9 = . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
1 + 3 + 5 + 7 + 9 + 11 + · · · + (2 n − 1) = . . . . . . .
.
.
. 1
4
9
16
25
.
.
.
n2
.
.
. Observe that, in the ﬁrst ﬁve lines of the table, the sum of the ﬁrst
n odd natural numbers really does add up to n2 . Notice also that these
ﬁrst ﬁve lines indicate that the n th odd natural number (the last number
in each sum) is 2n − 1. (For instance, when n = 2, the second odd natural
number is 2 · 2 − 1 = 3; when n = 3, the third odd natural number is 2 · 3 − 1 = 5, etc.)
The table begs a question. For any n, does the sum 1+3+5+7+· · ·+(2 n −1)
really always equal n2 ? In other words, is the conjecture true?
Let’s rephrase this as follows. For each natural number n (i.e. for each
line of the table) we have a statement S n , as follows: 143
S 1 : 1 = 12
S 2 : 1 + 3 = 22
S 3 : 1 + 3 + 5 = 32
.
.
.
S n : 1 + 3 + 5 + 7 + · · · + (2 n − 1) = n2
.
.
. Our question is: Are all of these statements true?
Mathematical induction is designed to answer just this kind of question.
It is used when we have a set of statements S1 , S2 , S3 , . . . , S n , . . ., and we
need to prove that they are all true. The method is really quite simple.
To visualize it, think of the statements as dominoes, lined up in a row.
Imagine you can prove the ﬁrst statement S1 , and symbolize this as domino
S 1 being knocked down. Additionally, imagine that you can prove that any
statement S k being true (falling) forces the next statement S k+1 to be true
(to fall). Then S1 falls, and knocks down S2 . Next S2 falls and knocks
down S3 , then S3 knocks down S4 , and so on. The inescapable conclusion
is that all the statements are knocked down (proved true).
The Simple Idea Behind Mathematical Induction ··· Sk Sk+1 Sk+2 Sk+3 Sk+4 · · · ··· S6 Sk Sk+1 Sk+2 Sk+3 Sk+4 · · · Sk+2 Sk+3 Sk+4 · · · S k+ S5 S4 S k+ S3 S2 S k+ S1 Statements are lined up like dominoes. S1 S3 S2 S5 S4 S6 (1) Suppose the ﬁrst statement falls (i.e. is proved true); S6 S5 S4 S3 S2 S1 ··· Sk Sk+ 1 (2) Suppose the k th falling always causes the ( k + 1) th to fall; 3 2 1 Then all must fall (i.e. all statements are proved true). Sk S6 S5 S4 S3 S2 S1 ··· ··· Mathematical Induction 144 This picture gives our outline for proof by mathematical induction. Proposition Outline for Proof by Induction
The statements S1 , S2 , S3 , S3 , . . . are all true. Proof. (Induction)
(1) Prove that the ﬁrst statement S1 is true.
(2) Given any integer k ≥ 1, prove that the statement S k ⇒ S k+1 is true.
It follows by mathematical induction that every S n is true.
In this setup, the ﬁrst step (1) is called the basis step. Because S1 is
usually a very simple statement, the basis step is often quite easy to do.
The second step (2) is called the inductive step. In the inductive step
direct proof is most often used to prove S k ⇒ S k+1 , so this step is usually
carried out by assuming S k is true and showing this forces S k+1 to be true.
The assumption that S k is true is called the inductive hypothesis.
Now let’s apply this technique to our original conjecture that the sum
of the ﬁrst n odd natural numbers equals n2 . This involves showing that
for each n ∈ N, the statement S n : 1 + 3 + 5 + 7 +· · ·+ (2n − 1) = n2 is true. Before
getting started, observe that S k is obtained from S n by plugging k in for n.
Thus S k is the statement S k : 1 + 3 + 5 + 7 +· · ·+ (2k − 1) = k2 . Also, we get S k+1
by plugging in k + 1 for n, so that S k+1 : 1 + 3 + 5 + 7 +· · ·+ (2( k + 1) − 1) = (k + 1)2 .
Example 10.1 Prove the following proposition.
Proposition If n ∈ N, then 1 + 3 + 5 + 7 + · · · + (2 n − 1) = n2 . Proof. We will prove this with mathematical induction.
(1) Observe that if n = 1, this statement is 1 = 12 , which is obviously true.
(2) We must now prove S k ⇒ S k+1 for any k ≥ 1. That is, we must show
that if 1+3+5+7+· · ·+(2 k −1) = k2 , then 1+3+5+7+· · ·+(2( k +1)−1) = (k +1)2 .
We use direct proof. Suppose 1 + 3 + 5 + 7 + · · · + (2 k − 1) = k2 . Then
1 + 3 + 5 + 7 + · · · · · · · · · · · · · · · + (2( k + 1) − 1) =
1 + 3 + 5 + 7 + · · · + (2 k − 1) + (2( k + 1) − 1) =
1 + 3 + 5 + 7 + · · · + (2 k − 1) + (2( k + 1) − 1) = k2 + (2( k + 1) − 1) = k2 + 2 k + 1 = ( k + 1)2 . Thus 1 + 3 + 5 + 7 + · · · + (2( k + 1) − 1) = (k + 1)2 . This proves that S k ⇒ S k+1 .
It follows by induction that 1 + 3 + 5 + 7 + · · · + (2n − 1) = n2 for every n ∈ N. 145
In induction proofs it is usually the case that the ﬁrst statement S1 is
indexed by the natural number 1, but this need not always be the case.
Depending on the problem, the ﬁrst statement could be S0 , or S m for any
other integer m. In the next example the statements are S0 , S1 , S2 , S3 , . . ..
The same outline is used, except that the basis step veriﬁes S0 , not S1 .
Example 10.2
Proposition Prove the following proposition.
If n is a non-negative integer, then 5 |( n5 − n). Proof. We will prove this with mathematical induction. Observe that the
ﬁrst non-negative integer is 0, so the basis step involves n = 0.
(1) If n = 0, this statement is 5 |(05 − 0) or 5 | 0, which is obviously true.
(2) Let k ≥ 0. We need to prove that if 5 |( k5 − k), then 5 |((k + 1)5 − ( k + 1)).
We use direct proof. Suppose 5 |(k5 − k). Thus k5 − k = 5a for some a ∈ Z.
Observe that
( k + 1)5 − ( k + 1) = k5 + 5 k4 + 10 k3 + 10 k2 + 5 k + 1 − k − 1 = ( k5 − k) + 5 k4 + 10 k3 + 10 k2 + 5 k
= 5a + 5 k4 + 10 k3 + 10 k2 + 5 k
= 5(a + k4 + 2 k3 + 2 k2 + k). This shows (k + 1)5 − (k + 1) is an integer multiple of 5, so 5 |((k + 1)5 − ( k + 1)).
We have now shown that 5 |( k5 − k) implies 5 |((k + 1)5 − ( k + 1)).
It follows by induction that 5 |( n5 − n) for all non-negative integers n.
Example 10.3 Prove the following proposition.
n Proposition If n ∈ Z and n ≥ 0, then i · i ! = ( n + 1)! − 1.
i =0 Proof. We will prove this with mathematical induction.
(1) If n = 0, this statement is
0 i · i ! = (0 + 1)! − 1.
i =0 Since the left-hand side is 0 · 0! = 0, and the right-hand side is 1! − 1 = 0,
the equation 0=0 i · i ! = (0 + 1)! − 1 is obviously valid, as both sides are
i
zero. Mathematical Induction 146 (2) Consider any integer k ≥ 0. We must show that S k implies S k+1 . That
is, we must show that
k i · i ! = ( k + 1)! − 1
i =0 implies
k+1 i · i ! = (( k + 1) + 1)! − 1.
i =0 We use direct proof. Suppose
k i · i ! = ( k + 1)! − 1.
i =0 Observe that
k k+1 i · i! = i · i ! + ( k + 1)( k + 1)! i =0 i =0 ( k + 1)! − 1 + ( k + 1)( k + 1)! = = ( k + 1)! + ( k + 1)( k + 1)! − 1 1 + ( k + 1) ( k + 1)! − 1 = = ( k + 2)( k + 1)! − 1
= ( k + 2)! − 1
= (( k + 1) + 1)! − 1. Therefore k+1 i · i ! = (( k + 1) + 1)! − 1.
i =0
n i · i ! = ( n + 1)! − 1 for every integer n ≥ 0. It follows by induction that
i =0 The next example illustrates a trick that you may occasionally ﬁnd
useful. You know that you can add equal quantities to both sides of an
equation without violating equality. But don’t forget that you can add
unequal quantities to both sides of an inequality, as long as the quantity
you add to the bigger side is bigger than the quantity you add to the
smaller side. For example, if x ≤ y and a ≤ b, then x + a ≤ y + b. Similarly, if
x ≤ y and b is positive, then x ≤ y + b. This oft-forgotten fact is used in the
next proof. 147
Example 10.4 Prove the following proposition.
For each n ∈ N, it follows that 2n ≤ 2n+1 − 2n−1 − 1. Proposition Proof. We will prove this with mathematical induction.
(1) If n = 1, this statement is 21 ≤ 21+1 − 21−1 − 1, which simpliﬁes to
2 ≤ 4 − 1 − 1, which is obviously true.
(2) Now suppose k ≥ 1 and 2k ≤ 2k+1 − 2k−1 − 1. We need to show that
this implies 2k+1 ≤ 2(k+1)+1 − 2(k+1)−1 − 1. We use direct proof. Suppose
2k ≤ 2k+1 − 2k−1 − 1. Now reason as follows.
2k ≤ 2k+1 − 2k−1 − 1 2(2k ) ≤ 2(2k+1 − 2k−1 − 1)
2 k+1 2 k+1 2 k+1 2 k+1 ≤2 k+2 −2 −2 ≤2 k+2 − 2k − 2 + 1 ≤2 k+2 (multiply both sides by 2) k −2 −1 ( k+1)+1 ≤2 (add 1 to the bigger side) k − 2(k+1)−1 − 1 It follows by induction that 2n ≤ 2n+1 − 2n−1 − 1 for each n ∈ N.
In the next proposition we’ll prove that if n ∈ N, then (1 + x)n ≥ 1 + nx for
all x ∈ R with x > −1. Thus we will need to prove that the statement
S n : (1 + x)n ≥ 1 + nx for every x ∈ R with x > −1 is true for every natural number n. (For the record, the inequality (1 + x)n ≥
1 + nx is sometimes known as Bernoulli’s Inequality.)
Example 10.5
Proposition Prove the following proposition.
If n ∈ N, then (1 + x)n ≥ 1 + nx for all x ∈ R with x > −1. Proof. We will prove this with mathematical induction.
(1) For the basis step, notice that when n = 1 the statement is (1 + x)1 ≥
1 + 1 · x , and this is true because both sides equal 1 + x.
(2) Assume that for some k ≥ 1, the statement (1 + x)k ≥ 1 + kx is true for
all x ∈ R with x > −1. From this we need to prove (1 + x)k+1 ≥ 1 + ( k + 1) x.
Notice that 1 + x is positive because x > −1, so we can multiply both sides
of (1 + x)k ≥ 1 + kx by (1 + x) without changing the direction of the ≥. Doing
this gives: Mathematical Induction 148 (1 + x)k (1 + x) ≥ (1 + kx)(1 + x)
(1 + x)k+1 ≥ 1 + x + kx + kx2 (1 + x)k+1 ≥ 1 + ( k + 1) x + kx2 Notice that the term kx2 is positive, so removing it from the right-hand
side will only make that side even smaller. Thus we get (1 + x)k+1 ≥
1 + ( k + 1) x.
This completes the proof.
As you now know, induction is used to prove statements of the form
∀ n ∈ N, S n . But notice the outline does not work for statements of form
∀ n ∈ Z, S n (where n is in Z, not N). The reason is that if you are trying to
prove ∀ n ∈ Z, S n by induction, and you’ve shown S1 is true and S k ⇒ S k+1 ,
then it only follows from this that S n is true for n ≥ 1. You haven’t proved
that any of the statements S0 , S−1 , S−2 , . . . are true. If you ever want to
prove ∀ n ∈ Z, S n by induction, you have to show that some S a is true and
S k ⇒ S k+1 and S k ⇒ S k−1 . 10.1 Proof by Strong Induction
This section discusses a useful variation on induction. However, for you to
fully understand and appreciate it, it’s probably necessary that you ﬁrst
work some of the Section 10.3 exercises. If you haven’t done this yet, then
it’s worth your while to do it now.
Occasionally it happens in induction proofs that it is diﬃcult to show
that S k forces S k+1 to be true. Instead you may ﬁnd that you need to use
the fact that some “lower” statement S m (with m < k) forces S k+1 to be
true. For these situations you can use a slight variant of induction called
strong induction. Strong induction works just like regular induction,
except that in Step (2) instead of assuming S k is true and showing this
forces S k+1 to be true, we assume that all the statements S1 , S2 , . . . , S k are
true and show this forces S k+1 to be true. The idea is that if it always
happens that the ﬁrst k dominoes falling makes the ( k + 1) th domino fall,
then all the dominoes must fall. Here is the outline.
Outline for Proof by Strong Induction
Proposition The statements S1 , S2 , S3 , S4 , . . . are all true.
Proof. (Strong induction)
(1) Prove the ﬁrst statement S1 . (Or the ﬁrst several S n .)
(2) Given any integer k ≥ 1, prove (S1 ∧ S2 ∧ S3 ∧ · · · ∧ S k ) ⇒ S k+1 . Proof by Strong Induction 149 Strong induction can be useful in situations where assuming S k is true
does not neatly lend itself to forcing S k+1 to be true. You might be better
served by showing some other statement (S k−1 or S k−2 for instance) forces
S k to be true. Strong induction says you are allowed to use any (or all) of
the statements S1 , S2 , . . . , S k to prove S k+1 .
As our ﬁrst example of strong induction, we are going to prove that
12 |( n4 − n2 ) for any n ∈ N. But ﬁrst, let’s look at how regular induction
would be problematic. In regular induction we would start by showing
12 |( n4 − n2 ) is true for n = 1. This part is easy because it reduces to 12 | 0,
which is clearly true. Next we would assume that 12 |(k4 − k2 ) and try to
show this implies 12 |((k + 1)4 − ( k + 1)2 ). Now, 12 |( k4 − k2 ) means k4 − k2 = 12a
for some a ∈ Z. Next we use this to try to show (k + 1)4 − ( k + 1)2 = 12b for
some integer b. Working out ( k + 1)4 − ( k + 1)2 , we get
( k + 1)4 − ( k + 1)2 = ( k4 + 4 k3 + 6 k2 + 4 k + 1) − ( k2 + 2 k + 1)
= ( k4 − k2 ) + 4 k3 + 6 k2 + 6 k
= 12a + 4 k3 + 6 k2 + 6 k. At this point we’re stuck because we can’t factor out a 12. Now let’s see
how strong induction can get us out of this bind.
The idea is to show S k−5 ⇒ S k+1 instead of S k ⇒ S k+1 . Thus our basis
step involves checking that S1 , S2 , S3 , S4 , S5 , S6 are all true so that S k−5
doesn’t “run oﬀ the end.”
Example 10.6
Proposition Prove the following proposition.
If n ∈ N, then 12 |( n4 − n2 ). Proof. We will prove this with strong induction.
(1) First note that the statement is true for the ﬁrst six positive integers:
If n = 1, 12 divides n4 − n2 = 14 − 12 = 0.
If n = 2, 12 divides n4 − n2 = 24 − 22 = 12.
If n = 3, 12 divides n4 − n2 = 34 − 32 = 72.
If n = 4, 12 divides n4 − n2 = 44 − 42 = 240.
If n = 5, 12 divides n4 − n2 = 54 − 52 = 600.
If n = 6, 12 divides n4 − n2 = 64 − 62 = 1260.
(2) Let k ≥ 6 and assume 12 |(m4 − m2 ) for 1 ≤ m ≤ k. (That is, assume
statements S1 , S2 , . . . , S k are all true.) We must show 12 |((k + 1)4 − ( k + 1)2 ).
(That is, we must show that S k+1 is true.) Since S k−5 is true, we have
12 |(( k − 5)4 − ( k − 5)2 ). For simplicity, let’s set m = k − 5, so we know Mathematical Induction 150
12 |( m4 − m2 ), meaning
( k + 1)4 − ( k + 1)2 m4 − m2 = 12a for some integer a. Observe that: = ( m + 6)4 − ( m + 6)2
= m4 + 24 m3 + 216 m2 + 864 m + 1296 − ( m2 + 12 m + 36) = ( m4 − m2 ) + 24 m3 + 216 m2 + 852 m + 1260
= 12a + 24 m3 + 216 m2 + 852 m + 1260
= 12 a + 2 m3 + 18 m2 + 71 m + 105 . Since (a + 2m3 + 18 m2 + 71m + 105) is an integer, we get 12 |((k + 1)4 − (k + 1)2 ).
Therefore, by strong induction it follows that 12 |(n4 − n2 ) for every positive
integer n.
Our next example involves mathematical objects called graphs. In
mathematics, the word graph is used in two contexts. One context involves
the graphs of equations and functions that you studied in algebra and
calculus. In the other context, a graph is a geometrical conﬁguration
consisting of points (called vertices) and edges which are lines connecting
the vertices. Following are some pictures of graphs. Let’s agree that all of
our graphs will be in “one piece,” that is you can travel from any vertex of
a graph to any other vertex by traversing a route of edges from one vertex
to the other.
v0
v1 v4 v2 v3 Figure 10.1. Examples of Graphs
A cycle in a graph is a sequence of distinct vertices in the graph that
forms a route that ends where it began. For example, the graph on the
far left of Figure 10.1 has a cycle that starts at vertex v1 , then goes to v2 ,
then to v3 , then v4 , and ﬁnally back to its starting point v1 . You can ﬁnd
cycles in both of the graphs on the left, but the two graphs on the right do
not have cycles. There is a special name for a graph that has no cycles;
it is called a tree. Thus the two graphs on the right of Figure 10.1 are
trees, but the two graphs on the left are not trees. Proof by Strong Induction 151 Our next example concerns trees. Observe that the trees in Figure
10.1 both have one fewer edge than vertices. For example the tree on the
far right has 5 vertices and 4 edges. The one next to it has 6 vertices and
5 edges. Try drawing a picture of any tree; you will ﬁnd that if it has n
vertices, then it has n − 1 edges. We now prove that this is true for any
tree.
Example 10.7
Proposition Prove the following proposition.
If a tree has n vertices, then it has n − 1 edges. Proof. Notice that this theorem asserts that for any n ∈ N, the following
statement is true: S n : A tree with n vertices has n − 1 edges. We use strong
induction to prove this.
(1) For the basis step, observe that if a tree has n = 1 vertices then it
consists of just one vertex and no edges. Thus it has n − 1 = 0 edges, so
the theorem is true when n = 1. Also, although it is not necessary for
the proof, notice that if a tree has n = 2 vertices, then the tree must have
just one edge (joining the two vertices), so the tree has n − 1 = 1 edge,
and thus the theorem is true for a tree with n = 2 vertices.
(2) Now take an integer k ≥ 1. We must show (S1 ∧ S2 ∧ · · · ∧ S k ) ⇒ S k+1 .
In words, we must show that if it is true that any tree with m vertices
has m − 1 edges, where 1 ≤ m ≤ k, then any tree with k + 1 vertices has
( k + 1) − 1 = k edges. We will use direct proof.
Suppose that for each integer m with 1 ≤ m ≤ k, any tree with m vertices
has m − 1 edges. Now let T be a tree with k + 1 vertices. Single out an
edge of T and label it e, as illustrated below.
···
··· T e
···
··· ··· T1
··· T2
···
··· Now remove the edge e from T , but leave the two endpoints of e. This
forms two smaller trees which we call T1 and T2 . Let’s say T1 has a
vertices and T2 has b vertices. Now, since each of these two smaller
trees has fewer than k + 1 vertices, our inductive hypothesis guarantees
that T1 has a − 1 edges, and T2 has b − 1 edges. Now think about our
original tree T . It has a + b vertices. Think about its number of edges. It
has a − 1 edges that belong to T1 and b − 1 edges that belong to T2 , plus Mathematical Induction 152 it has the additional edge e that belongs to neither T1 nor T2 . Thus, all
together, the number of edges that T has is (a 1) + ( b − 1) + 1 = (a + b) − 1.
In other words, T has one fewer edges than it has vertices. Thus it has
( k + 1) − 1 = k edges.
Thus it follows by strong induction that any tree with n vertices must have
n − 1 edges.
Notice that it was absolutely essential that we used strong induction
in the above proof because the two trees T1 and T2 will not both have k
vertices. At least one will have fewer than k vertices. Thus the statement
S k is not enough to imply S k+1 . We need to use the assumption that S m
will be true whenever m ≤ k, and strong induction allows us to do this.
10.2 Proof by Smallest Counterexample
This section introduces yet another proof technique, called proof by smallest counterexample. It is a hybrid of induction and proof by contradiction.
It has the nice feature that it leads you straight to a contradiction, and
is therefore more “automatic” than the proof by contradiction that was
introduced in Chapter 6. Here is the outline.
Outline for Proof by Smallest Counterexample
Proposition The statements S1 , S2 , S3 , S4 , . . . are all true.
Proof. (Smallest counterexample)
(1) Check that the ﬁrst statement S1 is true.
(2) For the sake of contradiction, suppose not every S n is true.
(3) Let k > 1 be the smallest integer for which S k is false.
(4) Then S k−1 is true and S k is false. Use this to get a contradiction.
Notice that this set-up leads you to a point where S k+1 is true and
S k is false. It is here, where true and false collide, that you will ﬁnd a
contradiction. Let’s look at an example.
Example 10.8
Proposition Prove the following proposition.
If n ∈ N, then 4 |(5n − 1). Proof. What follows is a proof by smallest counterexample. (We will
number the steps to match the outline, but that is not usually done in
practice.)
(1) If n = 1, then the statement is 4 |(51 − 1), or 4 | 4, which is true.
(2) For sake of contradiction, suppose it’s not true that 4 |(5n − 1) for all n. Fibonacci Numbers 153 (3) Let k > 1 be the smallest integer for which 4 | (5k − 1).
(4) Then 4 |(5k−1 − 1), so there is an integer a for which 5k−1 − 1 = 4a. Then:
5k−1 − 1 = 4a
5(5k−1 − 1) = 5 · 4a
5k − 5 = 20a
5k − 1 = 20a + 4
5k − 1 = 4(5a + 1) This means 4 |(5k − 1), a contradiction, because 4 | (5k − 1) in Step 3. Thus
we were wrong in Step 2 to assume that it is untrue that 4 |(5n − 1) for
every n. Therefore 4 |(5n − 1) is true for every n.
One word of warning about proof by smallest counterexample. When
you read proofs in other textbooks or in mathematical papers, it often
happens that the writer doesn’t tell you up front that proof by smallest
counterexample is being used. Instead, you will have to read through the
proof to glean from context that this technique is being used. In fact, the
same warning applies to all of our proof techniques. If you continue with
mathematics, you will gradually gain through experience the ability to
analyze a proof and understand exactly what approach is being used when
it is not stated explicitly. Frustrations await you, but do not be discouraged
by them. Frustration is a natural part of anything that’s worth doing.
10.3 Fibonacci Numbers
Leonardo Pisano, now known as Fibonacci, was an Italian mathematician
born around 1175. His most signiﬁcant work was a book Liber Abaci,
which is recognized as a catalyst in medieval Europe’s slow transition
from Roman numbers to the Hindu-Arabic number system. But he is
best known today for a number sequence which he described in his book
and which bears his name (although he did not discover it himself). The
Fibonacci Sequence is
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, . . . The numbers that appear in this sequence are called Fibonacci numbers.
The ﬁrst two terms in the sequence are 1 and 1, and thereafter any term
is the sum of the previous two terms. For example 8 = 3 + 5, and 13 = 5 + 8,
etc. We denote the n th term of this sequence as F n . Thus F1 = 1, F2 = 1, F3 = Mathematical Induction 154 2, F4 = 3, F7 = 13, and so on. Notice that the Fibonacci Sequence is entirely
determined by the rules F1 = 1, F2 = 1, and F n = F n−1 + F n−2 . We introduce Fibonacci’s sequence here partly because it is something
that everyone should know about, but also because it is a great source of
induction problems. This sequence, which appears with surprising frequency in nature, is ﬁlled with mysterious patterns and hidden structures.
Some of these structures will be revealed to you in the examples and
exercises.
We emphasize that the condition F n = F n−1 + F n−2 (or equivalently F n+1 =
F n + F n−1 ) is the perfect set-up for induction. It suggests that we can
determine something about F n by looking at earlier terms of the sequence.
In using induction to prove something about the Fibonacci Sequence, you
should expect to use the equation F n = F n−1 + F n−2 somewhere.
2
2
For our ﬁrst example we will prove that F n+1 − F n+1 F n − F n = (−1)n for
2
2
any natural number n. For example, if n = 5 we have F6 − F6 F5 − F5 =
82 − 8 · 5 − 52 = 64 − 40 − 25 = −1 = (−1)5 .
Example 10.9
Proposition Prove the following proposition.
2
2
The Fibonacci sequence obeys F n+1 − F n+1 F n − F n = (−1)n . Proof. We will prove this with mathematical induction.
2
2
2
2
(1) If n = 1 we have F n+1 − F n+1 F n − F n = F2 − F2 F1 − F1 = 12 − 1 · 1 − 12 = −1 =
2
2
(−1)1 = (−1)n , so indeed F n+1 − F n+1 F n − F n = (−1)n is true when n = 1.
2
2
(2) Take any integer k ≥ 1. We must show that if F k+1 − F k+1 F k − F k = (−1)k ,
2
2
then F k+2 − F k+2 F k+1 − F k+1 = (−1)k+1 . We use direct proof. Suppose
2
2
F k+1 − F k+1 F k − F k = (−1)k . Now we are going to carefully work out the
2
2
expression F k+2 − F k+2 F k+1 − F k+1 and show that it really does equal
(−1)k+1 . In doing this we will use the fact that F k+2 = F k+1 + F k .
2
2
F k+2 − F k+2 F k+1 − F k+1 2
= (F k+1 + F k )2 − (F k+1 + F k )F k+1 − F k+1 = 2
2
2
2
F k+1 + 2F k+1 F k + F k − F k+1 − F k F k+1 − F k+1 2
2
= −F k+1 + F k+1 F k + F k
2
2
= −(F k+1 − F k+1 F k − F k ) = −(−1)k
1 = (−1) (−1) (inductive hypothesis)
k = (−1)k+1
2
2
Therefore F k+2 − F k+2 F k+1 − F k+1 = (−1)k+1 .
2
2
It follows by induction that F n+1 − F n+1 F n − F n = (−1)n for every n ∈ N. Fibonacci Numbers 155 Let’s pause for a moment and think about what the result we just
2
2
2
proved means. Dividing both sides of F n+1 − F n+1 F n − F n = (−1)n by F n gives
F n+1
Fn 2 − (−1)n
F n+1
.
−1 =
2
Fn
Fn For large values of n the right-hand side is very close to zero, and the
left-hand side is F n+1 /F n plugged into the polynomial x2 − x − 1. Thus, as
n increases, the ratio F n+1 /F n approaches a root of x2 − x − 1 = 0. By the
quadratic formula, the roots of x2 − x − 1 are 1±2 5 . As F n+1 /F n > 1 and
1− 5
2 < 1, this ratio must be approaching the root lim n→∞ 1+ 5
2. Therefore F n+1 1 + 5
=
.
Fn
2 (10.1) (For a quick spot check, note that F13 /F12 ≈ 1.618025, while 1+2 5 ≈ 1.618033,
thus even for the relatively small value n = 12 the two numbers match
to four decimal places.) The number Φ = 1+2 5 is sometimes called the
Golden Ratio, and there has been much speculation about its occurrence
in nature as well as in classical art and architecture. One theory holds
that the Parthenon and the Great Pyramids of Egypt were designed in
accordance with this number.
But we are here concerned with things that can be proved. We close
by observing how the Fibonacci Sequence in many ways resembles a
geometric sequence. Recall that a geometric sequence with ﬁrst term a
and common ratio r has the form
a, ar , ar 2 , ar 3 , ar 4 , ar 5 , ar 6 , ar 7 , ar 8 , . . . where any term is obtained by multiplying the previous term by r . In
general its n th term is G n = ar n , and G n+1 /G n = r . Equation (10.1) tells
us that F n+1 /F n ≈ Φ. Thus even though it is not a geometric sequence,
the Fibonacci Sequence tends to behave like a geometric sequence with
common ratio Φ, and the further “out” you go, the higher the resemblance.
Exercises for Chapter 10
Prove the following statements with either induction, strong induction or proof
by smallest counterexample.
1. For every integer n ∈ N, it follows that 1 + 2 + 3 + 4 + · · · + n = n2 + n
.
2 Mathematical Induction 156 n( n + 1)(2 n + 1)
.
6
n2 ( n + 1)2
3. For every integer n ∈ N, it follows that 13 + 23 + 33 + 43 + · · · + n3 =
.
4
n( n + 1)( n + 2)
4. If n ∈ N, then 1 · 2 + 2 · 3 + 3 · 4 + 4 · 5 + · · · + n(n + 1) =
.
3
5. If n ∈ N, then 21 + 22 + 23 + · · · + 2n = 2n+1 − 2. 2. For every integer n ∈ N, it follows that 12 + 22 + 32 + 42 + · · · + n2 = n 6. For every natural number n, it follows that (8 i − 5) = 4 n2 − n. i =1 7. If n ∈ N, then 1 · 3 + 2 · 4 + 3 · 5 + 4 · 6 + · · · + n(n + 2) = n( n + 1)(2 n + 7)
.
6 1
2
3
n
1
+ + +···+
= 1−
2! 3! 4!
( n + 1)!
( n + 1)!
9. For any integer n ≥ 0, it follows that 24 |(52n − 1). 8. If n ∈ N, then 10. For any integer n ≥ 0, it follows that 3 |(52n − 1).
11. For any integer n ≥ 0, it follows that 3 |(n3 + 5n + 6).
12. For any integer n ≥ 0, it follows that 9 |(43n + 8).
13. For any integer n ≥ 0, it follows that 6 |(n3 − n).
14. Suppose a ∈ Z. Prove that 5 | 2n a implies 5 | a for any n ∈ N.
1
1
1
1
1
1
+
+
+
+···+
= 1−
.
1·2 2·3 3·4 4·5
n( n + 1)
n+1
16. For every natural number n, it follows that 2n + 1 ≤ 3n . 15. If n ∈ N, then 17. Suppose A 1 , A 2 , . . . A n are sets in some universal set U , and n ≥ 2. Prove that
A1 ∩ A2 ∩ · · · ∩ A n = A1 ∪ A2 ∪ · · · ∪ A n .
18. Suppose A 1 , A 2 , . . . A n are sets in some universal set U , and n ≥ 2. Prove that
A1 ∪ A2 ∪ · · · ∪ A n = A1 ∩ A2 ∩ · · · ∩ A n .
1
111
1
+ + +···+ 2 ≤ 2− .
149
n
n
20. Prove that (1 + 2 + 3 + · · · + n)2 = 13 + 23 + 33 + · · · + n3 for every n ∈ N.
1
1
n
11111
+
≥ 1+ .
21. If n ∈ N, then + + + + + · · · + n
12345
2 − 1 2n
2
(Note: This problem asserts that the sum of the ﬁrst n2 terms of the harmonic
series is at least 1 + n/2. It thus implies that the harmonic series diverges.)
1
1
1
1
1
1
1
1−
1−
1−
· · · 1 − n ≥ + n+1 .
22. If n ∈ N, then 1 −
2
4
8
16
2
42 19. Prove that 23. Use mathematical induction to prove the Binomial Theorem (Theorem 3.1).
You may ﬁnd that you need Equation (3.2).
n 24. Prove that k n
k = n2n−1 for each natural number n. k=1 25. Concerning the Fibonacci Sequence, prove that F1 + F2 + F3 + F4 + . . . + F n = F n+2 − 1.
n 26. Concerning the Fibonacci Sequence, prove that
k=1 2
F k = F n F n+1 . Fibonacci Numbers 157 27. Concerning the Fibonacci Sequence, prove that F1 + F3 + F5 + F7 + . . . + F2n−1 = F2n .
28. Concerning the Fibonacci Sequence, prove that F2 + F4 + F6 + F8 + . . . + F2n =
F2n+1 − 1.
29. In this problem n ∈ N and F n is the n th Fibonacci number. Prove that
n
0 (For example, 6
0 30. Here F n is the n 5
1 +
th +
+ n−1
1
4
2 + + n−2
2 3
3 + 2
4 +
+ n−3
3
1
5 0
n +···+ + 0
6 = F n+1 . = 1+5+6+1+0+0+0 = 13 = F6+1 .) Fibonacci number. Prove that
Fn = 1+ 5
2 n − 1− 5
2 n 5 n
k
r 31. Prove that
k=0 = n+1
r +1 , where 1 ≤ r ≤ n. 32. Prove that the number of n-digit binary numbers that have no consecutive
1’s is the Fibonacci number F n+2 . For example, for n = 2 there are three such
numbers (00, 01, and 10), and 3 = F2+2 = F4 . Also for n = 3 there are ﬁve such
numbers (000, 001, 010, 100, 101), and 5 = F3+2 = F5 .
33. Suppose n (inﬁnitely long) straight lines lie on a plane in such a way that no
two of the lines are parallel, and no three of the lines intersect at a single
2
point. Show that this arrangement divides the plane into n +2n+2 regions.
34. Prove that 31 + 32 + 33 + 34 + · · · + 3n = 3n+1 − 3
for every n ∈ N.
2 35. If n, k ∈ N, and n is even and k is odd, then
k n
k is even. 36. If n = 2 − 1 for some k ∈ N, then every entry in the nth row of Pascal’s Triangle
is odd. Part IV
Relations, Functions and
Cardinality CHAPTER 11 Relations I n mathematics there are endless ways that two entities can be related
to each other. Consider the following mathematical statements. 5 < 10
a ≡ b ( mod n) 5≤5
6∈Z 6 = 30
5
X ⊆Y 5 | 80
π ≈ 3.14 7>4
0 ≥ −1 x= y
2∉Z 83
Z⊆N In each case two entities appear on either side of a symbol, and we
interpret the symbol as expressing some relationship between the two
entities. Symbols such as <, ≤, =, |, |, ≥, >, ∈ and ⊂, etc., are called relations
because they convey relationships among things.
Relations are signiﬁcant. In fact, you would have to admit that there
would be precious little left of mathematics if we took away all the relations.
Therefore it is important to have a ﬁrm understanding of relations, and
this chapter is intended to develop that understanding.
Rather than focusing on each relation individually (an impossible task
anyway since there are inﬁnitely many diﬀerent relations) we will develop a
general theory that encompasses all relations. Understanding this general
theory will give us the conceptual framework and language needed to
understand and discuss any speciﬁc relation.
Before stating the theoretical deﬁnition of a relation, let’s look at a
motivational example. This example will lead us naturally to our deﬁnition.
Consider the set A = 1, 2, 3, 4, 5 . (There’s nothing special about this
particular set; any set of numbers would do for this example.) Elements of
A can be compared to each other by the symbol “<.” For example, 1 < 4,
2 < 3, 2 < 4, and so on. You have no trouble understanding this because the
notion of numeric order is so ingrained. But imagine you had to explain
it to an idiot savant, one with an obsession for detail but absolutely no
understanding of the meaning of (or relationships between) integers. You
might consider writing down for your student the following set:
R = (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5) Relations 162 The set R encodes the meaning of the < relation for elements in A . An
ordered pair (a, b) appears in the set if and only if a < b. If asked whether
or not it is true that 3 < 4, your student could look through R until he
found the ordered pair (3, 4); then he would know 3 < 4 is true. If asked
about 5 < 2, he would see that (5, 2) does not appear in R , so 5 < 2. The set
R , which is a subset of A × A , completely describes the relation < for A .
Though it may seem simple-minded at ﬁrst, this is exactly the idea
we will use for our main deﬁnition. This deﬁnition is general enough to
describe not just the relation < for the set A = 1, 2, 3, 4, 5 , but any relation
for any set A .
Deﬁnition 11.1 A relation on a set A is a subset R ⊆ A × A . We often
abbreviate the statement ( x, y) ∈ R as xR y. The statement ( x, y) ∉ R is
abbreviated as x R y.
Notice that a relation is a set, so we can use what we know about sets
to understand and explore relations. But before getting deeper into the
theory of relations, let’s look at some examples of Deﬁnition 11.1.
Example 11.1 Let A = 1, 2, 3, 4 , and consider the following set: R = (1, 1), (2, 1), (2, 2), (3, 3), (3, 2), (3, 1), (4, 4), (4, 3), (4, 2), (4, 1) ⊆ A × A The set R is a relation on A , by Deﬁnition 11.1. Since (1, 1) ∈ R , we have
1R 1. Similarly 2R 1 and 2R 2, and so on. However notice that (for example)
(3, 4) ∉ R , so 3 R 4. Observe that R is the familiar relation ≥ for the set A .
Chapter 1 proclaimed that all of mathematics can be described with
sets. Just look at how successful this program has been! The greater-than
relation is now a set R . (We might even express this in the rather cryptic
form ≥= R .)
Example 11.2 Let A = 1, 2, 3, 4 , and consider the following set:
S = (1, 1), (1, 3), (3, 1), (3, 3), (2, 2), (2, 4), (4, 2), (4, 4) ⊆ A × A Here we have 1S 1, 1S 3, 4S 2, etc., but 3 S 4 and 2 S 1. What does S mean?
Think of it as meaning “has the same parity as.” Thus 1S 1 reads “1 has
the same parity as 1,” and 4S 2 reads 4 has the same parity as 2.” 163
Example 11.3 Let B = 0, 1, 2, 3, 4, 5 , and consider the following set:
U = (1, 3), (3, 3), (5, 2), (2, 5), (4, 2) ⊆ B × B Then U is a relation on B because U ⊆ B × B. You may be hard-pressed
to invent any “meaning” for this particular relation. A relation does not
have to have any meaning. Any random subset of B × B is a relation on B,
whether or not it describes anything familiar.
Some relations can be described with pictures. For example, we can
depict the above relation U on B by drawing points labeled by elements of
B. The statement ( x, y) ∈ U is then represented by an arrow pointing from
x to y, a graphic symbol meaning “x relates to y.” Here’s a picture of U :
0 1 2 3 4 5 The next picture illustrates the relation R on the set A = a, b, c, d , where
xR y means x comes before y in the alphabet. According to Deﬁnition 11.1,
as a set this relation is R = (a, b), (a, c), (a, d ), ( b, c), (b, d ), ( c, d ) . You may
feel that the picture conveys the relation better than the set does. They
are two diﬀerent ways of expressing the same thing. In some instances
pictures are more convenient than sets for discussing relations.
a d b c Although pictures can help us visualize relations, they do have their
limitations. If A and R were inﬁnite, then the diagram would be impossible
to draw, but the set R might be easily expressed in set-builder notation.
For instance here is a relation that is too big to be described by a picture.
Example 11.4 Consider the set R = ( x, y) ∈ Z × Z : x − y ∈ N ⊆ Z × Z. This
is the > relation on the set A = Z. It is inﬁnite because there are inﬁnitely
many ways to have x > y where x and y are integers.
Example 11.5 The set R = ( x, x) : x ∈ R ⊆ R × R is the relation = on the
set R, because xR y means the same thing as x = y. Thus R is a set that
expresses the notion of equality of real numbers. Relations 164 Exercises for Section 11.0
1. Let A = 0, 1, 2, 3, 4, 5 . Write out the relation R that expresses > on A . Then
illustrate it with a diagram.
2. Let A = 1, 2, 3, 4, 5, 6 . Write out the relation R that expresses | (divides) on A .
Then illustrate it with a diagram.
3. Let A = 0, 1, 2, 3, 4, 5 . Write out the relation R that expresses ≥ on A . Then
illustrate it with a diagram.
4. The following diagram represents a relation R on a set A . Write the sets A
and R .
0 1 2 3 4 5 5. The following diagram represents a relation R on a set A . Write the sets A
and R .
0 1 2 3 4 5 6. Congruence modulo 5 is a relation on the set A = Z. In this relation xR y means
x ≡ y (mod 5). Write out the set R in set-builder notation.
7. Write the relation < on the set A = Z as a subset R of Z × Z. This is an inﬁnite
set, so you will have to use set-builder notation.
8. Let A = 1, 2, 3, 4, 5, 6 . Observe that
a diagram for this relation. ⊆ A × A , so R = is a relation on A . Draw 9. Let A = 1, 2, 3, 4, 5, 6 . How many diﬀerent relations are there on the set A ? 11.1 Properties of Relations
Some relations have special properties that other relations don’t have. For
example, the relation ≤ on Z has the property that x ≤ x for every x ∈ Z,
but the relation < on Z does not have this property, for x < x is never true.
The next deﬁnition lays out three particularly signiﬁcant properties that
relations may have. Properties of Relations 165 Deﬁnition 11.2 Suppose R is a relation on a set A .
1. Relation R is reﬂexive if xRx for every x ∈ A .
(That is, R is reﬂexive if ∀ x ∈ A , xRx.)
2. Relation R is symmetric if xR y implies yRx for all x, y ∈ A
(That is, R is symmetric if ∀ x, y ∈ A , xR y ⇒ yRx.)
3. Relation R is transitive if whenever xR y and yR z, then also xR z.
(That is, R is transitive if ∀ x, y, z ∈ A , ( xR y) ∧ ( yR z) ⇒ xR z. )
To illustrate this, let’s consider the set A = Z. Examples of reﬂexive
relations on Z include ≤, =, and |, for x ≤ x, x = x and x | x are all true for
any x ∈ Z. On the other hand, >, <, = and are not reﬂexive for none of
the statements x < x, x > x, x = x and x x is true.
The relation = is symmetric, for if x = y, then surely y = x also. Also,
the relation = is symmetric because x = y always implies y = x.
The relation ≤ is not symmetric, as x ≤ y does not necessarily imply
y ≤ x. For instance 5 ≤ 6 is true but 6 ≤ 5 is false. Notice that ( x ≤ y) ⇒ ( y ≤ x)
is true for some x and y (for example, it is true when x = 2 and y = 2) but
still ≤ is not symmetric because it is not the case that ( x ≤ y) ⇒ ( y ≤ x) is
true for all integers x and y.
The relation ≤ is transitive because whenever x ≤ y and y ≤ z, it also
is true that x ≤ z. Likewise <, ≥, > and = are all transitive. Examine the
following table and be sure you understand why it is labeled as it is.
Relations on Z:
Reﬂexive
Symmetric
Transitive < ≤ = | no
no
yes yes
no
yes yes
yes
yes yes
no
yes = no
no
no no
yes
no Example 11.6 Here A = b, c, d , e and R is the following relation on A :
R = ( b, b), ( b, c), ( c, b), ( c, c), ( d , d ), ( b, d ), ( d , b), ( c, d ), ( d , c) . Relation R is not reﬂexive, for although bRb, cR c and dRd , it is not true
that eR e. For a relation to be reﬂexive, xRx must be true for all x ∈ A .
The relation R is symmetric, because whenever we have xR y, it follows
that yRx too. Observe that bR c and cRb; bRd and dRb; dR c and cRd . If
we took away the ordered pair ( c, b) from R , then R would no longer be
symmetric.
The relation R is transitive, but it takes some work to check it. We
must check that the statement ( xR y ∧ yR z) ⇒ xR z is true for all x, y, z ∈ A .
In other words, we must check that whenever xR y and yR z, then also xR z. Relations 166 Notice that bR c and cRd and also bRd , so the statement ( bR c ∧ cRd ) ⇒
bRd is true. Likewise, bRd , dR c and also bR c, so ( bRd ∧ dR c) ⇒ bR c is
true, and so on. Moreover, note that ( bR c ∧ cRb) ⇒ bRb ﬁts the pattern
( xR y ∧ yR z) ⇒ xR z, where x = b, y = c and z = b; and ( bR c ∧ cRb) ⇒ bRb
is true because ( bR c ∧ cRb) and bRb are both true. We emphasize that
for R to be transitive, it is necessary that ( xR y ∧ yR z) ⇒ xR z is true for
all choices of x, y, z from A . Even if we took x = b, y = e and z = c, then
( bR e ∧ eR c) is false and bR c is true, but the statement ( bR e ∧ eR c) ⇒ bR c
is true. It’s not much fun, but going through all the combinations, you can
verify that ( xR y ∧ yR z) ⇒ xR z is true for all choices x, y, z ∈ A . (You should
try at least a few of them.)
The relation R from Example 11.6 has a meaning. You can think of
xR y as meaning that x and y are both consonants. Thus bR c because b
and c are both consonants; but b R e because it’s not true that b and e are
both consonants. Once we look at it this way, it’s immediately clear that R
has to be transitive. If x and y are both consonants and y and z are both
consonants, then surely x and z are both consonants. This illustrates a
point that we will see again later in this section: Knowing the meaning of
a relation can help us understand it and prove things about it.
Here is a picture of the relation from Example 11.6. Notice that we
can immediately spot several properties of R that may not have been so
clear from its set description. For instance, we see that R is not reﬂexive
because it lacks a loop at e, hence e R e.
d e b c Figure 11.1. The relation R from Example 11.6 In what follows, we summarize how to spot the various properties of a
relation from its diagram. Compare these with Figure 11.1. Properties of Relations 1. 2. 167 A relation is
reﬂexive if
for each point x ...
A relation is
symmetric if
whenever there is an
arrow from x to y ...
A relation is
transitive if
whenever there are
arrows from x to y
and y to z ... ...there is a
loop at x. x y x y x z ...there is also
an arrow from
y back to x: ...there is also
an arrow from
x to z: x y x y x z 3.
(This also means that
whenever there is an
arrow from x to y
and from y to x ... y ...there is also
a loop from
x back to x.) x y x Although these visual aids can be illuminating, their use is limited because many relations are too large and complex to be adequately described
as diagrams. For example, it would be impossible to draw a diagram
for the relation ≡ (mod n), where n ∈ N. Such a relation would best be
explained in a more theoretical (and less visual) way.
In the next example we prove that ≡ (mod n) is reﬂexive, symmetric
and transitive. Obviously we will not glean this from a drawing. Instead
we will prove it from the properties of ≡ (mod n) and Deﬁnition 11.2. Pay
special attention to this example. It illustrates how to prove things about
relations.
Example 11.7 Prove the following proposition.
Proposition Let n ∈ N. The relation ≡ (mod n) on the set Z is reﬂexive,
symmetric and transitive.
Proof. First we will show that ≡ (mod n) is reﬂexive. Take any integer
x ∈ Z and observe that n | 0, so n |( x − x). By deﬁnition of congruence modulo
n, we have x ≡ x (mod n). This shows x ≡ x (mod n) for every x ∈ Z, so ≡
(mod n) is reﬂexive. Relations 168 Next, we will show that ≡ (mod n) is symmetric. For this, we must
show that for all x, y ∈ Z, the condition x ≡ y (mod n) implies that y ≡ x
(mod n). We will use direct proof. Suppose x ≡ y (mod n). Thus n |( x − y)
by deﬁnition of congruence modulo n. Then x − y = na for some a ∈ Z by
deﬁnition of divisibility. Multiplying both sides by −1 gives y − x = n(−a).
Therefore n |( y − x), and this means y ≡ x (mod n). We’ve shown that x ≡ y
(mod n) implies that y ≡ x (mod n), and this means ≡ (mod n) is symmetric.
Finally we will show that ≡ (mod n) is transitive. For this we must
show that if x ≡ y (mod n) and y ≡ z (mod n), then x ≡ z (mod n). Again
we use direct proof. Suppose x ≡ y (mod n) and y ≡ z (mod n). This
means n |( x − y) and n |( y − z). Therefore there are integers a and b for
which x − y = na and y − z = nb. Adding these two equations, we obtain
x − z = na + nb. Consequently, x − z = n(a + b), so n |( x − z), hence x ≡ z (mod n).
This completes the proof that ≡ (mod n) is transitive.
The past three paragraphs have shown that ≡ (mod n) is reﬂexive,
symmetric and transitive, so the proof is complete.
As you continue your mathematical education, you will ﬁnd that the
reﬂexive, symmetric and transitive properties take on special signiﬁcance
in a variety of settings. In preparation for this, the next section explores
further consequences of these relations. But ﬁrst you should work some of
the following exercises.
Exercises for Section 11.1
1. Consider the relation R = (a, a), (b, b), ( c, c), ( d , d ), (a, b), (b, a) on set A = a, b, c, d .
Say whether R is reﬂexive, symmetric and transitive. If a property does not
hold, say why.
2. Consider the relation R = (a, b), (a, c), ( c, c), (b, b), ( c, b), (b, c) on the set A = a, b, c .
Say whether R is reﬂexive, symmetric and transitive. If a property does not
hold, say why.
3. Consider the relation R = (a, b), (a, c), ( c, b), (b, c) on the set A = a, b, c . Say
whether R is reﬂexive, symmetric and transitive. If a property does not hold,
say why.
4. Let A = a, b, c, d . Suppose R is the relation
R = (a, a), ( b, b), ( c, c), ( d , d ), (a, b), ( b, a), (a, c), ( c, a),
(a, d ), ( d , a), ( b, c), ( c, b), ( b, d ), ( d , b), ( c, d ), ( d , c) . Say whether R is reﬂexive, symmetric and transitive. If a property does not
hold, say why. Equivalence Relations 169 5. Consider the relation R = (0, 0), ( 2, 0), (0, 2), ( 2, 2) on R. Say whether R is
reﬂexive, symmetric and transitive. If a property does not hold, say why.
6. Consider the relation R = ( x, x) : x ∈ Z on Z. Say whether R is reﬂexive,
symmetric and transitive. If a property does not hold, say why. What familiar
relation is this?
7. There are 16 possible diﬀerent relations R on the set A = a, b . Describe all of
them. (A picture for each one will suﬃce, but don’t forget to label the nodes.)
8. Deﬁne a relation on Z as xR y if and only if | x − y | < 1. Say whether R is reﬂexive,
symmetric and transitive. If a property does not hold, say why. What familiar
relation is this?
9. Deﬁne a relation on Z by declaring xR y if and only if x and y have the same
parity. Say whether R is reﬂexive, symmetric and transitive. If a property does
not hold, say why. What familiar relation is this?
10. Suppose A = . Since ⊆ A × A , the set R = is a relation on A . Say whether
R is reﬂexive, symmetric and transitive. If a property does not hold, say why.
11. Suppose A = a, b, c, d and R = (a, a), (b, b), ( c, c), (d , d ) . Say whether R is reﬂexive, symmetric and transitive. If a property does not hold, say why.
12. Prove that the relation | (divides) on the set Z is reﬂexive and transitive. (Use
Example 11.7 as a guide if you are unsure of how to proceed.)
13. Consider the relation R = ( x, y) ∈ R × R : x − y ∈ Z on R. Prove that this relation
is reﬂexive, symmetric and transitive.
14. Suppose R is a symmetric and transitive relation on a set A , and there is an
element a ∈ A for which aRx for every x ∈ A . Prove that R is reﬂexive.
15. Prove or disprove: If a relation is symmetric and transitive, then it is also
reﬂexive.
16. Deﬁne a relation R on Z by declaring that xR y if and only if x2 ≡ y2 (mod 4).
Prove that R is reﬂexive, symmetric and transitive. 11.2 Equivalence Relations
The relation = on the set Z (or on any set A ) is reﬂexive, symmetric and
transitive. There are many other relations that are also reﬂexive, symmetric and transitive. Relations which have all three of these properties occur
very frequently in mathematics and often play quite signiﬁcant roles. (For
instance, this is certainly true of the relation =.) Such relations are given
a special name. They are called equivalence relations.
Deﬁnition 11.3 A relation R on a set A is an equivalence relation if
it is reﬂexive, symmetric and transitive. Relations 170 As an example, Figure 11.2 illustrates four diﬀerent equivalence relations R1 , R2 , R3 and R4 on the set A = − 1, 1, 2, 3, 4 . Each one has its
own meaning, as labeled. For example, in the second row the relation R2
literally means “has the same parity as.” So 1R2 3 means “1 has the same
parity as 3,” etc. Relation R “is equal to” (=) Diagram −1 R 4 = (−1, −1), (1, 1), (2, 2), (3, 3), (4, 4),
(1, 3), (3, 1), (2, 4), (4, 2) 1 2 4 1 2 3 4 1 2 3 −1 , 1 , 4 3 −1 4 − 1, 1, 3 , 2, 4 −1 R 3 = (−1, −1), (1, 1), (2, 2), (3, 3), (4, 4),
(1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1),
(2, 3), (3, 2), (2, 4), (4, 2), (1, 3), (3, 1) “has same parity and sign as” 2 2, 3, 4 R 2 = (−1, −1), (1, 1), (2, 2), (3, 3), (4, 4),
(−1, 1), (1, −1), (−1, 3), (3, −1),
(1, 3), (3, 1), (2, 4), (4, 2) “has same sign as” 1 3 R 1 = (−1, −1), (1, 1), (2, 2), (3, 3), (4, 4) “has same parity as” Equivalence
classes (see
next page) −1 , 1, 2, 3, 4 −1 − 1 , 1, 3 , 2, 4 Figure 11.2. Examples of equivalence relations on the set A = − 1, 1, 2, 3, 4 The diagrams in Figure 11.2 make it easy to check that each relation
is reﬂexive, symmetric and transitive, i.e. that each is an equivalence
relation. As you can see from these examples, equivalence relations on a
set tend to express some measure of “sameness” among the elements of
the set, whether it is true equality or something weaker (like having the
same parity). Equivalence Relations 171 It’s time to introduce an important deﬁnition. Whenever you have
an equivalence relation R on a set A , it divides A into subsets called
equivalence classes. Here is the deﬁnition.
Deﬁnition 11.4 Suppose R is an equivalence relation on a set A . Given
any element a ∈ A , the equivalence class containing a is the subset
x ∈ A : xRa of A consisting of all the elements of A that relate to a. This
set is denoted as [a]. Thus the equivalence class containing a is the set
[a] = x ∈ A : xRa .
Example 11.8 Consider the relation R1 in Figure 11.2. The equivalence
class containing 2 is the set [2] = x ∈ A : xR 1 2 . Since in this relation 2
relates to itself and nothing else, we have [2] = 2 . Other equivalence
classes for R1 are [−1] = − 1 , [1] = 1 , [3] = 3 and [4] = 4 . Thus this
relation has ﬁve separate equivalence classes.
Example 11.9 Consider the relation R2 in Figure 11.2. The equivalence
class containing 2 is the set [2] = x ∈ A : xR 2 2 . Since 2 relates only to itself
and 4, we have [2] = 2, 4 . Observe that we also have [4] = x ∈ A : xR 2 4 =
2, 4 , so [2] = [4]. Another equivalence class for R 2 is [1] = x ∈ A : xR 2 1
= − 1, 1, 3 . In addition, note that [1] = [−1] = [3] = − 1, 1, 3 . Thus this
relation has just two equivalence classes.
Example 11.10 The relation R4 in Figure 11.2 has three equivalence
classes. They are [−1] = − 1 , and [1] = [3] = 1, 3 , and [2] = [4] = 2, 4 .
Don’t be misled by Figure 11.2. It’s important to realize that not
every equivalence relation can be drawn as a diagram involving nodes
and arrows. Even the simple relation R = ( x, x) : x ∈ R which expresses
equality in the set R is too big to be drawn. Its picture would involve a
point for every real number, and a loop at each point. Clearly that’s too
many points and loops to draw.
We close this section with several other examples of equivalence relations on inﬁnite sets.
Example 11.11 Let P be the set of all polynomials. Deﬁne a relation R on
P as follows. Given two polynomials f ( x), g( x) ∈ P , let f ( x) R g( x) mean that
f ( x) and g( x) have the same degree. Thus, for example ( x2 + 3 x − 4) R (3 x2 − 2),
and ( x3 + 3 x2 − 4) R (3 x2 − 2). It takes just a quick mental check to see that
R is an equivalence relation. (Do it.) It’s easy to describe the equivalence
classes of R . For example [3 x2 + 2] is the set of all polynomials that have
the same degree as 3 x2 + 2, that is the set of all polynomials of degree 2.
We can write this as [3 x2 + 2] = a x2 + bx + c : a, b, c ∈ R, a = 0 . Relations 172 Recall that in Example 11.7 we proved that for a given n ∈ N the relation
≡ (mod n) is reﬂexive, symmetric and transitive. Thus, in our new parlance,
≡ (mod n) is an equivalence relation on Z. Consider the case n = 3. Let’s
ﬁnd the equivalence classes of the equivalence relation ≡ (mod 3). The
equivalence class containing 0 seems like a reasonable place to start.
Observe that
[0] = x ∈ Z : x ≡ 0 (mod 3)
= x ∈ Z : 3 |( x − 0) = x ∈ Z : 3 | x = . . . , −3, 0, 3, 6, 9, . . . . Thus the class [0] consists of all the multiples of 3. (Or, said diﬀerently,
[0] consists of all integers that have a remainder of 0 when divided by 3).
Note that [0] = [3] = [6] = [9], etc. The number 1 does not show up in the
set [0] so let’s next look at the equivalence class [1]:
[1] = x ∈ Z : x ≡ 1 (mod 3) = x ∈ Z : 3 |( x − 1) = . . . , −5, −2, 1, 4, 7, 10, . . . The equivalence class [1] consists of all integers that give a remainder of
1 when divided by 3. The number 2 is in neither of the sets [0] or [1], so
we next look at the equivalence class [2].
[2] = x ∈ Z : x ≡ 2 (mod 3) = x ∈ Z : 3 |( x − 2) = . . . , −4, −1, 2, 5, 8, 11, . . . The equivalence class [2] consists of all integers that give a remainder of
2 when divided by 3. Observe that any integer is in one of the sets [0], [1]
or [2], so we have listed all of the equivalence classes. Thus ≡ (mod 3) has
exactly three equivalence classes, as described above.
Similarly, you can show that the equivalence relation ≡ (mod n) has n
equivalence classes [0], [1], [2], . . . , [ n − 1].
Exercises for Section 11.2
1. Let A = 1, 2, 3, 4, 5, 6 , and consider the following equivalence relation on A :
R = (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 3),
(3, 2), (4, 5), (5, 4), (4, 6), (6, 4), (5, 6), (6, 5) . List the equivalence classes of R .
2. Let A = a, b, c, d , e . Suppose R is an equivalence relation on A . Suppose R has
two equivalence classes. Also aRd , bR c and eRd . Write out R as a set. Equivalence Classes and Partitions 173 3. Let A = a, b, c, d , e . Suppose R is an equivalence relation on A . Suppose R has
three equivalence classes. Also aRd and bR c. Write out R as a set.
4. Let A = a, b, c, d , e . Suppose R is an equivalence relation on A . Suppose also
that aRd and bR c, eRa and cR e. How many equivalence classes does R have?
5. There are two diﬀerent equivalence relations on the set A = a, b . Describe
them. Diagrams will suﬃce.
6. There are ﬁve diﬀerent equivalence relations on the set A = a, b, c . Describe
them all. Diagrams will suﬃce.
7. Deﬁne a relation R on Z as xR y if and only if 3 x − 5 y is even. Prove R is an
equivalence relation. Describe its equivalence classes.
8. Deﬁne a relation R on Z as xR y if and only if x2 + y2 is even. Prove R is an
equivalence relation. Describe its equivalence classes.
9. Deﬁne a relation R on Z as xR y if and only if 4 |( x +3 y). Prove R is an equivalence
relation. Describe its equivalence classes.
10. Suppose R and S are two equivalence relations on a set A . Prove that R ∩ S
is also an equivalence relation. (For an example of this, look at Figure 11.2.
Observe that for the equivalence relations R2 , R3 and R4 , we have R2 ∩ R3 = R4 .)
11. Prove or disprove: If R is an equivalence relation on an inﬁnite set A , then R
has inﬁnitely many equivalence classes.
12. Prove or disprove: If R and S are two equivalence relations on a set A , then
R ∪ S is also an equivalence relation on A . 11.3 Equivalence Classes and Partitions
This section collects several properties of equivalence classes.
Our ﬁrst result proves that [a] = [b] if and only if aRb. This is useful
because it assures us that whenever we are in a situation where [a] = [ b],
we also have aRb, and vice versa. Being able to switch back and forth
between these two pieces of information can be helpful in a variety of
situations, and you may ﬁnd yourself using this result a lot. Be sure to
notice that the proof uses all three properties (reﬂexive, symmetric and
transitive) of equivalence relations. Notice also that we have to use some
Chapter 8 techniques in dealing with the sets [a] and [ b].
Theorem 11.1 Suppose R is an equivalence relation on a set A . Suppose
also that a, b ∈ A . Then [a] = [ b] if and only if aRb.
Proof. Suppose [a] = [b]. Note that a ∈ [a], because [a] = x ∈ A : xRa and
aRa by the reﬂexive property of R . But since [a] = [ b], we also have Relations 174 a ∈ [ b] = x ∈ A : xRb . Then since a belongs to the set x ∈ A : xRb , it
follows that aRb. This completes the ﬁrst part of the if-and-only-if proof.
Conversely, suppose aRb. We need to show [a] = [b]. This will be
accomplished by showing [a] ⊆ [ b] and [b] ⊆ [a].
First we show [a] ⊆ [ b]. Suppose c ∈ [a]. By deﬁnition of [a] (recall
[a] = x ∈ A : xRa ), it follows that cRa. Now we have cRa and aRb, and
since R is transitive we obtain cRb. But cRb implies c ∈ x ∈ A : xRb = [ b].
This demonstrates that c ∈ [a] implies c ∈ [ b], so [a] ⊆ [b].
Next we show [ b] ⊆ [a]. Suppose c ∈ [b]. By deﬁnition of [ b] (recall
[ b] = x ∈ A : xRb ), it follows that cRb. Remember that we are assuming
aRb, and since R is symmetric, it follows that bRa. Now we have cRb
and bRa, and since R is transitive we obtain cRa. But cRa implies
c ∈ x ∈ A : xRa = [a]. This demonstrates that c ∈ [ b] implies c ∈ [a], so
[ b ] ⊆ [ a ].
The previous two paragraphs imply that [a] = [b]. To illustrate Theorem 11.1, recall how we worked out the equivalence
classes of ≡ (mod 3) at the end of Section 11.2. We observed that [−3] =
[9] = . . . , −3, 0, 3, 6, 9, . . . . Note that [−3] = [9] and −3 ≡ 9 (mod 3), just as
Theorem 11.1 predicts. The theorem assures us that this will work for any
equivalence relation. In this course and beyond you may ﬁnd yourself using
the result of Theorem 11.1 quite often. Over time it may become natural
and familiar, and you will use it automatically, without even thinking of it
as a theorem.
Our next topic addresses the fact that an equivalence relation on a set
A divides A into various equivalence classes. There is a special word for
this kind of situation. We address it now, as you are likely to encounter it
in subsequent mathematics classes.
Deﬁnition 11.5 A partition of a set A is a set of subsets of A , such
that the union of all the subsets equals A , and the intersection of any two
diﬀerent subsets is .
Example 11.12 Let A = a, b, c, d . One partition of A is a, b , c , d .
This is a set of three subsets a, b , c and d of A . The union of the
three subsets equals A ; the intersection of any two subsets is .
Other partitions of A are
a, b , c , d , a, c , b , d , a,b, c d , and a, b , c , d , to name a few. Intuitively, a partition is just a dividing up of A into pieces. Equivalence Classes and Partitions 175 Example 11.13 Consider the equivalence relations in Figure 11.2. Each
of these is a relation on the set A = − 1, 1, 2, 3, 4 . The equivalence classes
of each relation are listed on the right side of the ﬁgure. Observe that,
in each case, the set of equivalence classes forms a partition of A . For
example, the relation R1 yields the partition − 1 , 1 , 2 , 3 , 4 of A .
Likewise the equivalence classes of R2 form the partition − 1, 1, 3 , 2, 4 .
Example 11.14 Recall that we worked out the equivalence classes of the
equivalence relation ≡ (mod 3) on the set Z. These equivalence classes
give the following partition of Z: . . . , −3, 0, 3, 6, 9, . . . , . . . , −2, 1, 4, 7, 10, . . . ,
. . . , −1, 2, 5, 8, 11, . . . . We can write it more compactly as [0], [1], [2] .
Our examples and experience suggest that the equivalence classes of
an equivalence relation on a set form a partition of that set. This is indeed
the case, and we now prove it.
Theorem 11.2 Suppose R is an equivalence relation on a set A . Then
the set [a] : a ∈ A of equivalence classes of R forms a partition of A .
Proof. To show that [a] : a ∈ A is a partition of A we need to show two
things: We need to show that the union of all the sets [a] equals A , and
we need to show that if [a] = [ b], then [a] ∩ [ b] = .
Notationally, the union of all the sets [a] is a∈ A [a], so we need to prove
a∈ A [a] = A . Suppose x ∈ a∈ A [a]. This means x ∈ [a] for some a ∈ A . Since
[a] ⊆ A , it then follows that x ∈ A . Thus a∈ A [a] ⊆ A . On the other hand,
if x ∈ A , then x ∈ [ x], so x ∈ [a] for some a ∈ A . Therefore x ∈ a∈ A [a], and
this shows A ⊆ a∈ A [a]. Since a∈ A [a] ⊆ A and A ⊆ a∈ A [a], it follows that
a∈ A [ a ] = A .
Next we need to show that if [a] = [b] then [a] ∩ [b] = . Let’s use
contrapositive proof. Suppose it’s not the case that [a] ∩ [ b] = , so there is
some element c with c ∈ [a] ∩ [ b]. Thus c ∈ [a] and c ∈ [ b]. Now, c ∈ [a] means
cRa, and then aR c since R is symmetric. Also c ∈ [ b] means cRb. Now we
have aR c and cRb, so aRb (because R is transitive). Theorem 11.1 now
implies [a] = [b], so [a] = [ b] is not true.
Theorem 11.2 says the equivalence classes of any equivalence relation
on a set A form a partition of A . Conversely, any partition of A describes
an equivalence relation R where xR y if and only if x and y belong to the
same set in the partition. Thus equivalence relations and partitions are
really just two diﬀerent ways of looking at the same thing. Relations 176 Exercises for Section 11.3
1. List all the partitions of the set A = a, b . Compare your answer to the answer
to Exercise 5 of Section 11.2.
2. List all the partitions of the set A = a, b, c . Compare your answer to the
answer to Exercise 6 of Section 11.2.
3. Describe the partition of Z resulting from the equivalence relation ≡ (mod 4). 11.4 The Integers Modulo n
Example 11.7 proved that for a given n ∈ N, the relation ≡ (mod n) is
reﬂexive, symmetric and transitive, so it is an equivalence relation. This
is a particularly signiﬁcant equivalence relation in mathematics, and in
the present section we deduce some of its properties.
To make matters simpler, let’s pick a concrete n, say n = 5. Let’s begin
by looking at the equivalence classes of the relation ≡ (mod 5). There are
ﬁve equivalence classes, as follows.
[0]
[1]
[2]
[3]
[4] =
=
=
=
= x∈Z:x≡0
x∈Z:x≡1
x∈Z:x≡2
x∈Z:x≡3
x∈Z:x≡4 (mod
(mod
(mod
(mod
(mod 5))
5))
5))
5))
5)) =
=
=
=
= x ∈ Z : 5 |( x − 0)
x ∈ Z : 5 |( x − 1)
x ∈ Z : 5 |( x − 2)
x ∈ Z : 5 |( x − 3)
x ∈ Z : 5 |( x − 4) =
=
=
=
= . . . , −10, −5, 0, 5, 10, 15, . . .
. . . , −9, −4, 1, 6, 11, 16, . . .
. . . , −8, −3, 2, 7, 12, 17, . . .
. . . , −7, −2, 3, 8, 13, 18, . . .
. . . , −6, −1, 4, 9, 14, 19, . . . Notice how these equivalence classes form a partition of the set Z.
We label the ﬁve equivalence classes as [0], [1], [2], [3], and [4], but you
know of course that there are other ways to label them. For example,
[0] = [5] = [10] = [15], and so on; and [1] = [6] = [−4], etc. Still, for this
discussion we denote the ﬁve classes as [0], [1], [2], [3], and [4].
These ﬁve classes form a set, which we shall denote as Z5 . Thus
Z5 = [0], [1], [2], [3], [4] is a set of ﬁve sets. The interesting thing about Z5 is that even though
its elements are sets (and not numbers) it is possible to add and multiply
them. In fact, we can deﬁne the following rules that tell how elements of The Integers Modulo n 177 Z5 can be added and multiplied. [ a] + [ b ] = [ a + b ]
[ a] · [ b ] = [a·b] For example, [2] + [1] = [2 + 1] = [3], and [2] · [2] = [2 · 2] = [4]. We stress
that in doing this we are adding and multiplying sets (more precisely
equivalence classes), not numbers. We added (or multiplied) two elements
of Z5 and obtained another element of Z5 .
Here is a trickier example. Observe that [2] + [3] = [5]. This time we
added elements [2], [3] ∈ Z5 , and got the element [5] ∈ Z5 . That was easy,
except where is our answer [5] in the set Z5 = [0], [1], [2], [3], [4] ? Since
[5] = [0], it is more appropriate to write [2] + [3] = [0].
In a similar vein, [2] · [3] = [6] would be written as [2] · [3] = [1] because
[6] = [1]. Test your skill with this by verifying the following addition and
multiplication tables for Z5 .
+ [0] [1] [2] [3] [4] · [0] [1] [2] [3] [4] [0] [0] [1] [2] [3] [4] [0] [0] [0] [0] [0] [0] [1] [1] [2] [3] [4] [0] [1] [0] [1] [2] [3] [4] [2] [2] [3] [4] [0] [1] [2] [0] [2] [4] [1] [3] [3] [3] [4] [0] [1] [2] [3] [0] [3] [1] [4] [2] [4] [4] [0] [1] [2] [3] [4] [0] [4] [3] [2] [1] We call the set Z5 = [0], [1], [2], [3], [4] the integers modulo 5. As our
tables suggest, Z5 is more than just a set: It is a little number system
with its own addition and multiplication. In this way it is like the familiar
set Z which also comes equipped with an addition and a multiplication.
Of course, there is nothing special about the number 5. We can also
deﬁne Zn for any natural number n. Here is the deﬁnition.
Deﬁnition 11.6 Let n ∈ N. The equivalence classes of the equivalence
relation ≡ (mod n) are [0], [1], [2], . . . , [n − 1]. The integers modulo n is the
set Zn = [0], [1], [2], . . . , [ n − 1] . Elements of Zn can be added by the rule
[a] + [ b] = [a + b] and multiplied by the rule [a] · [ b] = [ab].
Given a natural number n, the set Zn is a number system containing n
elements. It has many of the algebraic properties that Z, R and Q possess. Relations 178 For example, it is probably obvious to you already that elements of Zn
obey the commutative laws [a] + [ b] = [ b] + [a] and [a] · [ b] = [b] · [a]. You can
also verify that the distributive law [a] · ([b] + [ c]) = [a] · [ b] + [a] · [ c] holds, as
follows.
[a] · ([ b] + [ c]) = [a] · [ b + c]
= [a( b + c)]
= [ab + ac]
= [ab] + [ac]
= [ a] · [ b ] + [ a] · [ c ] The integers modulo n are signiﬁcant because they more closely ﬁt certain
applications than do other number systems such as Z or R. If you go on to
take a course in abstract algebra, then you will work extensively with Zn
as well as other, more exotic, number systems. (In such a course you will
also use all of the proof techniques that we have discussed, as well as the
ideas of equivalence relations.)
To close this section we take up an issue that may have bothered
you earlier. It has to do with our deﬁnitions of addition [a] + [b] = [a + b]
and multiplication [a] · [ b] = [ab]. These deﬁnitions deﬁne addition and
multiplication of equivalence classes in terms of representatives a and b
in the equivalence classes. Since there are many diﬀerent ways to choose
such representatives, we may well wonder if addition and multiplication
are consistently deﬁned. For example, suppose two people, Alice and Bob,
want to multiply the elements [2] and [3] in Z5 . Alice does the calculation
as [2] · [3] = [6] = [1], so her ﬁnal answer is [1]. Bob does it diﬀerently.
Since [2] = [7] and [3] = [8], he works out [2] · [3] as [7] · [8] = [56]. Since 56 ≡ 1
(mod 5), Bob’s answer is [56] = [1], and that agrees with Alice’s answer. Will
their answers always agree or did they just get lucky (with the arithmetic)?
The fact is that no matter how they do the multiplication in Zn , their
answers will agree. To see why, suppose Alice and Bob want to multiply
the elements [a], [ b] ∈ Zn , and suppose [a] = [a ] and [ b] = [b ]. Alice and Bob
do the multiplication as follows.
Alice:
Bob: [a] · [ b] = [ab]
[a ] · [ b ] = [a b ] We need to show that their answers agree, that is we need to show
[ab] = [a b ]. Since [a] = [a ], we know by Theorem 11.1 that a ≡ a (mod n). Relations Between Sets 179 Thus n |(a − a ), so a − a = nk for some integer k. Likewise, as [ b] = [b ], we
know b ≡ b (mod n), or n |( b − b ), so b − b = n for some integer . Thus we
get a = a + nk and b = b + n . Therefore:
ab = (a + nk)( b + n ) ab = a b + a n + nkb + n2 k ab − a b = n(a + kb + nk ) This shows n |(ab − a b ), so ab ≡ a b (mod n), and from that we conclude
[ab] = [a b ]. Consequently Alice and Bob really do get the same answer, so
we can be assured that the deﬁnition of multiplication in Zn is consistent.
In one of the exercises, you will be asked to show that addition in Zn
is similarly consistent.
Exercises for Section 11.4
1. Write the addition and multiplication tables for Z2 .
2. Write the addition and multiplication tables for Z3 .
3. Write the addition and multiplication tables for Z4 .
4. Write the addition and multiplication tables for Z6 .
5. Suppose [a], [b] ∈ Z5 and [a] · [b] = [0]. Is it necessarily true that either [a] = [0]
or [b] = [0]?
6. Suppose [a], [b] ∈ Z6 and [a] · [b] = [0]. Is it necessarily true that either [a] = [0]
or [b] = [0]?
7. Do the following calculations in Z9 , in each case expressing your answer as [a]
with 0 ≤ a ≤ 8.
(a) [8] + [8]
(b) [24] + [11]
(c) [21] · [15]
(d) [8] · [8]
8. Suppose [a], [b] ∈ Zn , and [a] = [a ] and [b] = [b ]. Alice adds [a] and [ b] as
[a] + [ b] = [a + b]. Bob adds them as [a ] + [ b ] = [a + b ]. Show that their answers
[a + b] and [a + b ] are the same. 11.5 Relations Between Sets
In the beginning of this chapter, we deﬁned a relation on a set A to
be a subset R ⊆ A × A . This created a framework that could model any
situation in which elements of A are compared to themselves. In this
setting, the statement xR y has elements x and y from A on either side
of the R because R compares elements from A . But there are other Relations 180 relational symbols that don’t work this way. Consider ∈. The statement
5 ∈ Z expresses a relationship between 5 and Z (namely that the element 5
is in the set Z) but 5 and Z are not in any way naturally regarded as both
elements of some set A . To overcome this diﬃculty, we generalize the idea
of a relation on A to a relation from A to B.
Deﬁnition 11.7 A relation from a set A to a set B is a subset R ⊆ A × B.
We often abbreviate the statement ( x, y) ∈ R as xR y. The statement ( x, y) ∉ R
is abbreviated as x R y.
Suppose A = 1, 2 and B = P ( A ) = , 1 , 2 , 1, 2 . Then
R = (1, 1 ), (2, 2 ), (1, 1, 2 ), (2, 1, 2 ) ⊆ A × B is a relation from A to B. Notice that we have 1R 1 , 2R 2 , 1R 1, 2 and 2R 1, 2 . The relation R is
the familiar relation ∈ for the set A , that is x R X means exactly the same
thing as x ∈ X .
Example 11.15 Diagrams for relations from A to B diﬀer from diagrams for relations
on A . Since there are two sets A and B in a relation from A to B, we have
to draw labeled nodes for each of the two sets. Then we draw arrows from
x to y whenever xR y. The following ﬁgure illustrates this for Example
11.15.
A
1
2 B
1
2
1, 2 Figure 11.3. A relation from A to B
The ideas from this chapter show that any relation (whether it is a
familiar one like ≥, ≤, =, |, ∈ or ⊆, or a more exotic one) is really just a set.
Therefore the theory of relations is a part of set theory. In the next chapter
we will see that this idea touches on another important mathematical
construction, namely functions. We will deﬁne a function to be a special
kind of relation from one set to another, and in this context we will see
that any function is really just a set. CHAPTER 12 Functions ou know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes
a relationship between two (or more) quantities. You certainly understand
and appreciate the fact that relationships between quantities are important in all scientiﬁc disciplines, so you do not need to be convinced that
functions are important. Still, you may not be aware of the full signiﬁcance
of functions. Functions are more than merely descriptions of numeric
relationships. In a more general sense, functions can compare and relate
diﬀerent kinds of mathematical structures. You will see this as your
understanding of mathematics deepens. In preparation of this deepening,
we will now explore a more general and versatile view of functions.
The concept of a relation between sets (Deﬁnition 11.7) plays a big role
here, so you may want to quickly review it. Y 12.1 Functions
Let’s start on familiar ground. Consider the function f ( x) = x2 from R to R.
Its graph is the set of points R = ( x, x2 ) : x ∈ R ⊆ R × R.
R ( x, x2 ) x R Figure 12.1. A familiar function
Having read Chapter 11, you may see f in a new light. Its graph
R ⊆ R × R is a relation on the set R. In fact, as we shall see, functions
are just special kinds of relations. Before stating the exact deﬁnition, we Functions 182 look at another example. Consider the function f (n) = |n| + 2 that converts
integers n into natural numbers | n| + 2. Its graph is R = ( n, | n| + 2) : n ∈ Z
⊆ Z × N.
N
6
5
4
3
2
1
−4 −3 −2 −1 0 1 2 3 4 Z Figure 12.2. The function f : Z → N, where f ( n) = |n| + 2
Figure 12.2 shows the graph R as darkened dots in the grid of points Z × N.
Notice that in this example R is not a relation on a single set. The set of
input values Z is diﬀerent from the set N of output values, so the graph
R ⊆ Z × N is a relation from Z to N.
This example illustrates three things. First, a function can be viewed
as sending elements from one set A to another set B. (In the case of f ,
A = Z and B = N.) Second, such a function can be regarded as a relation
from A to B. Third, for every input value n, there is exactly one output
value f ( n). In your high school algebra course, this was expressed by the
“vertical line test”: Any vertical line intersects a function’s graph at most
once. It means that for any input value x, the graph contains exactly one
point of form ( x, f ( x)). Our main deﬁnition, given below, incorporates all of
these ideas.
Deﬁnition 12.1 Suppose A and B are sets. A function f from A to B
(denoted as f : A → B) is a relation f ⊆ A × B from A to B, satisfying the
property that for each a ∈ A the relation f contains exactly one ordered
pair of form (a, b). The statement (a, b) ∈ f is abbreviated f (a) = b.
Example 12.1 Consider the function f graphed in Figure 12.2. According
to Deﬁnition 12.1, we regard f as the set of points in its graph, that is
f = ( n, | n| + 2) : n ∈ Z ⊆ Z × N. This is a relation from Z to N, and indeed
given any a ∈ Z the set f contains exactly one ordered pair (a, |a| + 2) whose
ﬁrst coordinate is a. Since (1, 3) ∈ f , we write f (1) = 3; and since (−3, 5) ∈ f
we write f (−3) = 5, etc. In general, (a, b) ∈ f means that f sends the input Functions 183 value a to the output value b, and we express this as f (a) = b. This function
can be expressed by a formula: for each input value n, the output value
is | n| + 2, so we may write f ( n) = |n| + 2. All this agrees with the way we
thought of functions in algebra and calculus; the only diﬀerence is that
now we also think of a function as a relation.
Deﬁnition 12.2 For a function f : A → B, the set A is called the domain
of f . (Think of the domain as the set of possible “input values” for f .) The
set B is called the codomain of f . The range of f is the set f (a) : a ∈ A =
b : (a, b) ∈ f . (Think of the range as the set of all possible “output values”
for f . Think of the codomain as a sort of “target” for the outputs.)
Continuing Example 12.1, the domain of f is Z and its codomain is
N. Its range is f (a) : a ∈ Z = |a| + 2 : a ∈ Z = 2, 3, 4, 5, . . . . Notice that the
range is a subset of the codomain, but it does not (in this case) equal the
codomain.
In our examples so far, the domains and codomains are sets of numbers,
but this needn’t be the case in general, as the next example indicates.
Example 12.2 Let A = p, q, r , s and B = 0, 1, 2 , and
f = ( p, 0), ( q, 1), ( r , 2), ( s, 2) ⊆ A × B. This is a function f : A → B because each element of A occurs exactly once
as a ﬁrst coordinate of an ordered pair in f . We have f ( p) = 0, f ( q) = 1,
f ( r ) = 2 and f ( s) = 2. The domain of f is p, q, r , s , and the codomain and
range are both 0, 1, 2 .
B A ( p, 0) ( q, 0) ( r , 0) ( s, 0) 1 ( p, 1) ( q, 1) ( r , 1) ( s, 1) 2 ( p, 2) ( q, 2) ( r , 2) ( s, 2) B s 0 0 r
1 q
p p q r (a) s 2 A (b) Figure 12.3. Two ways of drawing the function f = ( p, 0), ( q, 1), ( r , 2), ( s, 2) 184 Functions If A and B are not both sets of numbers it can be diﬃcult to draw
a graph of f : A → B in the traditional sense. Figure 12.3(a) shows an
attempt at a graph of f from Example 12.2. The sets A and B are aligned
roughly as x- and y-axes, and the Cartesian product A × B is ﬁlled in
accordingly. The subset f ⊆ A × B is indicated with dashed lines, and this
can be regarded as a “graph” of f . A more natural visual description of f
is shown in 12.3(b). The sets A and B are drawn side-by-side, and arrows
point from a to b whenever f (a) = b.
In general, if f : A → B is the kind of function you may have encountered
in algebra or calculus, then conventional graphing techniques oﬀer the
best visual description of it. On the other hand, if A and B are ﬁnite or if
we are thinking of them as generic sets, then describing f with arrows is
often a more appropriate way of visualizing it.
We emphasize that, according to Deﬁnition 12.1, a function is really
just a special kind of set. Any function f : A → B is a subset of A × B. By
contrast, your calculus text probably deﬁned a function as a certain kind of
“rule.” While that intuitive outlook is adequate for the ﬁrst few semesters
of calculus, it does not hold up well to the rigorous mathematical standards
necessary for further progress. The problem is that words like “rule” are
too nebulous. Phrasing the deﬁnition of a function in the language of sets
removes the ambiguity.
Still, in practice we tend to think of functions as rules. Given f : Z → N
where f ( x) = | x| + 2, we think of this as a rule that associates any number
n ∈ Z to the number | n| + 2 in N, rather than a set containing ordered
pairs ( n, | n| + 2). It is only when we have to understand or interpret the
theoretical nature of functions (as we do in this text) that Deﬁnition 12.1
comes to bear. The deﬁnition is a foundation that gives us license to think
about functions in a more informal way.
The next example brings up a point about notation. Consider a function
such as f : Z2 → Z whose domain is a Cartesian product. This function takes
as input an ordered pair ( m, n) ∈ Z2 and sends it to a number f ((m, n)) ∈ Z.
To simplify the notation, it is common to write f (m, n) instead of f ((n, m)),
even though this is like writing f x instead of f ( x). We also remark that
although we’ve been using the letters f , g and h to denote functions, any
other reasonable symbol could be used. Greek letters such as ϕ and θ are
common.
Example 12.3 Say a function ϕ : Z2 → Z is deﬁned as ϕ( m, n) = 6m − 9n.
Note that as a set, this function is ϕ = (m, n), 6m − 9n : ( m, n) ∈ Z2 ⊆ Z2 × Z.
Find the range of ϕ. Functions 185 To answer this, ﬁrst observe that for any ( m, n) ∈ Z2 , the value f (m, n) =
6 m − 9 n = 3(2 m − 3 n) is a multiple of 3. Thus every number in the range is
a multiple of 3, so the range is a subset of the set of all multiples of 3. On
the other hand if b = 3k is a multiple of 3 we have ϕ(−k, −k) = 6(−k) − 9(−k) =
3 k = b, which means any multiple of 3 is in the range of ϕ. Therefore the
range of ϕ is the set 3k : k ∈ Z of all multiples of 3.
To conclude this section, let’s use Deﬁnition 12.1 to help us understand
what it means for two functions f : A → B and g : C → D to be equal.
According to our deﬁnition, f ⊆ A × B and g ⊆ C × D . If the two functions
are to be equal, we require that the following sets be equal: f = g, A = C
and B = D .
Suppose for example, that A = 1, 2, 3 and B = a, b . The two functions f = (1, a), (2, a), (3, b) and g = (3, b), (2, a), (1, a) from A to B are equal
because the sets f and g are equal. Observe that the equality f = g
means f ( x) = g( x) for every x ∈ A . We repackage this idea in the following
deﬁnition.
Deﬁnition 12.3 Two functions f : A → B and g : C → D are equal if A = C ,
B = D and f ( x) = g( x) for every x ∈ A .
According to the deﬁnition, to show functions f and g are equal, we just
need to conﬁrm that their domains are equal, their codomains are equal,
and f ( x) = g( x) for every x in the domain. There is a shade of meaning to
watch out for here. Consider the functions f : Z → N and g : Z → Z deﬁned
as f ( x) = | x| + 2 and g( x) = | x| + 2. Even though their domains are the same
and f ( x) = g( x) for every x in the domain, they are technically not equal
because their codomains diﬀer.
Exercises for Section 12.1
1. Suppose A = 0, 1, 2, 3, 4 , B = 2, 3, 4, 5 and f = (0, 3), (1, 3), (2, 4), (3, 2), (4, 2) . State
the domain and range of f . Find f (2) and f (1).
2. Suppose A = a, b, c, d , B = 2, 3, 4, 5, 6 and f = (a, 2), (b, 3), ( c, 4), ( d , 5) . State the
domain and range of f . Find f (b) and f ( d ).
3. There are four diﬀerent functions f : a, b → 0, 1 . List them all. Diagrams
will suﬃce.
4. There are eight diﬀerent functions f : a, b, c → 0, 1 . List them all. Diagrams
will suﬃce.
5. Give an example of a relation from a, b, c, d to d , e that is not a function. Functions 186 6. Suppose f : Z → Z is deﬁned as f = ( x, 4 x + 5) : x ∈ Z . State the domain, codomain
and range of f . Find f (10).
7. Consider the set f = ( x, y) ∈ Z × Z : 3 x + y = 4 . Is this a function from Z to Z?
Explain.
8. Consider the set f = ( x, y) ∈ Z × Z : x + 3 y = 4 . Is this a function from Z to Z?
Explain.
9. Consider the set f = ( x2 , x) : x ∈ R . Is this a function from R to R? Explain.
10. Consider the set f = ( x3 , x) : x ∈ R . Is this a function from R to R? Explain.
11. Is the set θ = ( X , | X |) : X ⊆ Z5 a function? If so, what is its domain and range?
12. Is the set θ = ( x, y), (3 y, 2 x, x + y) : x, y ∈ R a function? If so, what is its domain,
codomain and range? 12.2 Injective and Surjective Functions
You may recall from algebra and calculus that a function may be oneto-one and onto, and these properties are related to whether or not the
function is invertible. We now review these important ideas. In advanced
mathematics, the word injective is often used instead of one-to-one, and
surjective is used instead of onto. Here are the exact deﬁnitions.
Deﬁnition 12.4 A function f : A → B is: 1. injective (or one-to-one) if for every x, y ∈ A , x = y implies f ( x) = f ( y);
2. surjective (or onto) if for every b ∈ B there is an a ∈ A with f (a) = b;
3. bijective if f is both injective and surjective.
Below is a visual description of Deﬁnition 12.4. In essence, injective
means that unequal elements in A always get sent to unequal elements in
B. Surjective means that every element of B has an arrow pointing to it,
that is, it equals f (a) for some a in the domain of f .
A Injective means that for any
two x, y ∈ A , this happens... A x B x ...and not this: y A Surjective means that for
any b ∈ B... B B
b y A ...this happens: B a b Injective and Surjective Functions 187 For more concrete examples, consider the following functions from R
to R. The function f ( x) = x2 is not injective because −2 = 2, but f (−2) = f (2).
Nor is it surjective, for if b = −1 (or if b is any negative number), then
there is no a ∈ R with f (a) = b. On the other hand, g( x) = x3 is both injective
and surjective, so it is also bijective.
There are four possible injective/surjective combinations that a function may posses. This is illustrated in the following ﬁgure showing four
functions from A to B. Functions in the ﬁrst column are injective, those
in the second column are not injective. Functions in the ﬁrst row are
surjective, those in the second row are not.
Injective Surjective A
a
b
c Not injective B A
a
b
c B A
a
b
c 1
2
3 B 1
2 (bijective) Not surjective A
a
b B
1
2
3 1
2
3 We note in passing that, according to the deﬁnitions, a function is
surjective if and only if its codomain equals its range.
Often it is necessary to prove that a particular function f : A → B
is injective. For this we must prove that for any two elements x, y ∈ A ,
the conditional statement ( x = y) ⇒ f ( x) = f ( y) is true. The two main
approaches for this are summarized below.
How to show a function f : A → B is injective:
Direct approach:
Suppose x, y ∈ A and x = y. Contrapositive approach:
Suppose x, y ∈ A and f ( x) = f ( y). Therefore f ( x) = f ( y). Therefore x = y. .
.
. .
.
. Of these two approaches, the contrapositive is often the easiest to use,
especially if f is deﬁned by an algebraic formula. This is because the
contrapositive approach starts with the equation f ( x) = f ( y) and proceeds Functions 188 to the equation x = y. In algebra, as you know, it is usually easier to work
with equations than inequalities.
To prove that a function is not injective, you must disprove the statement
( x = y) ⇒ f ( x) = f ( y) . For this it suﬃces to ﬁnd example of two elements
x, y ∈ A for which x = y and f ( x) = f ( y).
Next we examine how to prove that f : A → B is surjective. According
to Deﬁnition 12.4, we must prove the statement ∀ b ∈ B, ∃ a ∈ A , f (a) = b. In
words, we must show that for any b ∈ B, there is at least one a ∈ A (which
may depend on b) having the property that f (a) = b. Here is an outline.
How to show a function f : A → B is surjective:
Suppose b ∈ B.
[Prove there exists a ∈ A for which f (a) = b.]
In the second step, we have to prove the existence of an a for which
f (a) = b. For this, just ﬁnding an example of such an a would suﬃce. (How
to ﬁnd such an example depends on how f is deﬁned. If f is given as a
formula, we may be able to ﬁnd a by solving the equation f (a) = b for a.
Sometimes you can ﬁnd a by just plain common sense.) To show f is not
surjective, we must prove the negation of ∀ b ∈ B, ∃ a ∈ A , f (a) = b, that is we
must prove ∃ b ∈ B, ∀ a ∈ A , f (a) = b.
The following examples illustrate these ideas. (For the ﬁrst example,
note that the set R − 0 is R with the number 0 removed.)
Example 12.4 Show that the function f : R − 0 → R deﬁned as f ( x) = 1 + 1
x
is injective but not surjective.
We will use the contrapositive approach to show that f is injective.
Suppose x, y ∈ R − 0 and f ( x) = f ( y). This means 1 + 1 = 1 + 1. Subtracting
x
y
1 from both sides and inverting produces x = y. Therefore f is injective.
Function f is not surjective because there exists an element b = 1 ∈ R
for which f ( x) = 1 + 1 = 1 for every x ∈ R − 0 .
x
Example 12.5 Show that the function g : Z × Z → Z × Z deﬁned by the
formula g(m, n) = ( m + n, m + 2n), is both injective and surjective.
We will use the contrapositive approach to show that g is injective.
Thus we need to show that g( m, n) = g(k, ) implies (m, n) = (k, ). Suppose
( m, n), ( k, ) ∈ Z × Z and g( m, n) = g( k, ). Then ( m + n, m + 2 n) = ( k + , k + 2 ). It
follows that m + n = k + and m + 2n = k + 2 . Subtracting the ﬁrst equation
from the second gives n = . Next, subtract n = from m + n = k + to get
m = k. Since m = k and n = , it follows that ( m, n) = ( k, ). Therefore g is
injective. Injective and Surjective Functions 189 To see that g is surjective, consider an arbitrary element ( b, c) ∈ Z × Z.
We need to show that there is some ( x, y) ∈ Z × Z for which g( x, y) = (b, c). To
ﬁnd ( x, y), note that g( x, y) = ( b, c) means ( x + y, x + 2 y) = (b, c). This leads to
the following system of equations.
x+
y=
x + 2y = b
c Solving gives x = 2b − c and y = c − b. Then ( x, y) = (2 b − c, c − b). We now
have g(2b − c, c − b) = ( b, c), and it follows that g is surjective.
Example 12.6 Consider function h : Z × Z → Q deﬁned as h(m, n) = m
.
| n| + 1 Determine whether this is injective and whether it is surjective.
This function is not injective because of the unequal elements (1, 2) and
(1, −2) in Z × Z for which h(1, 2) = h(1, −2) = 1 . However, h is surjective: Take
3
c
any element b ∈ Q. Then b = d for some c, d ∈ Z. Notice we may assume d is
c
c
positive by making c negative, if necessary. Then h( c, d − 1) = |d −1|+1 = d = b.
Exercises for Section 12.2
1. Let A = 1, 2, 3, 4 and B = a, b, c . Give an example of a function f : A → B that
is neither injective nor surjective.
2. Consider the logarithm function ln : (0, ∞) → R. Decide whether this function is
injective and whether it is surjective.
3. Consider the cosine function cos : R → R. Decide whether this function is injective
and whether it is surjective. What if it had been deﬁned as cos : R → [−1, 1]?
4. A function f : Z → Z × Z is deﬁned as f (n) = (2n, n + 3). Verify whether this
function is injective and whether it is surjective.
5. A function f : Z → Z is deﬁned as f ( n) = 2n + 1. Verify whether this function is
injective and whether it is surjective.
6. A function f : Z × Z → Z is deﬁned as f (m, n) = 3 n − 4m. Verify whether this
function is injective and whether it is surjective.
7. A function f : Z × Z → Z is deﬁned as f (m, n) = 2 n − 4m. Verify whether this
function is injective and whether it is surjective.
8. A function f : Z × Z → Z × Z is deﬁned as f (m, n) = (m + n, 2 m + n). Verify whether
this function is injective and whether it is surjective.
5x + 1
is bijective.
x−2
x+1 3
10. Prove the function f : R − 1 → R − 1 deﬁned by f ( x) =
is bijective.
x−1 9. Prove that the function f : R − 2 → R − 5 deﬁned by f ( x) = Functions 190 11. Consider the function θ : 0, 1 × N → Z deﬁned as θ (a, b) = (−1)a b. Is θ injective?
Is it surjective? Explain.
12. Consider the function θ : 0, 1 ×N → Z deﬁned as θ (a, b) = a−2ab + b. Is θ injective?
Is it surjective? Explain.
13. Consider the function f : R2 → R2 deﬁned by the formula f ( x, y) = ( x y, x3 ). Is f
injective? Is it surjective?
14. Consider the function θ : P (Z) → P (Z) deﬁned as θ ( X ) = X . Is θ injective? Is it
surjective?
15. This question concerns functions f : A , B, C , D , E , F , G → 1, 2, 3, 4, 5, 6, 7 . How
many such functions are there? How many of these functions are injective?
How many are surjective? How many are bijective?
16. This question concerns functions f : A , B, C , D , E → 1, 2, 3, 4, 5, 6, 7 . How many
such functions are there? How many of these functions are injective? How
many are surjective? How many are bijective?
17. This question concerns functions f : A , B, C , D , E , F , G → 1, 2 . How many such
functions are there? How many of these functions are injective? How many
are surjective? How many are bijective? 12.3 The Pigeonhole Principle
Here is a simple but useful idea. Imagine there is a set A of pigeons and
a set B of pigeon holes, and all the pigeons ﬂy into the pigeon holes. You
can think of this as describing a function f : A → B, where Pigeon X ﬂies
into Pigeon hole f ( X ). Figure 12.4 illustrates this. Pigeons Pigeon holes Pigeons Pigeon holes f f (a) (b) Figure 12.4. The Pigeonhole Principle In Figure 12.4(a) there are more pigeons than pigeon holes, and it
is obvious that in such a case at least two pigeons have to ﬂy into the The Pigeonhole Principle 191 same pigeon hole, meaning that f is not injective. In Figure 12.4(b) there
are fewer pigeons than pigeon holes, so clearly at least one pigeon hole
remains empty, meaning that f is not surjective,
Although the underlying idea expressed by these ﬁgures has little to
do with pigeons, it is nonetheless called the Pigeonhole Principle:
The Pigeonhole Principle Suppose A and B are ﬁnite sets and f : A → B
is any function. Then:
1. If | A | > |B|, then f is not injective.
2. If | A | < |B|, then f is not surjective.
Though the Pigeonhole Principle is obvious, it can be used to prove
some things that are not so obvious.
Example 12.7 Prove the following proposition. Proposition If A is any set of 10 integers between 1 and 100, then there
exist subsets X ⊆ A and Y ⊆ A for which the sum of elements in X equals
the sum of elements in Y .
To illustrate what this proposition is saying, consider the random set
A = 5, 7, 12, 11, 17, 50, 51, 80, 90, 100 of 10 integers between 1 and 100. Notice that A has subsets X = 5, 80 and
Y = 7, 11, 17, 50 for which the sum of the elements in X equals the sum of
the elements in Y . If we tried to “mess up” A by changing the 5 to a 6, we
get
A = 6, 7, 12, 11, 17, 50, 51, 80, 90, 100 which has subsets X = 7, 12, 17, 50 and Y = 6, 80 both of whose elements
add up to the same number (86). The proposition asserts that this is always
possible, no matter what A is. Here is a proof.
Proof. Suppose A ⊆ 1, 2, 3, 4, . . . , 99, 100 and | A | = 10, as stated. Notice that
if X ⊆ A , then X has no more than 10 elements, each between 1 and 100,
and therefore the sum of all the elements of X is less than 100 · 10 = 1000.
Consider the function
f : P ( A ) → 0, 1, 2, 3, 4, . . . , 1000 where f ( X ) is the sum of the elements in X . (Examples: f 3, 7, 50 = 60;
f 1, 70, 80, 95 = 246.) As |P ( A )| = 210 = 1024 > 1001 = 0, 1, 2, 3, . . . , 1000 , Functions 192 it follows from the Pigeonhole Principle that f is not injective. Therefore
there are two unequal sets X , Y ∈ P ( A ) for which f ( X ) = f (Y ). In other
words, there are subsets X ⊆ A and Y ⊆ A for which the sum of elements
in X equals the sum of elements in Y .
Example 12.8 Prove the following proposition. Proposition There are at least two Texans with the same number of
hairs on their heads.
Proof. We will use two facts. First, the population of Texas is more than
twenty million. Second, it is a biological fact that every human head
has fewer than one million hairs. Let A be the set of all Texans, and
let B = 0, 1, 2, 3, 4, . . . , 1000000 . Let f : A → B be the function for which f ( x)
equals the number of hairs on the head of x. Since | A | > |B|, the Pigeonhole
Principle asserts that f is not injective. Thus there are two Texans x and
y for whom f ( x) = f ( y), meaning that they have the same number of hairs
on their heads.
Exercises for Section 12.3
1. Prove that if six numbers are chosen at random, then at least two of them will
have the same remainder when divided by 5.
2. If a is a natural number, then there exist two unequal natural numbers k and
for which a k − a is divisible by 10.
3. Prove that if six natural numbers are chosen at random, then the sum or
diﬀerence of two of them is divisible by 9.
4. Consider a square whose side-length is one unit. Select any ﬁve points from
inside this square. Prove that at least two of these points are within 22 units
of each other.
5. Prove that any set of seven distinct natural numbers contains a pair of numbers
whose sum or diﬀerence is divisible by 10.
6. Given a sphere S , a great circle of S is the intersection of S with a plane
through its center. Every great circle divides S into two parts. A hemisphere
is the union of the great circle and one of these two parts. Prove that if ﬁve
points are placed arbitrarily on S , then there is a hemisphere that contains
four of them. Composition 193 12.4 Composition
You should be familiar with the notion of function composition from algebra
and calculus. Still, it is worthwhile to revisit it now with our more
sophisticated ideas about functions.
Deﬁnition 12.5 Suppose f : A → B and g : B → C are functions with the
property that the codomain of f equals the domain of g. The composition
of f with g is another function, denoted as g ◦ f and deﬁned as follows: If
x ∈ A , then g◦ f ( x) = g( f ( x)). Therefore g◦ f sends elements of A to elements
of C , so g ◦ f : A → C .
The following ﬁgure illustrates the deﬁnition. Here f : A → B, g : B → C ,
and g ◦ f : A → C . We have, for example, g ◦ f (0) = g( f (0)) = g(2) = 4. Be very
careful with the order of the symbols. Even though g comes ﬁrst in the
symbol g◦ f , we work out g◦ f ( x) as g( f ( x)), with f acting on x ﬁrst, followed
by g acting on f ( x).
A
0
1
2
3 A
0
1
2
3 f B
1
2
3 g◦ f C
g 4
5
6
7 C
4
5
6
7 Figure 12.5. Composition of two functions
Notice that the composition g ◦ f also makes sense if the range of f
is a subset of the domain of g. You should take note of this fact, but to
keep matters simple we will continue to emphasize situations where the
codomain of f equals the domain of g.
Example 12.9 Suppose A = a, b, c , B = 0, 1 , C = 1, 2, 3 . Let f : A → B
be the function f = (a, 0), (b, 1), ( c, 0) , and let g : B → C be the function
g = (0, 3), (1, 1) . Then g ◦ f = (a, 3), ( b, 1), ( c, 3) .
Example 12.10 Suppose A = a, b, c , B = 0, 1 , C = 1, 2, 3 . Let f : A → B
be the function f = (a, 0), (b, 1), ( c, 0) , and let g : C → B be the function
g = (1, 0), (2, 1), (3, 1) . In this situation the composition g ◦ f is not deﬁned
because the codomain B of f is not the same set as the domain C of g. Functions 194 Remember: In order for g ◦ f to make sense, the codomain of f must equal
the domain of g. (Or at least be a subset of it.)
Example 12.11 Let f : R → R be deﬁned as f ( x) = x2 + x, and g : R → R be
deﬁned as g( x) = x + 1. Then g ◦ f : R → R is the function deﬁned by the
formula g ◦ f ( x) = g( f ( x)) = g( x2 + x) = x2 + x + 1.
Since the domains and codomains of g and f are the same, we can in
this case do the composition in the other order. Note that f ◦ g : R → R is the
function deﬁned as f ◦ g ( x) = f ( g( x)) = f ( x + 1) = ( x + 1)2 + ( x + 1) = x2 + 3 x + 2.
This example illustrates that even when g ◦ f and f ◦ g are both deﬁned,
they are not necessarily equal. We can express this fact by saying function
composition is not commutative.
We close this section by proving several facts about function composition
that you are likely to encounter in your future study of mathematics. First,
we note that, although it is not commutative, function composition is
associative.
Theorem 12.1 Composition of functions is associative. That is if f : A → B,
g : B → C and h : C → D , then ( h ◦ g) ◦ f = h ◦ ( g ◦ f ).
Proof. Suppose f , g, h are as stated. It follows from Deﬁnition 12.5 that
both ( h ◦ g) ◦ f and h ◦ ( g ◦ f ) are functions from A to D . To show that they
are equal, we just need to show
( h ◦ g ) ◦ f ( x) = h ◦ ( g ◦ f ) ( x) for every x ∈ A . Note that Deﬁnition 12.5 yields
( h ◦ g) ◦ f ( x) = ( h ◦ g)( f ( x)) = h( g( f ( x)). Also
h ◦ ( g ◦ f ) ( x) = h( g ◦ f ( x)) = h( g( f ( x))). Thus
( h ◦ g ) ◦ f ( x ) = h ◦ ( g ◦ f ) ( x ), as both sides equal h( g( f ( x))).
Theorem 12.2 Suppose f : A → B and g : B → C . If both f and g are
injective, then g ◦ f is injective. If both f and g are surjective, then g ◦ f is
surjective. Composition 195 Proof. First suppose both f and g are injective. To see that g ◦ f is injective,
we must show that g ◦ f ( x) = g ◦ f ( y) implies x = y. Suppose g ◦ f ( x) = g ◦ f ( y).
This means g( f ( x)) = g( f ( y)). It follows that f ( x) = f ( y). (For otherwise g
wouldn’t be injective.) But since f ( x) = f ( y) and f is injective, it must be
that x = y. Therefore g ◦ f is injective.
Next suppose both f and g are surjective. To see that g ◦ f is surjective,
we must show that for any element c ∈ C , there is a corresponding element
a ∈ A for which g ◦ f (a) = c. Thus consider an arbitrary c ∈ C . Because g
is surjective, there is an element b ∈ B for which g( b) = c. And because
f is surjective, there is an element a ∈ A for which f (a) = b. Therefore
g( f (a)) = g( b) = c, which means g ◦ f (a) = c. Thus g ◦ f is surjective.
Exercises for Section 12.4
1. Suppose A = 5, 6, 8 , B = 0, 1 , C = 1, 2, 3 . Let f : A → B be the function f =
(5, 1), (6, 0), (8, 1) , and g : B → C be g = (0, 1), (1, 1) . Find g ◦ f .
2. Suppose A = 1, 2, 3, 4 , B = 0, 1, 2 , C = 1, 2, 3 . Let f : A → B be
f = (1, 0), (2, 1), (3, 2), (4, 0) , and g : B → C be g = (0, 1), (1, 1), (2, 3) . Find g ◦ f .
3. Suppose A = 1, 2, 3 . Let f : A → A be the function f = (1, 2), (2, 2), (3, 1) , and let
g : A → A be the function g = (1, 3), (2, 1), (3, 2) . Find g ◦ f and f ◦ g.
4. Suppose A = a, b, c . Let f : A → A be the function f = (a, c), (b, c), ( c, c) , and let
g : A → A be the function g = (a, a), ( b, b), ( c, a) . Find g ◦ f and f ◦ g.
5. Consider the functions f , g : R → R deﬁned as f ( x) =
the formulas for g ◦ f and f ◦ g.
6. Consider the functions f , g : R → R deﬁned as f ( x) =
the formulas for g ◦ f and f ◦ g. 3 x + 1 and g( x) = x3 . Find 1
x2 +1 and g( x) = 3 x + 2. Find 7. Consider the functions f , g : Z × Z → Z × Z deﬁned as f (m, n) = (mn, m2 ) and
g( m, n) = ( m + 1, m + n). Find the formulas for g ◦ f and f ◦ g.
8. Consider the functions f , g : Z × Z → Z × Z deﬁned as f (m, n) = (3m − 4n, 2m + n)
and g(m, n) = (5 m + n, m). Find the formulas for g ◦ f and f ◦ g.
9. Consider the functions f : Z × Z → Z deﬁned as f (m, n) = m + n and g : Z → Z × Z
deﬁned as g(m) = (m, m). Find the formulas for g ◦ f and f ◦ g.
10. Consider the function f : R2 → R2 deﬁned by the formula f ( x, y) = ( x y, x3 ). Find
a formula for f ◦ f . Functions 196
12.5 Inverse Functions You may recall from calculus that if a function f is injective and surjective,
then it has an inverse function f −1 that “undoes” the eﬀect of f in the
sense that f −1 ( f ( x)) = x for every x in the domain. (For example, if f ( x) = x3 ,
then f −1 ( x) = 3 x.) We now review these ideas. Our approach uses two
ingredients, outlined in the following deﬁnitions.
Deﬁnition 12.6 Given a set A , the identity function on A is the function i A : A → A deﬁned as i A ( x) = x for every x ∈ A .
For example if A = 1, 2, 3 , then i A = (1, 1), (2, 2), (3, 3) . Also i Z = (n, n) :
n ∈ Z . The identity function on a set is the function that sends any element
of the set to itself.
Notice that for any set A , the identity function i A is bijective: It is
injective because i A ( x) = i A ( y) immediately reduces to x = y. It is surjective
because if we take any element b in the codomain A , then b is also in the
domain A , and i A ( b) = b.
Deﬁnition 12.7 Given a relation R from A to B, the inverse relation
of R is the relation from B to A deﬁned as R −1 = ( y, x) : ( x, y) ∈ R . In other
words, the inverse of R is the relation R −1 obtained by interchanging the
elements in every ordered pair in R .
For example, let A = a, b, c and B = 1, 2, 3 , and suppose f is the
relation f = (a, 2), ( b, 3), ( c, 1) from A to B. Then f −1 = (2, a), (3, b), (1, c)
and this is a relation from B to A . Notice that f is actually a function
from A to B, and f −1 is a function from B to A . These two relations are
drawn below. Notice the drawing for relation f −1 is just the drawing for f
with arrows reversed.
A
a
b
c B
1
2
3 f = (a, 2), ( b, 3), ( c, 1) A
a
b
c B
1
2
3 f −1 = (2, a), (3, b), (1, c) For another example, let A and B be the same sets as above, but consider
the relation g = (a, 2), ( b, 3), ( c, 3) from A to B. Then g−1 = (2, a), (3, b), (3, c)
is a relation from B to A . These two relations are sketched below. Inverse Functions 197
A
a
b
c B
1
2
3 g = (a, 2), ( b, 3), ( c, 3) A
a
b
c B
1
2
3 g−1 = (2, a), (3, b), (3, c) This time, even though the relation g is a function, its inverse g−1 is
not a function because the element 3 occurs twice as a ﬁrst coordinate of
an ordered pair in g−1 .
In the above examples, relations f and g are both functions, and f −1 is
a function and g−1 is not. This begs a question: What properties does f
have and g lack that makes f −1 a function and g−1 not a function? The
answer is not hard to see. Function g is not injective because g( b) = g( c) = 3,
and thus ( b, 3) and ( c, 3) are both in g. This causes a problem with g−1
because it means (3, b) and (3, c) are both in g−1 , so g−1 can’t be a function.
Thus, in order for g−1 to be a function, it would be necessary that g be
injective.
But that is not enough. Function g also fails to be surjective because
no element of A is sent to the element 1 ∈ B. This means g−1 contains no
ordered pair whose ﬁrst coordinate is 1, so it can’t be a function from B to
A . If g−1 were to be a function it would be necessary that g be surjective.
The previous two paragraphs suggest that if g is a function, then it
must be bijective in order for its inverse relation g−1 to be a function.
Indeed, this is easy to verify. Conversely, if a function is bijective, then its
inverse relation is easily seen to be a function. We summarize this in the
following theorem.
Theorem 12.3 Let f : A → B be a function. Then f is bijective if and only
if the inverse relation f −1 is a function from B to A .
Suppose f : A → B is bijective, so according to the theorem f −1 is a
function. Observe that the relation f contains all the pairs ( x, f ( x)) for x ∈ A ,
so f −1 contains all the pairs ( f ( x), x). But ( f ( x), x) ∈ f −1 means f −1 ( f ( x)) = x.
Therefore f −1 ◦ f ( x) = x for every x ∈ A . From this we get f −1 ◦ f = i A . Similar
reasoning produces f ◦ f −1 = i B . This leads to the following deﬁnitions.
Deﬁnition 12.8 If f : A → B is bijective then its inverse is the function
f −1 : B → A . Functions f and f −1 obey the equations f −1 ◦ f = i A and
f ◦ f −1 = i B . Functions 198 You probably recall from algebra and calculus at least one technique
for computing an inverse: Given f , to ﬁnd f −1 , start with the equation
y = f ( x). Then interchange variables to get x = f ( y). Solving this equation
for y (if possible) produces y = f −1 ( x). The next two examples illustrate
this.
Example 12.12 The function f : R → R deﬁned as f ( x) = x3 + 1 is bijective.
Find its inverse.
We begin by writing y = x3 + 1. Now interchange variables to obtain
3
x = y3 + 1. Solving for y produces y = x − 1. Thus
f −1 ( x) = 3 x − 1. (You can check your answer by computing
f −1 ( f ( x)) = 3 f ( x) − 1 = 3 x3 + 1 − 1 = x. Therefore f −1 ( f ( x)) = x. Any answer other than x indicates a mistake.)
We close with one ﬁnal example. Exercise 12.5 showed that the function
g : Z × Z → Z × Z deﬁned by the formula g( m, n) = ( m + n, m + 2 n) is bijective. Let’s ﬁnd its inverse. The approach outlined above should work, but we
need to be careful to keep track of coordinates in Z × Z. We begin by
writing ( x, y) = g(m, n), then interchanging the variables ( x, y) and (m, n) to
get ( m, n) = g( x, y). This gives
( m, n) = ( x + y, x + 2 y), from which we get the following system of equations.
x+
y=
x + 2y = m
n Solving this system using techniques from algebra with which you are
familiar, we get
x = 2m − n
y=
n−m Then ( x, y) = (2m − n, n − m), so g−1 (m, n) = (2 m − n, n − m). Image and Preimage 199 We can check our work by conﬁrming that g−1 ( g(m, n)) = ( m, n). Doing
the math,
g−1 ( g( m, n)) =
= g−1 ( m + n, m + 2 n)
2( m + n) − ( m + 2 n), ( m + 2 n) − ( m + n) = ( m, n). Exercises for Section 12.5
1. Check that the function f : Z → Z deﬁned by f ( n) = 6 − n is bijective. Then
compute f −1 .
2. In Exercise 9 of Section 12.2 you proved that f : R − 2 → R − 5 deﬁned by
5x + 1
is bijective. Now ﬁnd its inverse.
x−2
3. Let B = 2n : n ∈ Z = . . . , 1 , 1 , 1, 2, 4, 8, . . . . Show that the function f : Z → B
42
deﬁned as f (n) = 2n is bijective. Then ﬁnd f −1 . f ( x) = 4. The function f : R → (0, ∞) deﬁned as f ( x) = e x 3 +1 is bijective. Find its inverse. 5. The function f : R → R deﬁned as f ( x) = π x − e is bijective. Find its inverse.
6. The function f : Z × Z → Z × Z deﬁned by the formula f (m, n) = (5m + 4n, 4m + 3n)
is bijective. Find its inverse.
7. Show that the function f : R2 → R2 deﬁned by the formula f ( x, y) = (( x2 + 1) y, x3 )
is bijective. Then ﬁnd its inverse.
8. Is the function θ : P (Z) → P (Z) deﬁned as θ ( X ) = X bijective? If so, what is its
inverse?
9. Consider the function f : R × N → N × R deﬁned as f ( x, y) = ( y, 3 x y). Check that
this is bijective; ﬁnd its inverse. 12.6 Image and Preimage
It is time to take up a matter of notation that you will encounter in future
mathematics classes. Suppose we have a function f : A → B. If X ⊆ A , the
expression f ( X ) has a special meaning. It stands for the set f ( x) : x ∈ X .
Similarly, if Y ⊆ B then f −1 (Y ) has a meaning even if f is not invertible.
The expression f −1 (Y ) stands for the set x ∈ A : f ( x) ∈ Y . Here are the
precise deﬁnitions.
Deﬁnition 12.9 Suppose f : A → B is a function.
1. If X ⊆ A , the image of X is the set f ( X ) = f ( x) : x ∈ X ⊆ B.
2. If Y ⊆ B, the preimage of Y is the set f −1 (Y ) = x ∈ A : f ( x) ∈ Y ⊆ A . Functions 200 In words, the image f ( X ) of X is the set of all things in B that f sends
elements of X to. (Roughly speaking, you might think of f ( X ) as a kind of
distorted “copy” or “image” of X in B.) The preimage f −1 (Y ) of Y is the set
of all things in A that f sends into Y .
Example 12.13 Consider f : s, t, u, v, w, x, y, z → 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 , where
f = ( s, 4), ( t, 8), ( u, 8), (v, 1), (w, 2), ( x, 4), ( y, 6), ( z, 4) . Notice that f is neither injective nor surjective, so it certainly is not
invertible. Be sure you understand the following statements.
a. f s, t, u, z = 8, 4
b. f s, x, z = 4
c. f s, v, w, y = 1, 2, 4, 6
d. f −1 4 = s, x, z
e. f −1 4, 9 = s, x, z
f. f −1 9 =
g. f −1 1, 4, 8 = s, t, u, v, x, z
It is important to realize that the X and Y in Deﬁnition 12.9 are
subsets (not elements!) of A and B. Thus, in the above example we had
f −1 4 = s, x, z , though f −1 (4) has absolutely no meaning because the
inverse function f −1 does not exist. Likewise, there is a subtle diﬀerence
between f a = 4 and f (a) = 4. Be careful.
Example 12.14 Consider the function f : R → R deﬁned as f ( x) = x2 . Observe that f 0, 1, 2 = 0, 1, 4 and f −1 0, 1, 4 = − 2, −1, 0, 1, 2 . This shows
f −1 ( f ( X )) = X in general.
Now check your understanding of the following statements: f ([−2, 3]) =
[0, 9], and f −1 ([0, 9]) = [−3, 3]. Also f (R) = [0, ∞) and f −1 ([−2, −1]) = .
If you continue your mathematical studies, you are likely to encounter
the following result in the future. For now, you are asked to prove it in
the exercises
Theorem 12.4 Suppose f : A → B is a function. Let W , X ⊆ A , and Y , Z ⊆ B.
Then:
1.
2.
3.
4. f (W ∩ X ) ⊆ f (W ) ∩ f ( X )
f (W ∪ X ) = f (W ) ∪ f ( X )
f −1 (Y ∩ Z ) = f −1 (Y ) ∩ f −1 ( Z )
f −1 (Y ∪ Z ) = f −1 (Y ) ∪ f −1 ( Z ) Image and Preimage 201 Exercises for Section 12.6
1. Consider the function f : R → R deﬁned as f ( x) = x2 + 3. Find f ([−3, 5]) and
f −1 ([12, 19]).
2. Consider the function f : 1, 2, 3, 4, 5, 6, 7 → 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 given as
f = (1, 3), (2, 8), (3, 3), (4, 1), (5, 2), (6, 4), (7, 6) . Find: f 1, 2, 3 , f 4, 5, 6, 7 , f ( ), f −1 0, 5, 9 and f −1 0, 3, 5, 9 . 3. This problem concerns functions f : 1, 2, 3, 4, 5, 6, 7 → 0, 1, 2, 3, 4 . How many
such functions have the property that f −1 3 = 3?
4. This problem concerns functions f : 1, 2, 3, 4, 5, 6, 7, 8 → 0, 1, 2, 3, 4, 5, 6 . How
many such functions have the property that f −1 2 = 4?
5. Consider a function f : A → B and a subset X ⊆ A . We observed in Section 12.6
that f −1 ( f ( X )) = X in general. However X ⊆ f −1 ( f ( X )) is always true. Prove this.
6. Given a function f : A → B and any subset Y ⊆ B, is it always true that f ( f −1 (Y )) =
Y ? Prove or give a counterexample.
7. Given a function f : A → B and subsets W , X ⊆ A , prove f (W ∩ X ) ⊆ f (W ) ∩ f ( X ).
8. Given a function f : A → B and subsets W , X ⊆ A , then f (W ∩ X ) = f (W ) ∩ f ( X ) is
false in general. Produce a counterexample.
9. Given a function f : A → B and subsets W , X ⊆ A , prove f (W ∪ X ) = f (W ) ∪ f ( X )
10. Given f : A → B and subsets Y , Z ⊆ B, prove f −1 (Y ∩ Z ) = f −1 (Y ) ∩ f −1 ( Z ).
11. Given f : A → B and subsets Y , Z ⊆ B, prove f −1 (Y ∪ Z ) = f −1 (Y ) ∪ f −1 ( Z ). CHAPTER 13 Cardinality of Sets his chapter is all about cardinality of sets. At ﬁrst this looks like a
very simple concept. To ﬁnd the cardinality of a set, just count its
elements. If A = a, b, c, d , then | A | = 4; if B = n ∈ Z : −5 ≤ n ≤ 5 , then
|B| = 11. In this case | A | < |B|. What could be simpler than that?
Actually, the idea of cardinality becomes quite subtle when the sets
are inﬁnite. The main point of this chapter is to show you that there are
numerous diﬀerent kinds of inﬁnity, and some inﬁnities are bigger than
others. Two sets A and B can both have inﬁnite cardinality, yet | A | < |B|. T 13.1 Sets With Equal Cardinalities
We begin with a discussion of what it means for two sets to have the
same cardinality. Up until this point we’ve said | A | = |B| if A and B have
the same number of elements: Count the elements of A , then count the
elements of B. If you get the same number, then | A | = |B|.
Although this is a ﬁne strategy if the sets are ﬁnite (and not too big!),
it doesn’t apply to inﬁnite sets because we’d never be done counting their
elements. We need a new deﬁnition of cardinality that applies to both
ﬁnite and inﬁnite sets. Here it is.
Deﬁnition 13.1 Two sets A and B have the same cardinality, written
| A | = |B|, if there exists a bijective function f : A → B. If no such bijective
function exists, then A and B have unequal cardinalities, that is | A | = |B|.
A
a
b
c
d
e f B
0
1
2
3
4 The above picture illustrates our deﬁnition. There is a bijective function
f : A → B, so | A | = |B|. Function f matches up A with B. Think of f as
describing how to overlay A onto B so that they ﬁt together perfectly. Sets With Equal Cardinalities 203 On the other hand, if A and B are as indicated in either of the following
ﬁgures, then there can be no bijection f : A → B. (The best we can do is a
function that is either injective or surjective, but not both). Therefore the
deﬁnition says | A | = |B| in these cases.
B f a
b
c
d A 0
1
2
3
4 A a
b
c
d
e B f 0
1
2
3 Example 13.1 The sets A = n ∈ Z : 0 ≤ n ≤ 5 and B = n ∈ Z : −5 ≤ n ≤ 0
have the same cardinality because there is a bijective function f : A → B
given by the rule f ( n) = −n.
Several comments are in order. First, if | A | = |B|, there can be lots of
bijective functions from A to B. We only need to ﬁnd one of them in order to
conclude | A | = |B|. Second, as bijective functions play such a big role here,
we use the word bijection to mean bijective function. Thus the function
f ( n) = − n from Example 13.1 is a bijection. Also, an injective function is
called an injection and a surjective function is called a surjection.
We emphasize and reiterate that Deﬁnition 13.1 applies to ﬁnite as
well as inﬁnite sets. If A and B are inﬁnite, then | A | = |B| provided there
exists a bijection f : A → B. If no such bijection exists, then | A | = |B|.
Example 13.2 This example shows that |N| = |Z|. To see why this is true,
notice that the following table describes a bijection f : N → Z.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ... f ( n) 0 1 −1 2 −2 3 −3 4 −4 5 −5 6 −6 7 −7 ... Notice that f is described in such a way that it is both injective and
surjective. Every integer appears exactly once on the inﬁnitely long second
row. Thus, according to the table, given any b ∈ Z there is some natural
number n with f ( n) = b, so f is surjective. It is injective because the way
the table is constructed forces f (m) = f ( n) whenever m = n. Because of this
bijection f : N → Z, we must conclude from Deﬁnition 13.1 that |N| = |Z|.
You may ﬁnd Example 13.2 slightly unsettling. On one hand it makes
sense that |N| = |Z| because N and Z are both inﬁnite, so their cardinalities
are both “inﬁnity.” On the other hand, Z seems twice as large as N because Cardinality of Sets 204 Z has all the negative integers as well as the positive ones. Deﬁnition
13.1 settles the issue by producing |N| = |Z|. We summarize this with a theorem.
Theorem 13.1 There exists a bijection f : N → Z. Therefore |N| = |Z|. The fact that N and Z have the same cardinality might prompt us to ask
if other pairs of inﬁnite sets have the same cardinality. How, for example,
do N and R compare? Let’s turn our attention to this issue.
In fact, |N| = |R|. This was ﬁrst recognized by G. Cantor (1845–1918),
who devised an ingenious argument to show that there are no surjective
functions f : N → R, which implies there are no bijections f : N → R, so
|N| = |R| by Deﬁnition 13.1.
We will now describe Cantor’s argument for why there are no surjections
f : N → R. We will reason informally, rather than writing out an exact proof.
Take any arbitrary function f : N → R. The following reasoning shows why
f can’t be surjective.
Imagine making a table for f , where values of n in N are in the lefthand column and the corresponding values f (n) are on the right. The
ﬁrst few entries might look something as follows. In this table, the real
numbers f (n) are written with all their decimal places trailing oﬀ to the
right. Thus, even though f (1) happens to be the real number 0.4, we write
it as 0.40000000 . . . ., etc.
n 1
2
3
4
5
6
7
8
9
10
11
12
13
14
.
.
. f ( n) 0
8
7
5
6
6
6
8
0
0
2
6
8
8 .
.
.
.
.
.
.
.
.
.
.
.
.
. 4
5
5
5
9
8
5
7
5
5
9
5
8
5
.
.
. 0
6
0
0
0
2
0
2
5
0
0
0
9
0 0
0
5
7
0
8
5
0
0
0
0
2
0
0 0
6
0
0
2
0
0
8
0
2
0
8
0
0 0
0
0
4
6
9
5
0
0
0
0
0
8
8 0
7
9
0
0
5
5
6
0
7
8
0
0
7 0
0
4
0
0
8
5
4
8
2
8
0
2
4 0
8
0
8
0
2
0
0
8
2
0
8
4
2 0
6
0
0
0
0
6
0
8
0
0
0
0
0 0
6
4
4
0
5
5
0
8
7
0
0
0
8 0
6
4
8
0
0
5
0
0
8
0
9
8
0 0
9
1
0
5
0
8
4
0
0
9
6
0
2 0
0
0
5
0
2
0
4
7
5
0
7
5
2 0. . .
0. . .
1. . .
0. . .
6. . .
0. . .
8. . .
8. . .
7. . .
1. . .
0. . .
1. . .
0. . .
6. . . Sets With Equal Cardinalities 205 There is a diagonal shaded band in the table. For each n ∈ N, this band
covers the n th decimal place of f ( n):
The
The
The
The 1st decimal place of f (1) is 4, and it’s shaded.
2nd decimal place of f (2) is 6, and it’s shaded.
3rd decimal place of f (3) is 5, and it’s shaded.
4 th decimal place of f (4) is 0, and it’s shaded, etc. This shaded diagonal shows why f cannot be surjective, for it implies that
there is a real number b that does not equal any f (n). Just let b ∈ R be a
number whose n th decimal place always diﬀers from the n th decimal place
of f (n). For deﬁniteness, let’s deﬁne b to be the number between 0 and 1
whose n th decimal place is 6 if the n th decimal place of f (n) is not 6, and
whose n th decimal place is 2 if the n th decimal place of f ( n) is 6. Thus, for
the function f illustrated in the above table, we have
b = 0.62662666666262 . . . and b has been deﬁned so that, for any n ∈ N, the n th decimal place of b
is unequal to the n th decimal place of f ( n). Therefore f (n) = b for every
natural number n, meaning f is not surjective.
Since this argument applies to any function f : N → R (not just the one
in the above example) we conclude that there exist no bijections f : N → R,
so |N| = |R| by Deﬁnition 13.1. We summarize this as a theorem.
Theorem 13.2 There exist no bijections f : N → R. Therefore |N| = |R|.
Exercises for Section 13.1
Show that the two given sets have equal cardinality by describing a bijection from
one to the other. Describe your bijection with a formula (i.e. not as a table).
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. R and (0, ∞)
R and ( 2, ∞)
R and (0, 1) The set of even integers and the set of odd integers
A = 3 k : k ∈ Z and B = 7 k : k ∈ Z
N and S = 2
n :n∈N Z and S = . . . , 1 , 1 , 1 , 1, 2, 4, 8, 16, . . .
842
Z and S = x ∈ R : sin x = 1 0, 1 × N and N
0, 1 × N and Z 206 Cardinality of Sets 13.2 Countable and Uncountable Sets
Let’s summarize the main points from the previous section.
1. | A | = |B| if and only if there exists a bijection f : A → B.
2. |N| = |Z| because there exists a bijection f : N → Z.
3. |N| = |R| because there exists no bijection f : N → R.
Thus, even though N, Z and R are all inﬁnite sets, their cardinalities are
not all the same. Sets N, Z have the same cardinality, but R’s cardinality
is diﬀerent from that of both the other sets. This is our ﬁrst indication
of how inﬁnite sets can have diﬀerent sizes, and we will now make some
deﬁnitions to put words and symbols to this phenomenon.
In a certain sense you can count the elements of N; you can count its
elements oﬀ as 1, 2, 3, 4, . . ., but you’d have to continue this process forever
to count the whole set. Thus we will call N a countably inﬁnite set, and
the same term is used for any set whose cardinality equals that of N.
Deﬁnition 13.2 Suppose A is a set. Then A is countably inﬁnite
if |N| = | A |, that is if there exists a bijection f : N → A . The set A is
uncountable if A is inﬁnite and |N| = | A |, that is, if A is inﬁnite and there
exist no bijections f : N → A .
Thus Z is countably inﬁnite but R is uncountable. This section deals
mainly with countably inﬁnite sets. Uncountable sets are treated later.
If A is countably inﬁnite, then |N| = | A |, so there is a bijection f : N → A .
You can think of f as “counting” the elements of A . The ﬁrst element of A
is f (1), followed by f (2), then f (3), and so on. It makes sense to think of a
countably inﬁnite set as the smallest type of inﬁnite set, because if the
counting process stopped, the set would be ﬁnite, not inﬁnite; a countably
inﬁnite set has the fewest number of elements that a set can have and
still be inﬁnite. It is common to reserve the special symbol ℵ0 to stand for
the cardinality of countably inﬁnite sets.
Deﬁnition 13.3 The cardinality of the natural numbers is denoted as
|N| = ℵ0 . Thus any countably inﬁnite set has cardinality ℵ0 .
(The symbol ℵ is the ﬁrst letter in the Hebrew alphabet, and is pronounced
“aleph.” The symbol ℵ0 is pronounced “aleph naught.”) The summary of
facts at the beginning of this section shows |Z| = ℵ0 and |R| = ℵ0 .
Example 13.3 Let E = 2k : k ∈ Z be the set of even integers. The function
f : Z → E deﬁned as f ( n) = 2 n is easily seen to be a bijection, so we have
|Z| = |E |. Thus, as |N| = |Z| = |E |, the set E is countably inﬁnite and |E | = ℵ0 . Countable and Uncountable Sets 207 Here is a signiﬁcant fact. The elements of any countably inﬁnite set A
can be written in an inﬁnitely long list a 1 , a 2 , a 3 , a 4 , . . . that begins with some
element a 1 ∈ A and includes every element of A . For example, the set E in
the above example can be written in list form as 0, 2, −2, 4, −4, 6, −6, 8, −8, . . ..
The reason that this can be done is as follows. Since A is countably inﬁnite,
Deﬁnition 13.2 says there is a bijection f : N → A . This allows us to list
out the set A as an inﬁnite list f (1), f (2), f (3), f (4), . . .. Conversely, if the
elements of a A can be written in list form as a 1 , a 2 , a 3 , . . ., then the function
f : N → A deﬁned as f ( n) = a n is a bijection, so A is countably inﬁnite. We
summarize this as follows.
Theorem 13.3 A set A is countably inﬁnite if and only if its elements
can be arranged in an inﬁnite list a 1 , a 2 , a 3 , a 4 , . . ..
As an example of how this theorem might be used, let P denote the set
of all prime numbers. Since we can list its elements as 2, 3, 5, 7, 11, 13, . . ., it
follows that the set P is countably inﬁnite.
As another consequence of Theorem 13.3, note that we can interpret the
fact that the set R is not countably inﬁnite as meaning that it is impossible
to write out all the elements of R in an inﬁnite list.
This begs a question. Is it also impossible to write out all the elements
of Q in an inﬁnite list? In other words, is the set Q of rational numbers
countably inﬁnite or uncountable? If you start plotting the rational numbers on the number line, they seem to mostly ﬁll up R. Sure, some numbers
such as 2, π and e will not be plotted, but the dots representing rational
numbers seem to predominate. We might thus expect Q to be uncountable.
However it is a surprising fact that Q is countable. The proof of this fact
works by showing how to write out all the rational numbers in an inﬁnitely
long list.
Theorem 13.4 The set Q of rational numbers is countable.
Proof. To prove this, we just need to show how to write the set Q in list
form. Begin by arranging all rational numbers in an inﬁnite array. This is
done by making the following chart. The top row has a list of all integers,
beginning with 0, then alternating signs as they increase. Each column
headed by an integer k contains all the fractions (in reduced form) whose
numerator is k. For example, the column headed by 2 contains the fractions
2222
222
1 , 3 , 5 , 7 , . . .. It does not contain 2 , 4 , 6 , and so on, because those fractions
are not reduced, and in fact their reduced forms appear in the column
headed by 1. You should examine this table and convince yourself that it
contains all rational numbers in Q. Cardinality of Sets 208
0 1 −1 2 −2 3 −3 4 −4 5 −5 ··· 0
1 1
1 −1
1 2
1 −2
1 3
1 −3
1 4
1 −4
1 5
1 −5
1 ··· 1
2 −1
2 2
3 −2
3 3
2 −3
2 4
3 −4
3 5
2 −5
2 ··· 1
3 −1
3 2
5 −2
5 3
4 −3
4 4
5 −4
5 5
3 −5
3 ··· 1
4 −1
4 2
7 −2
7 3
5 −3
5 4
7 −4
7 5
4 −5
4 ··· 1
5 −1
5 2
9 −2
9 3
7 −3
7 4
9 −4
9 5
6 −5
6 ··· 1
6 −1
6 2
11 −2
11 3
8 −3
8 4
11 −4
11 5
7 −5
7 ··· 1
7 −1
7 2
13 −2
13 3
10 −3
10 4
13 −4
13 5
8 −5
8 ···
..
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. Next, draw an inﬁnite path in this array, beginning at 0 and snaking
1
back and forth as indicated below. Every rational number is on this path.
0 1 −1 2 −2 3 −3 4 −4 5 −5 ··· 0
1 1
1 −1
1 2
1 −2
1 3
1 −3
1 4
1 −4
1 5
1 −5
1 ··· 1
2 −1
2 2
3 −2
3 3
2 −3
2 4
3 −4
3 5
2 −5
2 ··· 1
3 −1
3 2
5 −2
5 3
4 −3
4 4
5 −4
5 5
3 −5
3 ··· 1
4 −1
4 2
7 −2
7 3
5 −3
5 4
7 −4
7 5
4 −5
4 ··· 1
5 −1
5 2
9 −2
9 3
7 −3
7 4
9 −4
9 5
6 −5
6 ··· 1
6 −1
6 2
11 −2
11 3
8 −3
8 4
11 −4
11 5
7 −5
7 ··· 1
7 −1
7 2
13 −2
13 3
10 −3
10 4
13 −4
13 5
8 −5
8 ··· 1
8 −1
8 2
15 −2
15 3
11 −3
11 4
15 −4
15 5
9 −5
9 ···
..
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. Countable and Uncountable Sets 209 Beginning at 0 and following the path, we get an inﬁnite list of all
1
rational numbers:
0, 1, 1
22
111
12
2
2
2
2
3
1
, − , −1, 2, , , − , , , − , , − , − , − , − , −2, 3, , . . .
2
2
35
334
47
7
5
3
3
2 Therefore, by Theorem 13.3, it follows that Q is countably inﬁnite.
It is also true that the Cartesian product of two countably inﬁnite sets
is itself countably inﬁnite, as our next theorem states.
Theorem 13.5 If A and B are both countably inﬁnite, then A × B is
countably inﬁnite.
Proof. Suppose A and B are both countably inﬁnite. By Theorem 13.3, we
know we can write A and B in list form as
A = a1 , a2 , a3 , a4 , . . . , B = b1 , b2 , b3 , b4 , . . . . Figure 13.1 shows how to form an inﬁnite path winding through all of A × B.
Therefore A × B can be written in list form, so it is countably inﬁnite.
As an example of a consequence of this theorem, notice that since Q is
countably inﬁnite, the set Q × Q is also countably inﬁnite.
Recall that the word “corollary” means a result that follows easily from
some other result. We have the following corollary of Theorem 13.5.
Corollary 13.1 Given n countably inﬁnite sets A 1 , A 2 , A 3 , . . . , A n , with
n ≥ 2, the Cartesian product A 1 × A 2 × A 3 × · · · × A n is also countably inﬁnite.
Proof. The proof is by induction. For the basis step, notice that when
n = 2 the statement asserts that for countably inﬁnite sets A 1 and A 2 , the
product A 1 × A 2 is countably inﬁnite, and this is true by Theorem 13.5.
Now assume that for k ≥ 2, any product A 1 × A 2 × A 3 ×· · · × A k of countably
inﬁnite sets is countably inﬁnite. Now consider a product A 1 × A 2 × A 3 × · · ·
× A k+1 of countably inﬁnite sets. Observe that
A 1 × A 2 × A 3 × · · · × A k+1 = A 1 × A 2 × A 3 × · · · × A k × A k+1 = ( A 1 × A 2 × A 3 × · · · × A k ) × A k+1 . By the induction hypothesis, this is a product of two countably inﬁnite
sets, so it is countably inﬁnite by Theorem 13.5. Cardinality of Sets 210
B
.
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. .
.
. b7 (a 1 , b 7 ) (a 2 , b 7 ) (a 3 , b 7 ) (a 4 , b 7 ) (a 5 , b 7 ) (a 6 , b 7 ) (a 7 , b 7 ) · · · b6 (a 1 , b 6 ) (a 2 , b 6 ) (a 3 , b 6 ) (a 4 , b 6 ) (a 5 , b 6 ) (a 6 , b 6 ) (a 7 , b 6 ) · · · b5 (a 1 , b 5 ) (a 2 , b 5 ) (a 3 , b 5 ) (a 4 , b 5 ) (a 5 , b 5 ) (a 6 , b 5 ) (a 7 , b 5 ) · · · b4 (a 1 , b 4 ) (a 2 , b 4 ) (a 3 , b 4 ) (a 4 , b 4 ) (a 5 , b 4 ) (a 6 , b 4 ) (a 7 , b 4 ) · · · b3 (a 1 , b 3 ) (a 2 , b 3 ) (a 3 , b 3 ) (a 4 , b 3 ) (a 5 , b 3 ) (a 6 , b 3 ) (a 7 , b 3 ) · · · b2 (a 1 , b 2 ) (a 2 , b 2 ) (a 3 , b 2 ) (a 4 , b 2 ) (a 5 , b 2 ) (a 6 , b 2 ) (a 7 , b 2 ) · · · b1 (a 1 , b 1 ) (a 2 , b 1 ) (a 3 , b 1 ) (a 4 , b 1 ) (a 5 , b 1 ) (a 6 , b 1 ) (a 7 , b 1 ) · · · a1 a3 a2 a4 a5 a6 a7 ··· A Figure 13.1. The product of countably inﬁnite sets is countably inﬁnite
Theorem 13.6 If A and B are both countably inﬁnite, then A ∪ B is
countably inﬁnite.
Proof. Suppose A and B are both countably inﬁnite. By Theorem 13.3, we
know we can write A and B in list form as
A = a1 , a2 , a3 , a4 , . . . , B = b1 , b2 , b3 , b4 , . . . . We can “shuﬄe” A and B into one inﬁnite list for A ∪ B as follows.
A ∪ B = a1 , b1 , a2 , b2 , a3 , b3 , a4 , b4 , . . . . (We agree not to list an element twice if it belongs to both A and B.)
Therefore, by Theorem 13.3, it follows that A ∪ B is countably inﬁnite. Comparing Cardinalities 211 Exercises for Section 13.2
1. Prove that the set A = ln(n) : n ∈ N ⊆ R is countably inﬁnite.
2. Prove that the set A = (m, n) ∈ N × N : m ≤ n is countably inﬁnite.
3. Prove that the set A = (5n, −3 n) : n ∈ Z is countably inﬁnite.
4. Prove that the set of all irrational numbers is uncountable. (Suggestion:
Consider proof by contradiction using theorems 13.4 and 13.6.)
5. Prove or disprove: There exists a countably inﬁnite subset of the set of irrational
numbers.
6. Prove or disprove: There exists a bijective function f : Q → R.
7. Prove or disprove: The set Q100 is countably inﬁnite.
8. Prove or disprove: The set Z × Q is countably inﬁnite.
9. Prove or disprove: The set 0, 1 × N is countably inﬁnite.
10. Prove or disprove: The set A = 2
n : n ∈ N countably inﬁnite. 11. Describe a partition of N that divides N into eight countably inﬁnite subsets.
12. Describe a partition of N that divides N into ℵ0 countably inﬁnite subsets.
13. Prove or disprove: If A = { X ⊆ N : X is ﬁnite}, then | A | = ℵ0 .
14. Suppose A = (m, n) ∈ N × R : n = π m . Is it true that |N| = | A |? 13.3 Comparing Cardinalities
At this point we know that there are at least two diﬀerent kinds of inﬁnity.
On one hand, there are countably inﬁnite sets such as N, of cardinality
ℵ0 . Then there is the uncountable set R. Are there other kinds of inﬁnity
beyond these two kinds? The answer is “yes,” but to see why we ﬁrst need
to introduce some new deﬁnitions and theorems.
Our ﬁrst task will be to formulate a deﬁnition for what we mean by
| A | < |B|. Of course if A and B are ﬁnite we know exactly what this means:
| A | < |B| means that when the elements of A and B are counted, A is found
to have fewer elements than B. But this process breaks down if A or B is
inﬁnite, for then the elements can’t be counted.
The language of functions helps us overcome this diﬃculty. Notice
that for ﬁnite sets A and B it is intuitively clear that | A | < |B| if and only
if there exists an injective function f : A → B but there are no surjective
functions f : A → B. The following diagram illustrates this. Cardinality of Sets 212
B A
f
a
b
c
d 0
1
2
3
4 We will use this idea to deﬁne what is meant by | A | < |B|. For emphasis,
the following deﬁnition also restates what is meant by | A | = |B|.
Deﬁnition 13.4 Suppose A and B are sets.
(1) | A | = |B| means there is a bijective function f : A → B.
(2) | A | < |B| means there is an injective function f : A → B, but no surjective
f : A → B.
For example, consider N and R. The function f : N → R deﬁned as f ( n) = n
is clearly injective, but it is not surjective because given the element 1 ∈ R,
2
we have f ( n) = 1 for every n ∈ N. In fact, recall that we proved in Section
2
13.1 that there exist no surjective functions N → R. Therefore Deﬁnition
13.4 implies |N| < |R|. Said diﬀerently, ℵ0 < |R|.
Is there a set X for which |R| < | X |? The answer is “yes,” and the next
theorem is a major key in understanding why. Recall that P ( A ) denotes
the power set of A .
Theorem 13.7 If A is any set, then | A | < |P ( A )|. Proof. Before beginning the proof, we remark that this statement is obvious
if A is ﬁnite, for then | A | < 2| A | = |P ( A )|. But our proof must apply to all
sets A , both ﬁnite and inﬁnite, so it must use Deﬁnition 13.4.
We will prove the theorem with direct proof. Suppose A is an arbitrary
set. According to Deﬁnition 13.4, to prove | A | < |P ( A )| we must show that
there exists an injective function f : A → P ( A ), but that there exist no
surjective functions f : A → P ( A ).
To see that there is an injective f : A → P ( A ), deﬁne f by the rule
f ( x) = x . In words, f sends any element x of A to the one-element set
x ∈ P ( A ). Then f : A → P ( A ) is injective, because if f ( x) = f ( y), then
x = y . Now, the only way that x and y can be equal is if x = y, so it
follows that x = y. Thus f is injective.
Next we need to show that there exist no surjections f : A → P ( A ).
Suppose for the sake of contradiction that there does exist a surjective
function f : A → P ( A ). Notice that for any element x ∈ A , we have f ( x) ∈ Comparing Cardinalities 213 P ( A ), so f ( x) is a subset of A . Thus f is a function that sends elements of
A to subsets of A . It follows that for any x ∈ A , either x is an element of
the subset f ( x) or it is not. We use this idea to deﬁne the following subset
B of A .
B = x ∈ A : x ∉ f ( x) ⊆ A Now since B ⊆ A we have B ∈ P ( A ), and since f is surjective there must
be some element a ∈ A for which f (a) = B. Now, either a ∈ B or a ∉ B. We
will consider these two cases separately, and show that each leads to a
contradiction.
Case 1.
we have
Case 2.
we have If a ∈ B, then the deﬁnition of B implies a ∉ f (a), and since f (a) = B
a ∉ B, which is a contradiction.
If a ∉ B, then the deﬁnition of B implies a ∈ f (a), and since f (a) = B
a ∈ B, again a contradiction. Since the assumption that there is a surjective function f : A → P ( A )
leads to a contradiction, we conclude that there are no such surjective
functions.
In conclusion, we have seen that there exists an injective function
A → P ( A ) but no surjective function A → P ( A ), so deﬁnition 13.4 implies
that | A | < |P ( A )|.
Beginning with the set the set A = N and applying Theorem 13.7 over
and over again, we get the following chain of inﬁnite cardinalities.
ℵ0 = |N| < |P (N)| < |P (P (N))| < |P (P (P (N)))| < · · · Thus there is an inﬁnite sequence of diﬀerent types of inﬁnity, starting
with ℵ0 and becoming ever larger. The set N is countable, and all the sets
P (N), P (P (N)), etc. are uncountable.
Although we shall not do it here, it is not hard to prove that |R| = |P (N)|,
so |N| and |R| are just two relatively tame inﬁnities in a long list of other
wild and exotic inﬁnities.
Unless you plan on studying advanced set theory or the foundations
of mathematics, you are unlikely to ever encounter any types of inﬁnity
beyond ℵ0 and |R|. Still you will in future mathematics courses need to
distinguish between countably inﬁnite and uncountable sets, so we close
with two ﬁnal theorems that can help you do this.
Theorem 13.8 If A is an inﬁnite subset of a countably inﬁnite set, then
A is countably inﬁnite. 214 Cardinality of Sets Proof. Suppose A is an inﬁnite subset of the countably inﬁnite set B. Since
B is countably inﬁnite, its elements can be written in a list b 1 , b 2 , b 3 , b 4 , . . ..
Then we can also write A ’s elements in list form by proceeding through
the elements of B, in order, and selecting those that belong to A . Thus A
can be written in list form, and since A is inﬁnite, its list will be inﬁnite.
Consequently A is countably inﬁnite.
Theorem 13.9 If U ⊆ A , and U is uncountable, then A is uncountable
too.
Proof. Suppose for the sake of contradiction that U ⊆ A , and U is uncountable but A is not uncountable. Then since U ⊆ A and U is inﬁnite, then A
must be inﬁnite too. Since A is inﬁnite, and not uncountable, it must be
countably inﬁnite. Then U is an inﬁnite subset of a countably inﬁnite set
A , so U is countably inﬁnite by Theorem 13.8. Thus U is both uncountable
and countably inﬁnite, a contradiction.
Theorems 13.8 and 13.9 are often useful when we need to decide
whether a set A is countably inﬁnite or uncountable. The theorems
sometimes allow us to decide its cardinality by comparing it to a set whose
cardinality is known.
For example, suppose we want to decide whether or not the set A = R2
is uncountable. Since the uncountable set U = R can be regarded as the
x-axis of the plane R2 (and thus a subset of R2 ), Theorem 13.9 implies that
R2 is uncountable. Other examples can be found in the exercises.
Exercises for Section 13.3
1. Suppose B is an uncountable set and A is a set. Given that there is a surjective
function f : A → B, what can be said about the cardinality of A ?
2. Prove that the set C of complex numbers is uncountable.
3. Prove or disprove: If A is uncountable, then | A | = |R|.
4. Prove or disprove: If A ⊆ B ⊆ C and A and C are countably inﬁnite, then B is
countably inﬁnite.
5. Prove or disprove: The set 0, 1 × R is uncountable.
6. Prove or disprove: Every inﬁnite set is a subset of a countably inﬁnite set.
7. Prove or disprove: If A ⊆ B and A is countably inﬁnite and B is uncountable,
then B − A is uncountable. Conclusion f you have internalized the ideas in this book, then you have a set
of rhetorical tools for deciphering and communicating mathematics.
These tools are indispensable at any level. But of course it takes more
than mere tools to build something. Planning, creativity, inspiration, skill,
talent and passion are also vitally important. It is safe to say that if you
have come this far, then you probably possess a suﬃcient measure of these
traits.
The quest to understand mathematics has no end, but you are well
equipped for the journey. It is my hope that the things you have learned
from this book will lead you to a higher plane of understanding, creativity
and expression.
Good luck and best wishes. I R.H. Solutions Chapter 1 Exercises
Section 1.1
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27. {5 x − 1 : x ∈ Z} = {. . . − 11, −6, −1, 4, 9, 14, 19, 24, 29, . . .}
{ x ∈ Z : −2 ≤ x < 7} = {−2, −1, 0, 1, 2, 3, 4, 5, 6}
x ∈ R : x2 = 3 = − 3, 3
x ∈ R : x2 + 5 x = −6 = {−2, −3}
x ∈ R : sin π x = 0} = {. . . , −2, −1, 0, 1, 2, 3, 4, . . .} = Z
{
{ x ∈ Z : | x| < 5} = {−4, −3, −2, −1, 0, 1, 2, 3, 4}
{ x ∈ Z : |6 x| < 5} = {0}
{5a + 2 b : a, b ∈ Z} = {. . . , −2, −1, 0, 1, 2, 3, . . .} = Z
{2, 4, 8, 16, 32, 64 . . .} = {2 x : x ∈ N}
{. . . , −6, −3, 0, 3, 6, 9, 12, 15, . . .} = {3 x : x ∈ Z}
{0, 1, 4, 9, 16, 25, 36, . . .} = x2 : x ∈ Z
{3, 4, 5, 6, 7, 8} = { x ∈ Z : 3 ≤ x ≤ 8} = { x ∈ N : 3 ≤ x ≤ 8}
1
1
. . . , 8 , 1 , 2 , 1, 2, 4, 8, . . . = {2n : n ∈ Z}
4
π
π
π
. . . , −π, − π , 0, π , π, 32 , 2π, 52 , . . . = k2 : k ∈ Z
2
2 29. |{{1} , {2, {3, 4}} , }| = 3
33. |{ x ∈ Z : | x| < 10}| = 19
31. |{{{1} , {2, {3, 4}} , }}| = 1 35. | x ∈ Z : x2 < 10 | = 7
39. {( x, y) : x ∈ [1, 2], y ∈ [1, 2]} 37. | x ∈ N : x2 < 0 | = 0 43. {( x, y) : | x| = 2, y ∈ [0, 1]} 2 2 1 1 −3 −2 −1
−1 1 2 3 −2 −3 −2 −1
−1 1 2 3 −2 41. {( x, y) : x ∈ [−1, 1], y = 1} 45. ( x, y) : x, y ∈ R, x2 + y2 = 1 2 2 1 1 −3 −2 −1
−1
−2 1 2 3 −3 −2 −1
−1
−2 1 2 3 217
47. ( x, y) : x, y ∈ R, y ≥ x2 − 1 49. {( x, x + y) : x ∈ R, y ∈ Z} 3 3 2 2 1 1 −3 −2 −1
−1 1 2 −3 −2 −1
−1 3 1 2 3 −2 −2 −3 −3 51. {( x, y) ∈ R2 : ( y − x)( y + x) = 0}
3
2
1
−3 −2 −1
−1 1 2 3 −2
−3 Section 1.2
1. Suppose A = {1, 2, 3, 4} and B = {a, c}.
(a) A × B = {(1, a), (1, c), (2, a), (2, c), (3, a), (3, c), (4, a), (4, c)}
(b) B × A = {(a, 1), (a, 2), (a, 3), (a, 4), ( c, 1), ( c, 2), ( c, 3), ( c, 4)}
(c) A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)} (d) B × B = {(a, a), (a, c), ( c, a), ( c, c)}
(e) × B = {(a, b) : a ∈ , b ∈ B} = (There are no ordered pairs (a, b) with a ∈ .) (f) ( A × B) × B =
{((1, a), a), ((1, c), a), ((2, a), a), ((2, c), a), ((3, a), a), ((3, c), a), ((4, a), a), ((4, c), a),
((1, a), c), ((1, c), c), ((2, a), c), ((2, c), c), ((3, a), c), ((3, c), c), ((4, a), c), ((4, c), c)} (g) A × (B × B) =
(1, (a, a)), (1, (a, c)), (1, ( c, a)), (1, ( c, c)),
(2, (a, a)), (2, (a, c)), (2, ( c, a)), (2, ( c, c)),
(3, (a, a)), (3, (a, c)), (3, ( c, a)), (3, ( c, c)),
(4, (a, a)), (4, (a, c)), (4, ( c, a)), (4, ( c, c)) (h) B3 = {(a, a, a), (a, a, c), (a, c, a), (a, c, c), ( c, a, a), ( c, a, c), ( c, c, a), ( c, c, c)}
3. x ∈ R : x2 = 2 × {a, c, e} = (− 2, a), ( 2, a), (− 2, c), ( 2, c), (− 2, e), ( 2, e) 5. x ∈ R : x2 = 2 × { x ∈ R : | x| = 2} = (− 2, −2), ( 2, 2), (− 2, 2), ( 2, −2) 7. { } × {0, } × {0, 1} = {( , 0, 0), ( , 0, 1), ( , , 0), ( , , 1)} Solutions 218
Sketch the following Cartesian products on the x- y plane.
9. {1, 2, 3} × {−1, 0, 1} 15. {1} × [0, 1] 2 2 1 1 −3 −2 −1
−1 1 2 3 −3 −2 −1
−1 −2 1 2 3 1 2 3 −2 17. N × Z 11. [0, 1] × [0, 1]
2 2 1 1 −3 −2 −1
1 1 2 −3 −2 −1
−1 3 −2 −2 13. {1, 1.5, 2} × [1, 2] 19. [0, 1] × [0, 1] × [0, 1] 2 2 1 1 −3 −2 −1
−1 1 2 3 −2 −3 −2 −1
−1 1 2 3 −2 Section 1.3
A. List all the subsets of the following sets.
1. The subsets of {1, 2, 3, 4} are: {}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4},
{3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}
3. The subsets of {{R}} are: {} and {{R}}
5. The subsets of { } are {} and { }
7. The subsets of {R, {Q, N}} are: {}, {R},{{Q, N}}, {R, {Q, N}}
B. Write out the following sets by listing their elements between braces.
9. X : X ⊆ {3, 2, a} and | X | = 2 = {{3, 2} , {3, a} , {2, a}}
11. X : X ⊆ {3, 2, a} and | X | = 4 = {} =
C. Decide if the following statements are true or false.
13. R3 ⊆ R3 is true because any set is a subset of itself.
15. ( x, y) : x − 1 = 0 ⊆ ( x, y) : x2 − x = 0 This is true. (The even-numbered ones
are both false. You have to explain why.) 219
Section 1.4
A. Find the indicated sets.
1. P ({{a, b} , { c}}) = { , {{a, b}} , {{ c}} , {{a, b} , { c}}}
3. P ({{ } , 5}) = { , {{ }} , {5} , {{ } , 5}}
5. P (P ({2})) = { , { } , {{2}} , { , {2}}}
7. P ({a, b}) × P ({0, 1}) =
(,
({a} ,
({ b} ,
({a, b} , ),
),
),
), ( , {0}),
({a} , {0}),
({ b} , {0}),
({a, b} , {0}), ( , {1}),
({a} , {1}),
({ b} , {1}),
({a, b} , {1}), ( , {0, 1}),
({a} , {0, 1}),
({ b} , {0, 1}),
({a, b} , {0, 1}) 9. P ({a, b} × {0}) = { , {(a, 0)} , {(b, 0)} , {(a, 0), (b, 0)}}
11. { X ⊆ P ({1, 2, 3}) : | X | ≤ 1} =
{ , { } , {{1}} , {{2}} , {{3}} , {{1, 2}} , {{1, 3}} , {{2, 3}} , {{1, 2, 3}}} B. Suppose that | A | = m and |B| = n. Find the following cardinalities.
13. |P (P (P ( A )))| = 2 2(2 m) 15. |P ( A × B)| = 2mn
17. |{ X ∈ P ( A ) : | X | ≤ 1}| = m + 1
19. |P (P (P ( A × )))| = |P (P (P ( )))| = 4
Section 1.5
1. Suppose A = {4, 3, 6, 7, 1, 9}, B = {5, 6, 8, 4} and C = {5, 8, 4} . Find:
(a)
(b)
(c)
(d)
(e) A ∪ B = {1, 3, 4, 5, 6, 7, 8, 9} (f) A ∩ C = {4} A ∩ B = {4, 6} (g) B ∩ C = {5, 8, 4} A − B = {3, 7, 1, 9}
A − C = {3, 6, 7, 1, 9}
B − A = {5, 8} (h) B ∪ C = {5, 6, 8, 4}
(i) C − B = 3. Suppose A = {0, 1} and B = {1, 2}. Find:
(a) ( A × B) ∩ (B × B) = {(1, 1), (1, 2)}
(b) ( A × B) ∪ (B × B) = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}
(c) ( A × B) − (B × B) = {(0, 1), (0, 2)} (f) P ( A ) ∩ P (B) = { , {1}} (d) ( A ∩ B) × A = {(1, 0), (1, 1)} (g) P ( A ) − P (B) = {{0} , {0, 1}} (e) ( A × B) ∩ B = (h) P ( A ∩ B) = {{} , {1}} (i) , {(0, 1)}, {(0, 2)}, {(1, 1)}, {(1, 2)}, {(0, 1), (0, 2)}, {(0, 1), (1, 1)}, {(0, 1), (1, 2)}, {(0, 2), (1, 1)},
{(0, 2), (1, 2)}, {(1, 1), (1, 2)}, {(0, 2), (1, 1), (1, 2)}, {(0, 1), (1, 1), (1, 2)}, {(0, 1), (0, 2), (1, 2)},
{(0, 1), (0, 2), (1, 1)}, {(0, 1), (0, 2), (1, 1), (1, 2)} Solutions 220 5. Sketch the sets X = [1, 3] × [1, 3] and Y = [2, 4] × [2, 4] on the plane R2 . On separate
drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X . (Hint: X and Y are
Cartesian products of intervals. You may wish to review how you drew sets
like [1, 3] × [1, 3] in the Section 1.2.)
4 4 4 4 4 3 3 3 3 Y
3
2 X ∪Y 2 2 2 X ∩Y X
1 1
1 2 3 1 4 1 2 3 4
2 1
2 2 3 2 X −Y 1
4 Y −X 1 2 1
3 2 4 1 2 2 3 4 2 7. Sketch the sets X = ( x, y) ∈ R : x + y ≤ 1 and Y = ( x, y) ∈ R : x ≥ 0 on R . On
separate drawings, shade in the sets X ∪ Y , X ∩ Y , X − Y and Y − X .
2 2 Y X1 2 X ∪Y 1 −2 −1
−1 1 2 1 −2 −1
−1 −2 1 2 −2 −1
−1 −2 2 X ∩Y 1 1
2 2 X −Y −2 −1
−1 −2 −2 1 1
2 −2 −1
−1 Y −X 1 2 −2 9. The ﬁrst statement is true. (A picture should convince you; draw one if
necessary.) The second statement is false: notice for instance that (0.5, 0.5) is
in the right-hand set, but not the left-hand set.
Section 1.6
1. Suppose A = {4, 3, 6, 7, 1, 9} and B = {5, 6, 8, 4} have universal set U = {n ∈ Z : 0 ≤ n ≤ 10}.
(a)
(b)
(c)
(d)
(e) A = {0, 2, 5, 8, 10} (f) A − B = {4, 6} B = {0, 1, 2, 3, 7, 9, 10} (g) A − B = {5, 8} A∩A =
A ∪ A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = U (h) A ∩ B = {5, 8}
(i) A ∩ B = {0, 1, 2, 3, 4, 6, 7, 9, 10} A−A = A 3. Sketch the set X = [1, 3] × [1, 2] on the plane R2 . On separate drawings, shade in
the sets X , and X ∩ ([0, 2] × [0, 3]).
3 3
2 2 3
2 X
1
−1
−1 1
1 2 3 −1
−1 1 X
12 3 −1
−1 123
X ∩ ([0, 2] × [0, 3]) 5. Sketch the set X = ( x, y) ∈ R2 : 1 ≤ x2 + y2 ≤ 4 on the plane R2 . On a separate
drawing, shade in the set X . 221
3
2
1 3
2
1 X
123 X A U
123 A (shaded) Solution of 1.7, #1. Solution of 1.6, #5.
Section 1.7
1. Draw a Venn diagram for A . (Solution above right)
3. Draw a Venn diagram for ( A − B) ∩ C . Scratch work is shown on the right. The
set A − B is indicated with vertical shading.
C
C
The set C is indicated with horizontal shading. The intersection of A − B and C is thus
the overlapping region that is shaded with
both vertical and horizontal lines. The ﬁnal
BA
B
A
answer is drawn on the far right, where the
set ( A − B) ∩ C is shaded in gray.
5. Draw Venn diagrams for A ∪ (B ∩ C ) and ( A ∪ B) ∩ ( A ∪ C ). Based on your drawings,
do you think A ∪ (B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )?
If you do the drawings carefully, you will ﬁnd
that your Venn diagrams are the same for both
A ∪ (B ∩ C ) and ( A ∪ B) ∩ ( A ∪ C ). Each looks as
illustrated on the right. Based on this, we are
inclined to say that the equation A ∪ (B ∩ C ) =
( A ∪ B) ∩ ( A ∪ C ) holds for all sets A , B and C . C B A 7. Suppose sets A and B are in a universal set U . Draw Venn diagrams for A ∩ B
and A ∪ B. Based on your drawings, do you think it’s true that A ∩ B = A ∪ B?
The diagrams for A ∩ B and A ∪ B look exactly
alike. In either case the diagram is the shaded
region illustrated on the right. Thus we would
expect that the equation A ∩ B = A ∪ B is true
for any sets A and B.
9. Draw a Venn diagram for ( A ∩ B) − C .
C A B 11. The simplest answer is (B ∩ C ) − A .
13. One answer is ( A ∪ B ∪ C ) − ( A ∩ B ∩ C ). U
A B Solutions 222
Section 1.8 1. Suppose A 1 = {a, b, d , e, g, f }, A 2 = {a, b, c, d }, A 3 = {b, d , a} and A 4 = {a, b, h}.
4 4 (a) A i = { a, b , c , d , e , f , g , h} (b) i =1 A i = { a, b }
i =1 3. For each n ∈ N, let A n = {0, 1, 2, 3, . . . , n}.
(a)
A i = {0} ∪ N (b) 5. (a) [ i , i + 1] =[1, ∞) (b) R × [ i , i + 1] = {( x, y) : x, y ∈ R, y ≥ 1} (b) i ∈N i ∈N 7. (a)
i ∈N 9. (a)
X ∈P (N) X=N A i = {0, 1} i ∈N
i ∈N [ i , i + 1] = i ∈N R × [ i , i + 1] = (b)
X ∈P (N) X= 11. Yes, this is always true.
13. The ﬁrst is true, the second is false. Chapter 2 Exercises
Section 2.1
Decide whether or not the following are statements. In the case of a statement,
say if it is true or false.
1. Every real number is an even integer. (Statement, False)
3. If x and y are real numbers and 5 x = 5 y, then x = y. (Statement, True)
5. Sets Z and N are inﬁnite. (Statement, True)
7. The derivative of any polynomial of degree 5 is a polynomial of degree 6.
(Statement, False)
9. cos( x) = −1
This is not a statement. It is an open sentence because whether it’s true or
false depends on the value of x.
11. The integer x is a multiple of seven.
This is an open sentence, and not a statement.
13. Either x is a multiple of seven, or it is not.
This is a statement, for the sentence is true no matter what x is.
15. In the beginning God created the heaven and the earth.
This is a statement, for it is either deﬁnitely true or deﬁnitely false. There is
some controversy over whether it’s true or false, but no one claims that it is
neither true nor false. 223
Section 2.2
Express each statement as one of the forms P ∧ Q , P ∨ Q , or ∼ P . Be sure to
also state exactly what statements P and Q stand for.
1. The number 8 is both even and a power of 2.
P ∧Q
P : 8 is even
Q : 8 is a power of 2 3. x = y ∼ ( x = y) (Also ∼ P where P : x = y.) 5. y ≥ x ∼ ( y < x) (Also ∼ P where P : x < y.) 7. The number x equals zero, but the number y does not.
P∧ ∼ Q
P :x=0
Q: y=0 9. x ∈ A − B
( x ∈ A )∧ ∼ ( x ∈ B) 11. A ∈ X ∈ P (N) : | X | < ∞
( A ⊆ N) ∧ (| A | < ∞).
13. Human beings want to be good, but not too good, and not quite all the time.
P ∧ ∼ Q∧ ∼ R
P : Human beings want to be good.
Q : Human beings want to be too good.
R : Human beings want to be good all the time. Section 2.3
Without changing their meanings, convert each of the following sentences into a
sentence having the form “If P , then Q .”
1. A matrix is invertible provided that its determinant is not zero.
Answer: If a matrix has a determinant not equal to zero, then it is invertible.
3. For a function to be integrable, it is necessary that it is continuous.
Answer: If function is integrable, then it is continuous.
5. An integer is divisible by 8 only if it is divisible by 4.
Answer: If an integer is divisible by 8, then it is divisible by 4.
7. A series converges whenever it converges absolutely.
Answer: If a series converges absolutely, then it converges.
9. A function is integrable provided the function is continuous.
Answer: If a function is continuous, then that function is integrable.
11. You fail only if you stop writing.
Answer: If you fail, then you have stopped writing.
13. Whenever people agree with me I feel I must be wrong.
Answer: If people agree with me, then I feel I must be wrong. Solutions 224 Section 2.4
Without changing their meanings, convert each of the following sentences into a
sentence having the form “P if and only if Q .”
1. For a matrix to be invertible, it is necessary and suﬃcient that its determinant
is not zero.
Answer: A matrix is invertible if and only if its determinant is not zero.
3. If x y = 0 then x = 0 or y = 0, and conversely.
Answer: x y = 0 if and only if x = 0 or y = 0
5. For an occurrence to become an adventure, it is necessary and suﬃcient for
one to recount it.
Answer: An occurrence becomes an adventure if and only if one recounts it.
Section 2.5
1. Write a truth table for P ∨ (Q ⇒ R ) 5. Write a truth table for (P ∧ ∼ P ) ∨ Q P Q R Q⇒R P ∨ (Q ⇒ R ) P Q (P ∧ ∼ P ) (P ∧ ∼ P ) ∨ Q T T T T T T T F T T T F F T T F F F T F T T T F T F T F F F F T F F T T F T T T T F T F F F F F T T T F F F T T 7. Write a truth table for (P ∧ ∼ P ) ⇒ Q
P Q P ⇒Q T T T F F T T F T T F T F F F T F F T F T F F F F T T T ∼ (P ⇒ Q ) (P ∧ ∼ P ) ⇒ Q F
P (P ∧ ∼ P ) T 3. Write a truth table for ∼ (P ⇒ Q ) Q T F
9. Write a truth table for ∼ (∼ P ∧ ∼ Q ). P Q ∼P ∼Q ∼ P∨ ∼ Q ∼ (∼ P ∧ ∼ Q ) T T F F F T T F F T T F F T T F T F F F T T T F 11. Suppose P is false and that the statement (R ⇒ S ) ⇔ (P ∧ Q ) is true. Find the
truth values of R and S . (This can be done without a truth table.) 225
Answer: Since P is false, it follows that (P ∧ Q ) is false also. But then in order
for (R ⇒ S ) ⇔ (P ∧ Q ) to be true, it must be that (R ⇒ S ) is false. The only way
for (R ⇒ S ) to be false is if R is true and S is false.
Section 2.6
A. Use truth tables to show that the following statements are logically equivalent.
1. P ∧ (Q ∨ R ) = (P ∧ Q ) ∨ (P ∧ R )
P Q R Q∨R P ∧Q P ∧R P ∧ (Q ∨ R ) (P ∧ Q ) ∨ (P ∧ R ) T T T T T T T T T T F T T F T T T F T T F T T T T F F F F F F F F T T T F F F F F T F T F F F F F F T T F F F F F F F F F F F F Thus since their columns agree, the two statements are logically equivalent.
3. P ⇒ Q = (∼ P ) ∨ Q
P Q ∼P (∼ P ) ∨ Q P ⇒Q T T F T T T F F F F F T T T T F F T T T Thus since their columns agree, the two statements are logically equivalent.
5. ∼ (P ∨ Q ∨ R ) = (∼ P ) ∧ (∼ Q ) ∧ (∼ R )
P Q R P ∨Q ∨R ∼P ∼Q ∼R ∼ (P ∨ Q ∨ R ) ( P ) ∧ (∼ Q ) ∧ (∼ R ) T T T T F F F F F T T F T F F T F F T F T T F T F F F T F F T F T T F F F T T T T F F F F F T F T T F T F F F F T T T T F F F F F F F T T T T T Thus since their columns agree, the two statements are logically equivalent. Solutions 226
7. P ⇒ Q = (P ∧ ∼ Q ) ⇒ (Q ∧ ∼ Q )
P Q ∼Q P∧ ∼ Q Q∧ ∼ Q (P ∧ ∼ Q ) ⇒ (Q ∧ ∼ Q ) P ⇒Q T T F F F T T T F T T F F F F T F F F T T F F T F F T T Thus since their columns agree, the two statements are logically equivalent.
B. Decide whether or not the following pairs of statements are logically equivalent.
9. By DeMorgan’s Law, we have ∼ (∼ P ∨ ∼ Q ) =∼∼ P ∧ ∼∼ Q = P ∧ Q . Thus the
two statements are logically equivalent.
11. (∼ P ) ∧ (P ⇒ Q ) and ∼ (Q ⇒ P )
P Q ∼P P ⇒Q Q⇒P (∼ P ) ∧ (P ⇒ Q ) ∼ (Q ⇒ P ) T T F T T F F T F F F T F F F T T T F T T F F T T T T F The columns for the two statements do not quite agree, thus the two statements are not logically equivalent.
Section 2.7
Write the following as English sentences. Say whether the statements are
true or false.
1. ∀ x ∈ R, x2 > 0
Answer: For every real number x, x2 > 0,
Also: For every real number x, it follows that x2 > 0.
Also: The square of any real number is positive. (etc.)
This statement is FALSE. Reason: 0 is a real number but it’s not true that
02 > 0.
3. ∃ a ∈ R, ∀ x ∈ R, ax = x.
Answer: There exists a real number a for which ax = x for every real number
x.
This statement is TRUE. Reason: Consider a = 1.
5. ∀ n ∈ N, ∃ X ∈ P (N), | X | < n
Answer: For every natural number n, there is a subset X of N with | X | < n.
This statement is TRUE. Reason: Suppose n ∈ N. Let X = . Then | X | = 0 < n.
7. ∀ X ⊆ N, ∃ n ∈ Z, | X | = n
Answer: For any subset X of N, there exists an integer n for which | X | = n. 227
This statement is FALSE. For example, the set X = {2, 4, 6, 8, . . .} of all even
natural numbers is inﬁnite, so there does not exist any integer n for which
| X | = n.
9. ∀ n ∈ Z, ∃ m ∈ Z, m = n + 5
Answer: For every integer n there is another integer m such that m = n + 5.
This statement is TRUE.
Section 2.9
Translate each of the following sentences into symbolic logic.
1. If f is a polynomial and its degree is greater than 2, then f is not constant.
Translation: (P ∧ Q ) ⇒ R , where
P : f is a polynomial,
Q : f has degree greater than 2,
R : f is not constant.
3. If x is prime then x is not a rational number.
Translation: P ⇒∼ Q , where
P : x is prime,
Q : x is a rational number
5. For every positive number ε, there is a positive number δ for which | x − a| < δ
implies | f ( x) − f (a)| < ε.
Translation: ∀ ε ∈ R, ε > 0, ∃ δ ∈ R, δ > 0, (| x − a| < δ) ⇒ (| f ( x) − f (a)| < ε)
7. There exists a real number a for which a + x = x for every real number x.
Translation: ∃a ∈ R, ∀ x ∈ R, a + x = x
9. If x is a rational number and x = 0, then tan( x) is not a rational number.
Translation: (( x ∈ Q) ∧ ( x = 0)) ⇒ (tan( x) ∉ Q)
11. There is a Providence that protects idiots, drunkards, children and the United
States of America.
One translation is as follows. Let R be union of the set of idiots, the set of
drunkards, the set of children, and the set consisting of the USA. Let P be the
open sentence P ( x): x is a Providence. Let S be the open sentence S ( x, y): x
protects y. Then the translation is ∃ x, ∀ y ∈ R , P ( x) ∧ S ( x, y).
(Notice that, although this is mathematically correct, some humor has been
lost in the translation.)
13. Everything is funny as long as it is happening to Somebody Else.
Translation: ∀ x, ∼ H ( x) ⇒ F ( x),
where H ( x): x is happening to me, and F ( x) : x is funny.
Section 2.10
Negate the following sentences.
1. The number x is positive but the number y is not positive.
The “but” can be interpreted as “and.” Using DeMorgan’s Law, the negation is:
The number x is not positive or the number y is positive. Solutions 228 3. For every prime number p there is another prime number q with q > p.
Negation: There is a prime number p such that for every prime number q,
q ≤ p.
Also: There exists a prime number p for which q ≤ p for every prime number q.
(etc.)
5. For every positive number ε there is a positive number M for which | f ( x) − b| < ε
whenever x > M .
To negate this, it may be helpful to ﬁrst write it in symbolic form. The statement
is ∀ε ∈ (0, ∞), ∃ M ∈ (0, ∞), ( x > M ) ⇒ (| f ( x) − b| < ε).
Working out the negation, we have
∼ ∀ε ∈ (0, ∞), ∃ M ∈ (0, ∞), ( x > M ) ⇒ (| f ( x) − b| < ε) = ∃ε ∈ (0, ∞), ∼ ∃ M ∈ (0, ∞), ( x > M ) ⇒ (| f ( x) − b| < ε) = ∃ε ∈ (0, ∞), ∀ M ∈ (0, ∞), ∼ ( x > M ) ⇒ (| f ( x) − b| < ε) . Finally, using the idea from Example 2.14, we can negate the conditional
statement that appears here to get
∃ε ∈ (0, ∞), ∀ M ∈ (0, ∞), ∃ x, ( x > M )∧ ∼ (| f ( x) − b| < ε) . Negation: There exists a positive number ε with the property that for every
positive number M , there is a number x for which x > M and | f ( x) − b| ≥ ε.
7. I don’t eat anything that has a face.
Negation: I will eat some things that have a face.
(Note. If your answer was “I will eat anything that has a face.” then that is
wrong, both morally and mathematically.)
9. If sin( x) < 0, then it is not the case that 0 ≤ x ≤ π.
Negation: There exists a number x for which sin( x) < 0 and 0 ≤ x ≤ π.
11. You can fool all of the people all of the time.
There are several ways to negate this, including:
There is a person that you can’t fool all the time. or
There is a person x and a time y for which x is not fooled at time y.
(But Abraham Lincoln said it better.)
Chapter 3 Exercises
Section 3.1
1. Consider lists made from the letters , , , , , , with repetition allowed.
(a) How many length-4 lists are there? Answer: 6 · 6 · 6 · 6 = 1296.
(b) How many length-4 lists are there that begin with ?
Answer: 1 · 6 · 6 · 6 = 216.
(c) How many length-4 lists are there that do not begin with ?
Answer: 5 · 6 · 6 · 6 = 1080. 229
3. How many ways can you make a list of length 3 from symbols , , , , , if... (a) ... repetition is allowed. Answer: 6 · 6 · 6 = 216.
(b) ... repetition is not allowed. Answer: 6 · 5 · 4 = 120.
(c) ... repetition is not allowed and the list must contain the letter .
Answer: 5 · 4 + 5 · 4 + 5 · 4 = 60.
(d) ... repetition is allowed and the list must contain the letter .
Answer: 6 · 6 · 6 − 5 · 5 · 5 = 91.
(Note: See Example 3.2 if a more detailed explanation is required.)
5. Five cards are dealt oﬀ of a standard 52-card deck and lined up in a row. How
many such line-ups are there in which all ﬁve cards are of the same color? (i.e.
all black or all red.)
There are 26·25·24·23·22 = 7, 893, 600 possible black-card lineups and 26·25·24·23·
22 = 7, 893, 600 possible red-card lineups, so the answer is 7, 893, 600 + 7, 893, 600 =
15, 787, 200.
7. This problems involves 8-digit binary strings such as 10011011 or 00001010.
(i.e. 8-digit numbers composed of 0’s and 1’s.)
(a) How many such strings are there? Answer: 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2· = 256.
(b) How many such strings end in 0? Answer: 2 · 2 · 2 · 2 · 2 · 2 · 2 · 1· = 128.
(c) How many such strings have the property that their second and fourth
digits are 1’s? Answer: 2 · 1 · 2 · 1 · 2 · 2 · 2 · 2· = 64.
(d) How many such strings are such that their second or fourth digits are 1’s?
Answer: These strings can be divided into four types. Type 1 consists of
those strings of form ∗1 ∗ 0 ∗ ∗∗, Type 2 consist of strings of form ∗0 ∗ 1 ∗ ∗∗,
and Type 3 consists of those of form ∗1 ∗ 1 ∗ ∗ ∗ ∗. By the multiplication
principle there are 26 = 64 strings of each type, so there are 3 · 64 = 192
8-digit binary strings whose second or fourth digits are 1’s.
9. This problem concerns 4-letter codes that can be made from the letters , , , , ,
..., of the English Alphabet.
(a) How many such codes can be made? Answer: 26 · 26 · 26 · 26 = 456976
(b) How many such codes are there that have no two consecutive letters the
same?
To answer this we use the Multiplication Principle. There are 26 choices
for the ﬁrst letter. The second letter can’t be the same as the ﬁrst letter, so
there are only 25 choices for it. The third letter can’t be the same as the
second letter, so there are only 25 choices for it. The fourth letter can’t be
the same as the third letter, so there are only 25 choices for it. Thus there
are 26 · 25 · 25 · 25 = 406, 250 codes with no two consecutive letters the
same.
11. This problem concerns lists of length 6 made from the letters , , , , , , , .
How many such lists are possible if repetition is not allowed and the list
contains two consecutive vowels? Solutions 230 Answer: There are just two vowels and to choose from. The lists we want
to make can be divided into ﬁve types. They have one of the forms V V ∗ ∗ ∗ ∗,
or ∗V V ∗ ∗∗, or ∗ ∗ V V ∗ ∗, or ∗ ∗ ∗V V∗, or ∗ ∗ ∗ ∗ V V , where V indicates a
vowel and ∗ indicates a consonant. By the Multiplication Principle, there are
2 · 1 · 6 · 5 · 4 · 3 = 720 lists of form V V ∗∗∗∗. In fact, that for the same reason there
are 720 lists of each form. Thus the answer to the question is 5 · 720 = 3600
Section 3.2
1. What is the smallest n for which n! has more than 10 digits? Answer n = 14.
3. How many 5-digit positive integers are there in which there are no repeated
digits and all digits are odd? Answer: 5! = 120.
5. Using only pencil and paper, ﬁnd the value of
Answer: 120! = 120·119·118! = 120 · 119 = 14, 280.
118!
118! 120!
118! . 7. How many 9-digit numbers can be made from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 if
repetition is not allowed and all the odd digits occur ﬁrst (on the left) followed
by all the even digits?
Answer: 5!4! = 2880.
Section 3.3
1. Suppose a set A has 37 elements. How many subsets of A have 10 elements?
How many subsets have 30 elements? How many have 0 elements?
Answers: 37 = 348,330,136; 37 = 10,295,472; 307 = 1.
10
30
3. A set X has exactly 56 subsets with 3 elements. What is the cardinality of X ?
The answer will be n, where n = 56. After some trial and error, you will
3
discover 8 = 56, so | X | = 8.
3
5. How many 16-digit binary strings contain exactly seven 1’s?
Answer: Make such a string as follows. Start with a list of 16 blank spots.
Choose 7 of the blank spots for the 1’s and put 0’s in the other spots. There
are 176 = 11400 ways to do this.
7. |{ X ∈ P ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}) : | X | < 4}| = 10
0 + 10
1 + 10
2 + 10
3 = 1 + 10 + 45 + 120 = 176. 9. This problem concerns lists of length six made from the letters , , , , , ,
without repetition. How many such lists have the property that the occurs
before the ?
Answer: Make such a list as follows. Begin with six blank spaces and select
two of these spaces. Put the in the ﬁrst selected space and the in the
second. There are 6 = 15 ways of doing this. For each of these 15 choices
2
there are 4! = 24 ways of ﬁlling in the remaining spaces. Thus the answer to
the question is 15 × 24 = 360 such lists. 231
11. How many 10-digit integers contain no 0’s and exactly three 6’s?
Answer: Make such a number as follows: Start with 10 blank spaces and choose
three of these spaces for the 6’s. There are 130 = 120 ways of doing this. For
each of these 120 choices we can ﬁll in the remaining seven blanks with choices
from the digits 1, 2, 3, 4, 5, 7, 8, 9, and there are 87 to do this. Thus the answer to
the question is 130 · 87 = 251, 658, 240.
13. Assume n, k ∈ N with k ≤ n. Then n
k = n!
( n− k)! k! = n!
k!( n− k)! = n!
( n−( n− k))!( n− k)! = n
n− k . Section 3.4
1. Write out Row 11 of Pascal’s triangle.
Answer: 1 11 55 165 330 462 462 330 165 55 11 1 8 13 3. Use the Binomial Theorem to ﬁnd the coeﬃcient of x in ( x + 2) .
Answer: According to the Binomial Theorem, the coeﬃcient of x8 y5 in ( x + y)13
is 183 x8 y5 = 1287 x8 y5 . Now plug in y = 2 to get the ﬁnal answer of 41184 x8 .
5. Use the Binomial Theorem to show n=0 n = 2n . Hint: Observe that 2n = (1+1)n .
k
k
Now use the Binomial Theorem to work out ( x + y)n and plug in x = 1 and y = 1.
7. Use the Binomial Theorem to show n=0 3k n = 4n .
k
k
Hint: Observe that 4n = (1 + 3)n . Now look at the hint for the previous problem.
9. Use the Binomial Theorem to show n − n + n − n + n − n + . . . ±
0
3
5
1
2
4
Hint: Observe that 0 = 0n = (1 + (−1))n . Now use the Binomial Theorem. n
n = 0. 11. Use the Binomial Theorem to show 9n = n=0 (−1)k n 10n−k .
k
k
Hint: Observe that 9n = (10 + (−1))n . Now use the Binomial Theorem.
13. Assume n ≥ 3. Then n
3 = n−1
3 + n−1
2 = n−2
3 + n−2
2 + n−1
2 = ··· = 2
2 + 3
2 + · · ·+ n−1
2 . Section 3.5
1. At a certain university 523 of the seniors are history majors or math majors
(or both). There are 100 senior math majors, and 33 seniors are majoring in
both history and math. How many seniors are majoring in history?
Answer: Let A be the set of senior math majors and B be the set of senior
history majors. From | A ∪ B| = | A | + |B| − | A ∩ B| we get 523 = 100 + |B| − 33, so
|B| = 523 + 33 − 100 = 456. There are 456 history majors.
3. How many 4-digit positive integers are there that are even or contain no 0’s?
Answer: Let A be the set of 4-digit even positive integers, and let B be the
set of 4-digit positive integers that contain no 0’s. We seek | A ∪ B|. By the
Multiplication Principle | A | = 9 · 10 · 10 · 5 = 4500. (Note the ﬁrst digit cannot be 0
and the last digit must be even.) Also |B| = 9·9·9·9 = 6561. Further, A ∩ B consists
of all even 4-digit integers that have no 0’s. It follows that | A ∩ B| = 9 · 9 · 9 · 4 = 2916.
Then the answer to our question is | A ∪ B| = | A |+|B|−| A ∩ B| = 4500 + 6561 − 2916 =
8145. Solutions 232 5. How many 7-digit binary strings begin in 1 or end in 1 or have exactly four 1’s?
Answer: Let A be the set of such strings that begin in 1. Let B be the set of such
strings that end in 1. Let C be the set of such strings that have exactly four 1’s.
Then the answer to our question is | A ∪ B ∪ C |. Using Equation (3.4) to compute
this number, we have | A ∪ B ∪ C | = | A |+|B|+|C |−| A ∩ B|−| A ∩ C |−|B ∩ C |+| A ∩ B ∩ C | =
26 + 26 + 7
4 − 25 − 6
3 − 6
3 + 5
2 = 64 + 64 + 35 − 32 − 20 − 20 + 10 = 85. 7. This problem concerns 4-card hands dealt oﬀ of a standard 52-card deck. How
many 4-card hands are there for which all four cards are of the same suit or
all four cards are red?
Answer: Let A be the set of 4-card hands for which all four cards are of the
same suit. Let B be the set of 4-card hands for which all four cards are red.
Then A ∩ B is the set of 4-card hands for which the four cards are either all
hearts or all diamonds. The answer to our question is | A ∪ B| = | A |+|B|−| A ∩ B| =
4 13
4 + 26
4 −2 13
4 =2 13
4 + 26
4 = 1430 + 14950 = 16380. 9. A 4-letter list is made from the letters , , , , , according to the following
rule: Repetition is allowed, and the ﬁrst two letters on the list are vowels or
the list ends in .
Answer: Let A be the set of such lists for which the ﬁrst two letters are
vowels, so | A | = 2 · 2 · 6 · 6 = 144. Let B be the set of such lists that end in D, so
|B| = 6 · 6 · 6 · 1 = 216. Then A ∩ B is the set of such lists for which the ﬁrst two
entries are vowels and the list ends in . Thus | A ∩ B| = 2 · 2 · 6 · 1 = 24. The
answer to our question is | A ∪ B| = | A | + |B| − | A ∩ B| = 144 + 216 − 24 = 336.
Chapter 4 Exercises
1. If x is an even integer, then x2 is even.
Proof. Suppose x is even. Thus x = 2a for some a ∈ Z.
Consequently x2 = (2a)2 = 4a2 = 2(2a2 ).
Therefore x2 = 2b, where b is the integer 2a2 .
Thus x2 is even by deﬁnition of an even number.
3. If a is an odd integer, then a2 + 3a + 5 is odd.
Proof. Suppose a is odd.
Thus a = 2 c + 1 for some integer c, by deﬁnition of an odd number.
Then a2 + 3a + 5 = (2 c + 1)2 + 3(2 c + 1) + 5 = 4 c2 + 4 c + 1 + 6 c + 3 + 5 = 4 c2 + 10 c + 9
= 4 c2 + 10 c + 8 + 1 = 2(2 c2 + 5 c + 4) + 1.
This shows a2 + 3a + 5 = 2b + 1, where b = 2 c2 + 5 c + 4 ∈ Z.
Therefore a2 + 3a + 5 is odd.
5. Suppose x, y ∈ Z. If x is even, then x y is even.
Proof. Suppose x, y ∈ Z and x is even.
Then x = 2a for some integer a, by deﬁnition of an even number.
Thus x y = (2a)( y) = 2(a y).
Therefore x y = 2b where b is the integer a y, so x y is even. 233
7. Suppose a, b ∈ Z. If a | b, then a2 | b2 .
Proof. Suppose a | b.
By deﬁnition of divisibility, this means b = ac for some integer c.
Squaring both sides of this equation produces b2 = a2 c2 .
Then b2 = a2 d , where d = c2 ∈ Z.
By deﬁnition of divisibility, this means a2 | b2 .
9. Suppose a is an integer. If 7 | 4a, then 7 | a.
Proof. Suppose 7 | 4a.
By deﬁnition of divisibility, this means 4a = 7 c for some integer c.
Since 4a = 2(2a) it follows that 4a is even, and since 4a = 7 c, we know 7 c is even.
But then c can’t be odd, because that would make 7 c odd, not even.
Thus c is even, so c = 2d for some integer d .
Now go back to the equation 4a = 7 c and plug in c = 2d . We get 4a = 14 d .
Dividing both sides by 2 gives 2a = 7d .
Now, since 2a = 7d , it follows that 7d is even, and thus d cannot be odd.
Then d is even, so d = 2 e for some integer e.
Plugging d = 2 e back into 2a = 7d gives 2a = 14 e.
Dividing both sides of 2a = 14 e by 2 produces a = 7 e.
Finally, the equation a = 7 e means that 7 | a, by deﬁnition of divisibility.
11. Suppose a, b, c, d ∈ Z. If a | b and c | d , then ac | bd .
Proof. Suppose a | b and c | d .
As a | b, the deﬁnition of divisibility means there is an integer x for which b = ax.
As c | d , the deﬁnition of divisibility means there is an integer y for which d = c y.
Since b = ax, we can multiply one side of d = c y by b and the other by ax.
This gives bd = axc y, or bd = (ac)( x y).
Since x y ∈ Z, the deﬁnition of divisibility applied to bd = (ac)( x y) gives ac | bd .
13. Suppose x, y ∈ R. If x2 + 5 y = y2 + 5 x, then x = y or x + y = 5.
Proof. Suppose x2 + 5 y = y2 + 5 x.
Then x2 − y2 = 5 x − 5 y, and factoring gives ( x − y)( x + y) = 5( x − y).
Now consider two cases.
Case 1. If x − y = 0 we can divide both sides of ( x − y)( x + y) = 5( x − y) by the
non-zero quantity x − y to get x + y = 5.
Case 2. If x − y = 0, then x = y. (By adding y to both sides.)
Thus x = y or x + y = 5.
15. If n ∈ Z, then n2 + 3n + 4 is even.
Proof. Suppose n ∈ Z. We consider two cases.
Case 1. Suppose n is even. Then n = 2a for some a ∈ Z.
Therefore n2 + 3n + 4 = (2a)2 + 3(2a) + 4 = 4a2 + 6a + 4 = 2(2a2 + 3a + 2). Solutions 234 So n2 + 3 n + 4 = 2b where b = 2a2 + 3a + 2 ∈ Z, so n2 + 3n + 4 is even.
Case 2. Suppose n is odd. Then n = 2a + 1 for some a ∈ Z.
Therefore n2 + 3n + 4 = (2a + 1)2 + 3(2a + 1) + 4 = 4a2 + 4a + 1 + 6a + 3 + 4 = 4a2 + 10a + 8
= 2(2a2 + 5a + 4). So n2 + 3 n + 4 = 2 b where b = 2a2 + 5a + 4 ∈ Z, so n2 + 3 n + 4 is even.
In either case n2 + 3 n + 4 is even.
17. If two integers have opposite parity, then their product is even.
Proof. Suppose a and b are two integers with opposite parity. Thus one is even
and the other is odd. Without loss of generality, suppose a is even and b is
odd. Therefore there are integers c and d for which a = 2 c and b = 2 d + 1. Then
the product of a and b is ab = 2 c(2d + 1) = 2(2 cd + c). Therefore ab = 2k where
k = 2 cd + c ∈ Z. Therefore the product ab is even.
19. Suppose a, b, c ∈ Z. If a2 | b and b3 | c then a6 | c.
Proof. Since a2 | b we have b = ka2 for some k ∈ Z. Since b3 | c we have c = hb3
for some h ∈ Z. Thus c = h(ka2 )3 = hk3 a6 . Hence a6 | c.
21. If p is prime and 0 < k < p then p |
Proof. From the formula p
k = p
k . p!
( p− k)! k! , p! = we get p
( p − k)! k!.
k Now, since the prime number p is a factor of p! on the left, it must also be a
p
factor of k ( p − k)!k! on the right. Thus the prime number p appears in the
p
prime factorization of k ( p − k)! k!.
Now, k! is a product of numbers smaller than p, so its prime factorization
contains no p’s. Similarly the prime factorization of ( p − k)! contains no p’s.
p
But we noted that the prime factorization of k ( p − k)!k! must contain a p, so it
p
p
follows that the prime factorization of k contains a p. Thus k is a multiple
p
of p, so p divides k .
23. If n ∈ N then 2n
n is even. n
Proof. By deﬁnition, 2n is the number of n-element subsets of a set A with 2n
elements. For each subset X ⊆ A with | X | = n, the complement X is a diﬀerent
set, but it also has 2 n − n = n elements. Imagine listing out all the n-elements
subset of a set A . It could be done in such a way that the list has form X1, X1, X2, X2, X3, X3, X4, X4, X5, X5 . . . This list has an even number of items, for they are grouped in pairs. Thus
is even. 2n
n 235
25. If a, b, c ∈ N and c ≤ b ≤ a then a
b b
c = a
b− c a− b + c
c . Proof. Assume a, b, c ∈ N with c ≤ b ≤ a. Then we have
(a− b+ c)!
a!
(a− b+ c)!(a− b)! ( b− c)! c! = (a− b+ c)!
a!
( b− c)!(a− b+ c)! (a− b)! c! = a
b− c a− b + c
c a
b b
c = a!
b!
(a− b)! b! ( b− c)! c! = . Chapter 5 Exercises
1. Proposition Suppose n ∈ Z. If n2 is even, then n is even.
Proof. (Contrapositive) Suppose n is not even. Then n is odd, so n = 2a + 1 for
some integer a, by deﬁnition of an odd number. Thus n2 = (2a + 1)2 = 4a2 + 4a + 1 =
2(2a2 + 2a) + 1. Consequently n2 = 2 b + 1, where b is the integer 2a2 + 2a, so n2 is
odd. Therefore n2 is not even.
3. Proposition Suppose a, b ∈ Z. If a2 (b2 − 2b) is odd, then a and b are odd.
Proof. (Contrapositive) Suppose it is not the case that a and b are odd. Then,
by DeMorgan’s Law, at least one of a and b is even. Let us look at these cases
separately.
Case 1. Suppose a is even. Then a = 2 c for some integer c. Thus a2 (b2 − 2 b)
= (2 c)2 ( b2 − 2 b) = 2(2 c2 ( b2 − 2 b)), which is even.
Case 2. Suppose b is even. Then b = 2 c for some integer c. Thus a2 (b2 − 2 b)
= a2 ((2 c)2 − 2(2 c)) = 2(a2 (2 c2 − 2 c)), which is even.
(A third case involving a and b both even is unnecessary, for either of the two
cases above cover this case.) Thus in either case a2 (b2 − 2b) is even, so it is not
odd.
5. Proposition Suppose x ∈ R. If x2 + 5 x < 0 then x < 0.
Proof. (Contrapositive) Suppose it is not the case that x < 0, so x ≥ 0. Then
neither x2 nor 5 x is negative, so x2 + 5 x ≥ 0. Thus it is not true that x2 + 5 x < 0.
7. Proposition Suppose a, b ∈ Z. If both ab and a + b are even, then both a and
b are even.
Proof. (Contrapositive) Suppose it is not the case that both a and b are even.
Then at least one of them is odd. There are three cases to consider.
Case 1. Suppose a is even and b is odd. Then there are integers c and
d for which a = 2 c and b = 2 d + 1. Then ab = 2 c(2 d + 1), which is even; and
a + b = 2 c + 2 d + 1 = 2( c + d ) + 1, which is odd. Thus it is not the case that both
ab and a + b are even.
Case 2. Suppose a is odd and b is even. Then there are integers c and d for
which a = 2 c + 1 and b = 2d . Then ab = (2 c + 1)(2 d ) = 2( d (2 c + 1)), which is even;
and a + b = 2 c + 1 + 2d = 2( c + d ) + 1, which is odd. Thus it is not the case that
both ab and a + b are even.
Case 3. Suppose a is odd and b is odd. Then there are integers c and d for 236 Solutions which a = 2 c + 1 and b = 2d + 1. Then ab = (2 c + 1)(2d + 1) = 4 cd + 2 c + 2d + 1 =
2(2 cd + c + d ) + 1, which is odd; and a + b = 2 c + 1 + 2 d + 1 = 2( c + d + 1), which is
even. Thus it is not the case that both ab and a + b are even.
These cases show that it is not the case that ab and a + b are both even. (Note
that unlike Exercise 3 above, we really did need all three cases here, for each
case involved speciﬁc parities for both a and b.)
9. Proposition Suppose n ∈ Z. If 3 n2 , then 3 n.
Proof. (Contrapositive) Suppose it is not the case that 3 n, so 3 | n. This means
that n = 3a for some integer a. Consequently n2 = 9a2 , from which we get
n2 = 3(3a2 ). This shows that there in an integer b = 3a2 for which n2 = 3 b, which
means 3 | n2 . Therefore it is not the case that 3 n2 .
11. Proposition Suppose x, y ∈ Z. If x2 ( y + 3) is even, then x is even or y is odd.
Proof. (Contrapositive) Suppose it is not the case that x is even or y is odd.
Using DeMorgan’s Law, this means x is not even and y is not odd, which is to
say x is odd and y is even. Thus there are integers a and b for which x = 2a + 1
and y = 2 b. Consequently x2 ( y + 3) = (2a + 1)2 (2 b + 3) = (4a2 + 4a + 1)(2b + 3) =
8a2 b + 8ab + 2 b + 12a2 + 12a + 3 = 8a2 b + 8ab + 2 b + 12a2 + 12a + 2 + 1 =
2(4a2 b + 4ab + b + 6a2 + 6a + 1) + 1. This shows x2 ( y + 3) = 2 c + 1 for c = 4a2 b + 4ab +
b + 6a2 + 6a + 1 ∈ Z. Consequently, x2 ( y + 3) is not even. 13. Proposition Suppose x ∈ R. If x5 + 7 x3 + 5 x ≥ x4 + x2 + 8, then x ≥ 0.
Proof. (Contrapositive) Suppose it is not true that x ≥ 0. Then x < 0, that is
x is negative. Consequently, the expressions x5 , 7 x3 and 5 x are all negative
(note the odd powers) so x5 + 7 x3 + 5 x < 0. Similarly the terms x4 , x2 , and 8
are all positive (note the even powers), so 0 < x4 + x2 + 8. From this we get
x5 + 7 x3 + 5 x < x4 + x2 + 8, so it is not true that x5 + 7 x3 + 5 x ≥ x4 + x2 + 8.
15. Proposition Suppose x ∈ Z. If x3 − 1 is even, then x is odd.
Proof. (Contrapositive) Suppose x is not odd. Thus x is even, so x = 2a for some
integer a. Then x3 − 1 = (2a)3 − 1 = 8a3 − 1 = 8a3 − 2 + 1 = 2(4a3 − 1) + 1. Therefore
x3 − 1 = 2 b + 1 where b = 4a3 − 1 ∈ Z, so x3 − 1 is odd. Thus x3 − 1 is not even.
17. Proposition If n is odd, then 8 |(n2 − 1).
Proof. (Direct) Suppose n is odd, so n = 2a + 1 for some integer a. Then n2 − 1 =
(2a + 1)2 − 1 = 4a2 + 4a = 4(a2 + a) = 4a(a + 1). So far we have n2 − 1 = 4a(a + 1), but
we want a factor of 8, not 4. But notice that one of a or a + 1 must be even, so
a(a + 1) is even and hence a(a + 1) = 2 c for some integer c. Now we have n2 − 1 =
4a(a + 1) = 4(2 c) = 8 c. But n2 − 1 = 8 c means 8 |( n2 − 1). 237
19. Proposition Let a, b ∈ Z and n ∈ N. If a ≡ b (mod n) and a ≡ c (mod n), then
c ≡ b (mod n).
Proof. (Direct) Suppose a ≡ b (mod n) and a ≡ c (mod n).
This means n |(a − b) and n |(a − c).
Thus there are integers d and e for which a − b = nd and a − c = ne.
Subtracting the second equation from the ﬁrst gives c − b = nd − ne.
Thus c − b = n( d − e), so n |( c − b) by deﬁnition of divisibility.
Therefore c ≡ b (mod n) by deﬁnition of congruence modulo n.
21. Proposition Let a, b ∈ Z and n ∈ N. If a ≡ b (mod n), then a3 ≡ b3 (mod n).
Proof. (Direct) Suppose a ≡ b (mod n). This means n |(a − b), so there is an
integer c for which a − b = nc. Now multiply both sides of this equation by
a2 + ab + b2 .
a−b
2 2 = nc = nc(a2 + ab + b2 ) 3 = nc(a2 + ab + b2 ) a3 − b 3 = nc(a2 + ab + b2 ) (a − b)(a + ab + b )
3 2 2 2 2 a + a b + ab − ba − ab − b Since a2 + ab + b2 ∈ Z, the equation a3 − b3 = nc(a2 + ab + b2 ) implies n |(a3 − b3 ),
and therefore a3 ≡ b3 (mod n).
23. Proposition Let a, b, c ∈ Z and n ∈ N. If a ≡ b (mod n), then ca ≡ cb (mod n).
Proof. (Direct) Suppose a ≡ b (mod n). This means n |(a− b), so there is an integer
d for which a − b = nd . Multiply both sides of this by c to get ac − bc = ndc.
Consequently, there is an integer e = dc for which ac − bc = ne, so n |(ac − bc) and
consequently ac ≡ bc (mod n).
25. If n ∈ N and 2n − 1 is prime, then n is prime.
Proof. Assume n is not prime. Write n = ab for some a, b > 1. Then 2n − 1 =
2ab − 1 = (2b − 1)(2a(b−1) + 2a(b−2) + · · · + 1). Hence 2n − 1 is composite.
27. If a ≡ 0 (mod 4) or a ≡ 1 (mod 4) then a
2 is even. −
Proof. We prove this directly. Assume a ≡ 0 (mod 4). Then a = a(a2 1) . Since
2
k
a = 4 k for some k ∈ N, we have a = 4k(42 −1) = 2 k(4 k − 1). Hence a is even.
2
2
1)(4
Now assume a ≡ 1 (mod 4). Then a = 4k + 1 for some k ∈ N. Hence a = (4k−2 k) =
2
a
2 k(4 k − 1). Hence, 2 is even. This proves the result. Solutions 238
Chapter 6 Exercises
1. Suppose n is an integer. If n is odd, then n2 is odd. Proof. Suppose for the sake of contradiction that n is odd and n2 is not odd.
Then n2 is even. Now, since n is odd, we have n = 2a + 1 for some integer a.
Thus n2 = (2a + 1)2 = 4a2 + 4a + 1 = 2(2a2 + 2a) + 1. This shows n2 = 2 b + 1, where
b is the integer b = 2a2 + 2a. Therefore we have n2 is odd and n2 is even, a
contradiction.
3. Prove that 3 2 is irrational. Proof. Suppose for the sake of contradiction that 3 2 is not irrational. Therefore
it is rational, so there exist integers a and b for which 3 2 = a . Let us assume
b
that this fraction is reduced, so a and b are not both even. Now we have
3
3
33
2 = a , which gives 2 = a3 , or 2 b3 = a3 . From this we see that a3 is even,
b
b
from which we deduce that a is even. (For if a were odd, then a3 = (2 c + 1)3 =
8 c3 + 12 c2 + 6 c + 1 = 2(4 c3 + 6 c2 + 3 c) + 1 would be odd, not even.) Since a is even,
it follows that a = 2 d for some integer d . The equation 2b3 = a3 from above then
becomes 2 b3 = (2d )3 , or 2b3 = 8 d 3 . Dividing by 2, we get b3 = 4d 3 , and it follows
that b3 is even. Thus b is even also. (Using the same argument we used when
a3 was even.) At this point we have discovered that both a and b are even,
and this contradicts the fact (observed above) that the a and b are not both
even.
Here is an alternative proof.
Proof. Suppose for the sake of contradiction that 3 2 is not irrational. Therefore
3
there exist integers a and b for which 3 2 = a . Cubing both sides, we get 2 = a3 .
b
b
From this, a3 = b3 + b3 , which contradicts Fermat’s Last Theorem.
5. Prove that 3 is irrational. Proof. Suppose for the sake of contradiction that 3 is not irrational. Therefore
it is rational, so there exist integers a and b for which 3 = a . Let us assume
b
that this fraction is reduced, so a and b have no common factor. Notice that
2
2
2
3 = a , so 3 = a2 , or 3 b2 = a2 . This means 3 | a2 .
b
b
Now we are going to show that if a ∈ Z and 3 | a2 , then 3 | a. (This is a proofwithin-a-proof.) We will use contrapositive proof to prove this conditional
statement. Suppose 3 a. Then there is a remainder of either 1 or 2 when 3 is
divided into a.
Case 1. There is a remainder of 1 when 3 is divided into a Then a = 3m + 1
for some integer m. Consequently, a2 = 9m2 + 6m + 1 = 3(3m2 + 2m) + 1, and this
means 3 divides into a2 with a remainder of 1. Thus 3 a2 .
Case 2. There is a remainder of 2 when 3 is divided into a Then a = 3m + 2
for some integer m. Consequently, a2 = 9m2 + 12 m + 4 = 9 m2 + 12m + 3 + 1 = 239
3(3 m2 + 4 m + 1) + 1, and this means 3 divides into a2 with a remainder of 1. Thus
3 a2 .
In either case we have 3 a2 , so we’ve shown 3 a implies 3 a2 . Therefore, if
3 | a2 , then 3 | a.
Now go back to 3 | a2 in the ﬁrst paragraph. This combined with the result of
the second paragraph implies 3 | a, so a = 3d for some integer d . Now also in the
ﬁrst paragraph we had 3b2 = a2 , which now becomes 3b2 = (3d )2 or 3b2 = 9d 2 , so
b2 = 3 d 2 . But this means 3 | b2 , and the second paragraph implies 3 | b. Thus we
have concluded that 3 | a and 3 | b, but this contradicts the fact that the fraction
a
b is reduced. 7. If a, b ∈ Z, then a2 − 4b − 3 = 0.
Proof. Suppose for the sake of contradiction that a, b ∈ Z but a2 − 4b − 3 = 0. Then
we have a2 = 4b + 3 = 2(2b + 1) + 1, which means a2 is odd. Therefore a is odd also,
so a = 2 c + 1 for some integer c. Plugging this back into a2 − 4b − 3 = 0 gives us
(2 c + 1)2 − 4 b − 3
2 = 0 4 c + 4 c + 1 − 4b − 3 = 0 4 c2 + 4 c − 4 b = 2 2 c2 + 2 c − 2 b = 1 = 1 2 2( c + c − b) From this last equation, we conclude that 1 is an even number, a contradiction.
9. Suppose a, b ∈ R and a = 0. If a is rational and ab is irrational, then b is
irrational.
Proof. Suppose for the sake of contradiction that a is rational and ab is irrational and b is not irrational. Thus we have a and b rational, and ab irrational.
c
Since a and b are rational, we know there are integers c, d , e, f for which a = d
e
ce
and b = f . Then ab = d f , and since both ce and d f are integers, it follows
that ab is rational. But this is a contradiction because we started out with ab
irrational.
11. There exist no integers a and b for which 18a + 6b = 1.
Proof. Suppose for the sake of contradiction that there do exist integers a
and b for which 18a + 6b = 1. Then 1 = 2(9a + 3b), which means 1 is even, a
contradiction.
13. For every x ∈ [π/2, π], sin x − cos x ≥ 1.
Proof. Suppose for the sake of contradiction that x ∈ [π/2, π], but sin x − cos x < 1.
Since x ∈ [π/2, π], we know sin x ≥ 0 and cos x ≤ 0, so sin x − cos x ≥ 0. Therefore 240 Solutions we have 0 ≤ sin x − cos x < 1. Now the square of any number between 0 and
1 is still a number between 0 and 1, so we have 0 ≤ (sin x − cos x)2 < 1, or 0 ≤
sin2 x − 2 sin x cos x + cos2 x < 1. Using the fact that sin2 x + cos2 x = 1, this becomes
0 ≤ −2 sin x cos x + 1 < 1. Subtracting 1, we obtain −2 sin x cos x < 0. But above we
remarked that sin x ≥ 0 and cos x ≤ 0, and hence −2 sin x cos x ≥ 0. We now have
the contradiction −2 sin x cos x < 0 and −2 sin x cos x ≥ 0.
15. If b ∈ Z and b k for every k ∈ N, then b = 0.
Proof. Suppose for the sake of contradiction that b ∈ Z and b k for every k ∈ N,
but b = 0.
Case 1. Suppose b > 0. Then b ∈ N, so b| b, contradicting b k for every k ∈ N.
Case 2. Suppose b < 0. Then −b ∈ N, so b|(−b), again a contradiction
17. For every n ∈ Z, 4 (n2 + 2).
Proof. Assume there exists n ∈ Z with 4 | (n2 + 2). Then for some k ∈ Z, 4k = n2 + 2
or 2 k = n2 + 2(1 − k). If n is odd, this means 2k is odd, and we’ve reached a
contradiction. If n is even then n = 2 j and we get k = 2 j 2 + 1 − k for some j ∈ Z.
Hence 2(k − j 2 ) = 1, so 1 is even, a contradiction.
Remark. It is fairly easy to see that two more than a perfect square is always
either 2 (mod 4) or 3 (mod 4). This would end the proof immediately.
19. The product of 5 consecutive integers is a multiple of 120.
Proof. Given any collection of 5 consecutive integers, at least one must be a
multiple of two, at least one must be a mulitple of three, at least one must be
a multiple of four and at least one must be a multiple of 5. Hence the product
is a multiple of 5 · 4 · 3 · 2 = 120. In particular, the product is a multiple of 60.
21. Hints for Exercises 20–23. For Exercises 20, ﬁrst show that the equation
a2 + b2 = 3 c2 has no solutions (other than the trivial solution (a, b, c) = (0, 0, 0))
in the integers. To do this, investigate the remainders of a sum of squares
(mod 4). After you’ve done this, prove that the only solution is indeed the trivial
solution.
Now, assume that the equation x2 + y2 − 3 = 0 has a rational solution. Use the
deﬁnition of rational numbers to yield a contradiction. Chapter 7 Exercises
1. Suppose x ∈ Z. Then x is even if and only if 3 x + 5 is odd.
Proof. We ﬁrst use direct proof to show that if x is even, then 3 x + 5 is odd.
Suppose x is even. Then x = 2n for some integer n. Thus 3 x + 5 = 3(2n) + 5 =
6 n + 5 = 6 n + 4 + 1 = 2(3 n + 2) + 1. Thus 3 x + 5 is odd because it has form 2 k + 1,
where k = 3n + 2 ∈ Z. 241
Conversely, we need to show that if 3 x + 5 is odd, then x is even. We will
prove this using contrapositive proof. Suppose x is not even. Then x is odd, so
x = 2 n + 1 for some integer n. Thus 3 x + 5 = 3(2 n + 1) + 5 = 6 n + 8 = 2(3 n + 4). This
means says 3 x + 5 is twice the integer 3 n + 4, so 3 x + 5 is even, not odd.
3. Given an integer a, then a3 + a2 + a is even if and only if a is even.
Proof. First we will prove that if a3 + a2 + a is even then a is even. This is done
with contrapositive proof. Suppose a is not even. Then a is odd, so there is an
integer n for which a = 2n + 1. Then
a3 + a2 + a = (2 n + 1)3 + (2 n + 1)2 + (2 n + 1) = 8 n3 + 12 n2 + 6 n + 1 + 4 n2 + 4 n + 1 + 2 n + 1 = 8 n3 + 16 n2 + 12 n + 2 + 1 = 2(4 n3 + 8 n2 + 6 n + 1) + 1. This expresses a3 + a2 + a as twice an integer plus 1, so a3 + a2 + a is odd, not
even. We have now shown that if a3 + a2 + a is even then a is even.
Conversely, we need to show that if a is even, then a3 + a2 + a is even. We will use
direct proof. Suppose a is even, so a = 2n for some integer n. Then a3 + a2 + a =
(2 n)3 + (2 n)2 + 2 n = 8 n3 + 4 n2 + 2 n = 2(4 n3 + 2 n2 + n). Therefore, a3 + a2 + a is even
because it’s twice an integer.
5. An integer a is odd if and only if a3 is odd.
Proof. Suppose that a is odd. Then a = 2n + 1 for some integer n, and a3 =
(2 n + 1)3 = 8 n3 + 12 n2 + 6 n + 1 = 2(4 n3 + 6 n2 + 3 n) + 1. This shows that a3 is twice
an integer, plus 1, so a3 is odd. Thus we’ve proved that if a is odd then a3 is
odd.
Conversely we need to show that if a3 is odd, then a is odd. For this we employ
contrapositive proof. Suppose a is not odd. Thus a is even, so a = 2n for some
integer n. Then a3 = (2 n)3 = 8n3 = 2(4n3 ) is even (not odd).
7. Suppose x, y ∈ R. Then ( x + y)2 = x2 + y2 if and only if x = 0 or y = 0.
Proof. First we prove with direct proof that if ( x + y)2 = x2 + y2 , then x = 0 or
y = 0. Suppose ( x + y)2 = x2 + y2 . From this we get x2 + 2 x y + y2 = x2 + y2 , so 2 x y = 0,
and hence x y = 0. Thus x = 0 or y = 0.
Conversely, we need to show that if x = 0 or y = 0, then ( x + y)2 = x2 + y2 . This
will be done with cases.
Case 1. If x = 0 then ( x + y)2 = (0 + y)2 = y2 = 02 + y2 = x2 + y2 .
Case 2. If y = 0 then ( x + y)2 = ( x + 0)2 = x2 = x2 + 02 = x2 + y2 .
Either way, we have ( x + y)2 = x2 + y2 . 242 Solutions 9. Suppose a ∈ Z. Prove that 14 | a if and only if 7 | a and 2 | a.
Proof. First we prove that if 14 | a, then 7 | a and 2 | a. Direct proof is used.
Suppose 14 | a. This means a = 14m for some integer m. Therefore a = 7(2m),
which means 7 | a, and also a = 2(7m), which means 2 | a. Thus 7 | a and 2 | a.
Conversely, we need to prove that if 7 | a and 2 | a, then 14 | a. Once again direct
proof if used. Suppose 7 | a and 2 | a. Since 2 | a it follows that a = 2m for some
integer m, and that in turn implies that a is even. Since 7 | a it follows that
a = 7 n for some integer n. Now, since a is known to be even, and a = 7 n, it
follows that n is even (if it were odd, then a = 7 n would be odd). Thus n = 2 p for
an appropriate integer p, and plugging n = 2 p back into a = 7n gives a = 7(2 p),
so a = 14 p. Therefore 14 | a.
11. Suppose a, b ∈ Z. Prove that (a − 3) b2 is even if and only if a is odd or b is even.
Proof. First we will prove that if (a − 3) b2 is even, then a is odd or b is even.
For this we use contrapositive proof. Suppose it is not the case that a is
odd or b is even. Then by DeMorgan’s law, a is even and b is odd. Thus
there are integers m and n for which a = 2 m and b = 2n + 1. Now observe
(a −3) b2 = (2 m −3)(2 n +1)2 = (2 m −3)(4 n2 +4 n +1) = 8 mn2 +8 mn +2 m −12 n2 −12 n −3 =
8 mn2 + 8 mn + 2 m − 12 n2 − 12 n − 4 + 1 = 2(4 mn2 + 4 mn + m − 6 n2 − 6 n − 2) + 1. This
shows (a − 3)b2 is odd, so it’s not even. Conversely, we need to show that if a is odd or b is even, then (a − 3)b2 is even.
For this we use direct proof, with cases.
Case 1. Suppose a is odd. Then a = 2m + 1 for some integer m. Thus (a − 3) b2 =
(2 m + 1 − 3) b2 = (2 m − 2) b2 = 2( m − 1) b2 . Thus in this case (a − 3) b2 is even.
Case 2. Suppose b is even. Then b = 2n for some integer n. Thus (a − 3) b2 =
(a − 3)(2 n)2 = (a − 3)4 n2 = 2(a − 3)2 n2 =. Thus in this case (a − 3) b2 is even.
Therefore, in any event, (a − 3) b2 is even.
13. Suppose a, b ∈ Z. If a + b is odd, then a2 + b2 is odd.
Hint: Use direct proof. Suppose a + b is odd. Argue that this means a and b
have opposite parity. Then use cases.
15. Suppose a, b ∈ Z. Prove that a + b is even if and only if a and b have the same
parity.
Proof. First we will show that if a + b is even, then a and b have the same parity.
For this we use contrapositive proof. Suppose it is not the case that a and b
have the same parity. Then one of a and b is even and the other is odd. Without
loss of generality, let’s say that a is even and b is odd. Thus there are integers
m and n for which a = 2 m and b = 2 n + 1. Then a + b = 2 m + 2 n + 1 = 2( m + n) + 1,
so a + b is odd, not even.
Conversely, we need to show that if a and b have the same parity, then a + b is
even. For this, we use direct proof with cases. Suppose a and b have the same 243
parity.
Case 1. Both a and b are even. Then there are integers m and n for which
a = 2 m and b = 2 n, so a + b = 2 m + 2 n = 2( m + n) is clearly even.
Case 2. Both a and b are odd. Then there are integers m and n for which
a = 2 m + 1 and b = 2 n + 1, so a + b = 2 m + 1 + 2 n + 1 = 2( m + n + 1) is clearly even.
Either way, a + b is even. This completes the proof.
17. There is a prime number between 90 and 100.
Proof. Simply observe that 97 is prime.
19. If n ∈ N, then 20 + 21 + 22 + 23 + 24 + · · · + 2n = 2n+1 − 1.
Proof. We use direct proof. Suppose n ∈ N. Let S be the number
S = 20 + 21 + 22 + 23 + 24 + · · · + 2n−1 + 2n .
(1)
In what follows, we will solve for S and show S = 2n+1 − 1. Multiplying both
sides of (1) by 2 gives
2S = 21 + 22 + 23 + 24 + 25 + · · · + 2n + 2n+1 .
(2)
0
n+1
Now subtract Equation (1) from Equation (2) to obtain 2S − S = −2 + 2 ,
which simpliﬁes to S = 2n+1 − 1. Combining this with Equation (1) produces
20 + 21 + 22 + 23 + 24 + · · · + 2n = 2n+1 − 1, so the proof is complete.
21. Every real solution of x3 + x + 3 = 0 is irrational.
Proof. Suppose for the sake of contradiction that this polynomial has a rational
solution a . We may assume that this fraction is fully reduced, so a and b are
b
2
not both even. Observe that we have a + a + 3 = 0. Clearing the denominator
b
b
gives
a3 + ab2 + 3 b3 = 0. Consider two cases: First, if both a and b are odd, the left-hand side is a sum
of three odds, which is odd, meaning 0 is odd, a contradiction. Second, if one
of a and b is odd and the other is even, then the middle term of a3 + ab2 + 3b3
is even, while a3 and 3b2 have opposite parity. Then a3 + ab2 + 3 b3 is the sum
of two evens and an odd, which is odd, again contradicting the fact that 0 is
even.
23. Suppose a, b and c are integers. If a | b and a | (b2 − c), then a | c.
Proof. (Direct) Suppose a | b and a | ( b2 − c). This means that b = ad and
b2 − c = ae for some integers d and e. Squaring the ﬁrst equation produces
b2 = a2 d 2 . Subtracting b2 − c = ae from b2 = a2 d 2 gives c = a2 d 2 − ae = a(ad 2 − e).
As ad 2 − e ∈ Z, it follows that a | c.
25. If p > 1 is an integer and n p for each integer n for which 2 ≤ n ≤
prime. p, then p is Solutions 244 Proof. (Contrapositive) Suppose that p is not prime, so it factors as p = mn for
1 < m, n < p . Observe that it is not the case that both m > p and n > p, because if this were
true the inequalities would multiply to give mn > p p = p, which contradicts
p = mn.
Therefore m ≤ p or n ≤ p. Without loss of generality, say n ≤ p. Then the
equation p = mn gives n | p, with 1 < n ≤ p. Therefore it is not true that n p
for each integer n for which 2 ≤ n ≤ p.
27. Suppose a, b ∈ Z. If a2 + b2 is a perfect square, then a and b are not both odd.
Proof. (Contradiction) Suppose a2 + b2 is a perfect square, and a and b are both
odd. As a2 + b2 is a perfect square, say c is the integer for which c2 = a2 + b2 . As
a and b are odd, we have a = 2 m + 1 and b = 2 n + 1 for integers m and n. Then
c2 = a2 + b2 = (2 m + 1)2 + (2 n + 1)2 = 4( m2 + n2 + mn) + 2. This is even, so c is even also; let c = 2k. Now the above equation results in
(2 k)2 = 4( m2 + n2 + mn) + 2, which simpliﬁes to 2 k2 = 2( m2 + n2 + mn) + 1. Thus 2 k2
is both even and odd, a contradiction. Chapter 8 Exercises
1. Prove that {12 n : n ∈ Z} ⊆ {2n : n ∈ Z} ∩ {3n : n ∈ Z}.
Proof. Suppose a ∈ {12n : n ∈ Z}. This means a = 12 n for some n ∈ Z. Therefore
a = 2(6 n) and a = 3(4 n). From a = 2(6 n), it follows that a is multiple of 2, so
a ∈ {2 n : n ∈ Z}. From a = 3(4 n), it follows that a is multiple of 3, so a ∈ {3 n : n ∈ Z}.
Thus by deﬁnition of the intersection of two sets, we have a ∈ {2 n : n ∈ Z} ∩
{3 n : n ∈ Z}. Thus {12 n : n ∈ Z} ⊆ {2 n : n ∈ Z} ∩ {3 n : n ∈ Z}.
3. If k ∈ Z, then {n ∈ Z : n | k} ⊆ n ∈ Z : n | k2 .
Proof. Suppose k ∈ Z. We now need to show {n ∈ Z : n | k} ⊆ n ∈ Z : n | k2 .
Suppose a ∈ {n ∈ Z : n | k}. Then it follows that a | k, so there is an integer c for
which k = ac. Then k2 = a2 c2 . Therefore k2 = a(ac2 ), and from this the deﬁnition
of divisibility gives a | k2 . But a | k2 means that a ∈ n ∈ Z : n | k2 . We have now
shown {n ∈ Z : n | k} ⊆ n ∈ Z : n | k2 .
5. If p and q are integers, then { pn : n ∈ N} ∩ { qn : n ∈ N} = .
Proof. Suppose p and q are integers. Consider the integer pq. Observe that
pq ∈ { pn : n ∈ N} and pq ∈ { qn : n ∈ N}, so pq ∈ { pn : n ∈ N} ∩ { qn : n ∈ N}. Therefore
{ pn : n ∈ N} ∩ { qn : n ∈ N} = . 245
7. Suppose A , B and C are sets. If B ⊆ C , then A × B ⊆ A × C .
Proof. This is a conditional statement, and we’ll prove it with direct proof.
Suppose B ⊆ C . (Now we need to prove A × B ⊆ A × C .)
Suppose (a, b) ∈ A × B. Then by deﬁnition of the Cartesian product we have a ∈ A
and b ∈ B. But since b ∈ B and B ⊆ C , we have b ∈ C . Since a ∈ A and b ∈ C , it
follows that (a, b) ∈ A × C . Now we’ve shown (a, b) ∈ A × B implies (a, b) ∈ A × C , so
A × B ⊆ A × C.
In summary, we’ve shown that if B ⊆ C , then A × B ⊆ A × C . This completes the
proof.
9. If A , B and C are sets then A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ).
Proof. First we will show A ∩ (B ∪ C ) ⊆ ( A ∩ B) ∪ ( A ∩ C ). Suppose a ∈ A ∩ (B ∪ C ).
Then a ∈ A and a ∈ B ∪ C , by deﬁnition of intersection. Now (by deﬁnition of
union) a ∈ B ∪ C implies that a ∈ B or a ∈ C . Thus we have that a ∈ A and
a ∈ B, or a ∈ A and a ∈ C , and from this it follows that a ∈ A ∩ B or a ∈ A ∩ C .
Therefore a ∈ ( A ∩ B) ∪ ( A ∩ C ), by deﬁnition of union. This paragraph has shown
a ∈ A ∩ (B ∪ C ) implies a ∈ ( A ∩ B) ∪ ( A ∩ C ), so A ∩ (B ∪ C ) ⊆ ( A ∩ B) ∪ ( A ∩ C ).
Now we will show ( A ∩ B) ∪ ( A ∩ C ) ⊆ A ∩ (B ∪ C ). Suppose a ∈ ( A ∩ B) ∪ ( A ∩ C ).
Then by deﬁnition of union, a ∈ A ∩ B or a ∈ A ∩ C . In the ﬁrst case, if a ∈ A ∩ B,
then certainly a ∈ A ∩ (B ∪ C ). Likewise, in the second case a ∈ A ∩ C we have a ∈
A ∩ (B ∪ C ) also. Thus in either case a ∈ A ∩ (B ∪ C ). We’ve shown a ∈ ( A ∩ B) ∪ ( A ∩ C )
implies a ∈ A ∩ (B ∪ C ), so ( A ∩ B) ∪ ( A ∩ C ) ⊆ A ∩ (B ∪ C ).
Since A ∩ (B ∪ C ) ⊆ ( A ∩ B) ∪ ( A ∩ C ) and ( A ∩ B) ∪ ( A ∩ C ) ⊆ A ∩ (B ∪ C ), it follows that
A ∩ (B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ).
11. If A and B are sets in a universal set U , then A ∪ B = A ∩ B.
Proof. Just observe the following sequence of equalities.
A ∪ B = U − ( A ∪ B)
(def. of complement)
= { x : ( x ∈ U ) ∧ ( x ∉ A ∪ B)}
(def. of −)
=
=
=
=
=
=
=
=
= { x : ( x ∈ U )∧ ∼ ( x ∈ A ∪ B)}
{ x : ( x ∈ U )∧ ∼ (( x ∈ A ) ∨ ( x ∈ B))}
{ x : ( x ∈ U ) ∧ (∼ ( x ∈ A )∧ ∼ ( x ∈ B))}
{ x : ( x ∈ U ) ∧ ( x ∉ A ) ∧ ( x ∉ B)}
{ x : ( x ∈ U ) ∧ ( x ∈ U ) ∧ ( x ∉ A ) ∧ ( x ∉ B)}
{ x : (( x ∈ U ) ∧ ( x ∉ A )) ∧ (( x ∈ U ) ∧ ( x ∉ B))}
{ x : ( x ∈ U ) ∧ ( x ∉ A )} ∩ { x : ( x ∈ U ) ∧ ( x ∉ B)}
(U − A ) ∩ (U − B)
A∩B The proof is complete. (def. of ∪)
(DeMorgan)
(x ∈ U ) = (x ∈ U ) ∧ (x ∈ U ) (regroup)
(def. of ∩)
(def. of −)
(def. of complement) Solutions 246
13. If A , B and C are sets, then A − (B ∪ C ) = ( A − B) ∩ ( A − C ).
Proof. Just observe the following sequence of equalities.
A − (B ∪ C ) = { x : ( x ∈ A ) ∧ ( x ∉ B ∪ C )}
(def. of −)
=
=
=
=
=
=
=
= { x : ( x ∈ A )∧ ∼ ( x ∈ B ∪ C )}
{ x : ( x ∈ A )∧ ∼ (( x ∈ B) ∨ ( x ∈ C ))}
{ x : ( x ∈ A ) ∧ (∼ ( x ∈ B)∧ ∼ ( x ∈ C ))}
{ x : ( x ∈ A ) ∧ ( x ∉ B) ∧ ( x ∉ C )}
{ x : ( x ∈ A ) ∧ ( x ∈ A ) ∧ ( x ∉ B ) ∧ ( x ∉ C )}
{ x : (( x ∈ A ) ∧ ( x ∉ B)) ∧ (( x ∈ A ) ∧ ( x ∉ C ))}
{ x : ( x ∈ A ) ∧ ( x ∉ B)} ∩ { x : ( x ∈ A ) ∧ ( x ∉ C )}
( A − B) ∩ ( A − C ) (def. of ∪)
(DeMorgan)
(x ∈ A) = (x ∈ A) ∧ (x ∈ A) (regroup)
(def. of ∩)
(def. of −) The proof is complete.
15. If A , B and C are sets, then ( A ∩ B) − C = ( A − C ) ∩ (B − C ).
Proof. Just observe the following sequence of equalities.
( A ∩ B) − C = { x : ( x ∈ A ∩ B) ∧ ( x ∉ C )}
(def. of −)
= { x : ( x ∈ A ) ∧ ( x ∈ B ) ∧ ( x ∉ C )}
(def. of ∩)
= { x : ( x ∈ A ) ∧ ( x ∉ C ) ∧ ( x ∈ B) ∧ ( x ∉ C )}
(regroup)
= { x : (( x ∈ A ) ∧ ( x ∉ C )) ∧ (( x ∈ B) ∧ ( x ∉ C ))}
(regroup)
= { x : ( x ∈ A ) ∧ ( x ∉ C )} ∩ { x : ( x ∈ B) ∧ ( x ∉ C )} (def. of ∩)
= ( A − C ) ∩ (B − C )
(def. of ∩)
The proof is complete.
17. If A , B and C are sets, then A × (B ∩ C ) = ( A × B) ∩ ( A × C ).
Proof. See Example 8.12.
19. Prove that {9n : n ∈ Z} ⊆ {3n : n ∈ Z}, but {9n : n ∈ Z} = {3n : n ∈ Z}.
Proof. Suppose a ∈ {9n : n ∈ Z}. This means a = 9n for some integer n ∈ Z. Thus
a = 9n = (32 )n = 32n . This shows a is an integer power of 3, so a ∈ {3n : n ∈ Z}.
Therefore a ∈ {9n : n ∈ Z} implies a ∈ {3n : n ∈ Z}, so {9n : n ∈ Z} ⊆ {3n : n ∈ Z}.
But notice {9n : n ∈ Z} = {3n : n ∈ Z} as 3 ∈ {3n : n ∈ Z}, but 3 ∉ {9n : n ∈ Z}
21. Suppose A and B are sets. Prove A ⊆ B if and only if A − B = .
Proof. First we will prove that if A ⊆ B, then A − B = . Contrapositive proof is
used. Suppose that A − B = . Thus there is an element a ∈ A − B, which means
a ∈ A but a ∉ B. Since not every element of A is in B, we have A ⊆ B.
Conversely, we will prove that if A − B = , then A ⊆ B. Again, contrapositive
proof is used. Suppose A ⊆ B. This means that it is not the case that every
element of A is an element of B, so there is an element a ∈ A with a ∉ B.
Therefore we have a ∈ A − B, so A − B = . 247
23. For each a ∈ R, let A a = ( x, a( x2 − 1)) ∈ R2 : x ∈ R . Prove that
Proof. First we will show that {(−1, 0), (1, 0))} ⊆ a∈R a∈R A a = {(−1, 0), (1, 0))}. A a . Notice that for any a ∈ R, we have (−1, 0) ∈ A a because A a contains the ordered pair (−1, a((−1)2 − 1) = (−1, 0).
Similarly (1, 0) ∈ A a . Thus each element of {(−1, 0), (1, 0))} belongs to every set
A a , so every element of
A a , so {(−1, 0), (1, 0))} ⊆
Aa.
a∈R Now we will show a∈R a∈R A a ⊆ {(−1, 0), (1, 0))}. Suppose ( c, d ) ∈ a∈R A a . This means ( c, d ) is in every set A a . In particular ( c, d ) ∈ A 0 = ( x, 0( x2 − 1)) : x ∈ R = {( x, 0) : x ∈ R}.
It follows that d = 0. Then also we have ( c, d ) = ( c, 0) ∈ A 1 = ( x, 1( x2 − 1)) : x ∈ R =
( x, x2 − 1) : x ∈ R . Therefore ( c, 0) has the form ( c, c2 − 1), that is ( c, 0) = ( c, c2 − 1).
From this we get c2 − 1 = 0, so c = ±1. Therefore ( c, d ) = (1, 0) or ( c, d ) = (−1, 0),
so ( c, d ) ∈ {(−1, 0), (1, 0))}. This completes the demonstration that ( c, d ) ∈
Aa
a∈R implies ( c, d ) ∈ {(−1, 0), (1, 0))}, so it follows that
Now it’s been shown that {(−1, 0), (1, 0))} ⊆
follows that a∈R a∈R a∈R A a ⊆ {(−1, 0), (1, 0))}. A a and a∈R A a ⊆ {(−1, 0), (1, 0))}, so it A a = {(−1, 0), (1, 0))}. 25. Suppose A , B, C and D are sets. Prove that ( A × B) ∪ (C × D ) ⊆ ( A ∪ C ) × (B ∪ D ).
Proof. Suppose (a, b) ∈ ( A × B) ∪ (C × D ).
By deﬁnition of union, this means (a, b) ∈ ( A × B) or (a, b) ∈ (C × D ).
We examine these two cases individually.
Case 1. Suppose (a, b) ∈ ( A × B). By deﬁnition of ×, it follows that a ∈ A and
b ∈ B. From this, it follows from the deﬁnition of ∪ that a ∈ A ∪ C and b ∈ B ∪ D .
Again from the deﬁnition of ×, we get (a, b) ∈ ( A ∪ C ) × (B ∪ D ).
Case 2. Suppose (a, b) ∈ (C × D ). By deﬁnition of ×, it follows that a ∈ C and
b ∈ D . From this, it follows from the deﬁnition of ∪ that a ∈ A ∪ C and b ∈ B ∪ D .
Again from the deﬁnition of ×, we get (a, b) ∈ ( A ∪ C ) × (B ∪ D ).
In either case, we obtained (a, b) ∈ ( A ∪ C ) × (B ∪ D ),
so we’ve proved that (a, b) ∈ ( A × B) ∪ (C × D ) implies (a, b) ∈ ( A ∪ C ) × (B ∪ D ).
Therefore ( A × B) ∪ (C × D ) ⊆ ( A ∪ C ) × (B ∪ D ).
27. Prove {12a + 4b : a, b ∈ Z} = {4 c : c ∈ Z}.
Proof. First we show {12a + 4b : a, b ∈ Z} ⊆ {4 c : c ∈ Z}. Suppose x ∈ {12a + 4b : a, b ∈ Z}.
Then x = 12a + 4b for some integers a and b. From this we get x = 4(3a + b), so
x = 4 c where c is the integer 3a + b. Consequently x ∈ {4 c : c ∈ Z}. This establishes
that {12a + 4 b : a, b ∈ Z} ⊆ {4 c : c ∈ Z}.
Next we show {4 c : c ∈ Z} ⊆ {12a + 4b : a, b ∈ Z}. Suppose x ∈ {4 c : c ∈ Z}. Then x = 4 c
for some c ∈ Z. Thus x = (12 + 4(−2)) c = 12 c + 4(−2 c), and since c and −2 c are
integers we have x ∈ {12a + 4 b : a, b ∈ Z}.
This proves that {12a + 4b : a, b ∈ Z} = {4 c : c ∈ Z}. Solutions 248
29. Suppose A = . Prove that A × B ⊆ A × C , if and only if B ⊆ C . Proof. First we will prove that if A × B ⊆ A × C , then B ⊆ C . Using contrapositive,
suppose that B ⊆ C . This means there is an element c ∈ C with c ∉ B. Since
A = , there exists an element a ∈ A . Now consider the ordered pair (a, c). Note
that (a, c) ∈ A × C , but (a, c) ∈ A × B. This means A × B ⊆ A × C .
Conversely, we will now show that if B ⊆ C , then A × B ⊆ A × C . We use direct
proof. Suppose B ⊆ C . Assume that (a, b) ∈ A × B. This means a ∈ A and b ∈ B.
But, as B ⊆ C , we also have b ∈ C . From a ∈ A and b ∈ C , we get (a, b) ∈ A × C .
We’ve now shown (a, b) ∈ A × B implies (a, b) ∈ A × C , so A × B ⊆ A × C .
31. Suppose B = and A × B ⊆ B × C . Prove A ⊆ C . Proof. Suppose B = and A × B ⊆ B × C . In what follows, we show that A ⊆ C .
Let x ∈ A . Because B is not empty, it contains some element b. Observe that
( x, b) ∈ A × B. But as A × B ⊆ B × C , we also have ( x, b) ∈ B × C , so, in particular,
x ∈ B. As x ∈ A and x ∈ B, we have ( x, x) ∈ A × B. But as A × B ⊆ B × C , it follows
that ( x, x) ∈ B × C . This implies x ∈ C .
Now we’ve shown x ∈ A implies x ∈ C , so A ⊆ C . Chapter 9 Exercises
1. If x, y ∈ R, then | x + y| = | x| + | y|.
This is false.
Disproof: Here is a counterexample: Let x = 1 and y = −1. Then | x + y| = 0 and
| x| + | y| = 2, so it’s not true that | x + y| = | x| + | y|.
3. If n ∈ Z and n5 − n is even, then n is even.
This is false.
Disproof: Here is a counterexample: Let n = 3. Then n5 − n = 35 − 3 = 240, but
n is not even.
5. If A , B, C and D are sets, then ( A × B) ∪ (C × D ) = ( A ∪ C ) × (B ∪ D ).
This is false.
Disproof: Here is a counterexample: Let A = {1, 2}, B = {1, 2}, C = {2, 3} and
D = {2, 3}. Then ( A × B) ∪ (C × D ) = {(1, 1), (1, 2), (2, 1), (2, 2)} ∪ {(2, 2), (2, 3), (3, 2), (3, 3)} =
{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)} . Also ( A ∪ C ) × (B ∪ D ) = {1, 2, 3} × {1, 2, 3}=
{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, so you can see that ( A × B) ∪
(C × D ) = ( A ∪ C ) × (B ∪ D ).
7. If A , B and C are sets, and A × C = B × C , then A = B.
This is false.
Disproof: Here is a counterexample: Let A = {1}, B = {2} and C = . Then
A × C = B × C = , but A = B. 249
9. If A and B are sets, then P ( A ) − P (B) ⊆ P ( A − B).
This is false.
Disproof: Here is a counterexample: Let A = {1, 2} and B = {1}. Then P ( A ) −
P (B) = { , {1} , {2} , {1, 2}} − { , {1}} = {{2} , {1, 2}}. Also P ( A − B) = P ({2}) = { , {2}}. In
this example we have P ( A ) − P (B) ⊆ P ( A − B).
11. If a, b ∈ N, then a + b < ab.
This is false.
Disproof: Here is a counterexample: Let a = 1 and b = 1. Then a + b = 2 and
ab = 1, so it’s not true that a + b < ab.
13. There exists a set X for which R ⊆ X and ∈ X . This is true. Proof. Simply let X = R ∪ { }. If x ∈ R, then x ∈ R ∪ { } = X , so R ⊆ X . Likewise,
∈ R ∪ { } = X because ∈ { }.
15. Every odd integer is the sum of three odd integers. This is true.
Proof. Suppose n is odd. Then n = n + 1 + (−1), and therefore n is the sum of
three odd integers.
17. For all sets A and B, if A − B = , then B = .
This is false.
Disproof: Here is a counterexample: Just let A =
but it’s not true that B = . and B = . Then A − B = , 19. For every r , s ∈ Q with r < s, there is an irrational number u for which r < u < s.
This is true.
Proof. (Direct) Suppose r , s ∈ Q with r < s. Consider the number u = r + 2 s−r .
2
In what follows we will show that u is irrational and r < u < s. Certainly since
s − r is positive, it follows that r < r + 2 s−r = u. Also, since 2 < 2 we have
2
u= r+ 2 s−r
s−r
< r+2
= s,
2
2 and therefore u < s. Thus we can conclude r < u < s.
Now we just need to show that u is irrational. Suppose for the sake of contradiction that u is rational. Then u = a for some integers a and b. Since r and s
b
e
c
are rational, we have r = d and s = f for some c, d , e, f ∈ Z. Now we have
u
a
b
ad − bc
bd
(ad − bc)2 d f
bd ( ed − c f ) =
=
=
= s−r
2
e
c
c
f −d
+2
d
2
ed − c f
2
2d f r+ 2 2 Solutions 250
This expresses 2 as a quotient of two integers, so
Thus u is irrational. 2 is rational, a contradiction. In summary, we have produced an irrational number u with r < u < s, so the
proof is complete.
21. There exist two prime numbers p and q for which p − q = 97.
This statement is false.
Disproof: Suppose for the sake of contradiction that this is true. Let p and
q be prime numbers for which p − q = 97. Now, since their diﬀerence is odd, p
and q must have opposite parity, so one of p and q is even and the other is
odd. But there exists only one even prime number (namely 2), so either p = 2
or q = 2. If p = 2, then p − q = 97 implies q = 2 − 97 = −95, which is not prime.
On the other hand if q = 2, then p − q = 97 implies p = 99, but that’s not prime
either. Thus one of p or q is not prime, a contradiction.
23. If x, y ∈ R and x3 < y3 , then x < y. This is true.
Proof. (Contrapositive) Suppose x ≥ y. We need to show x3 ≥ y3 .
Case 1. Suppose x and y have opposite signs, that is one of x and y is positive
and the other is negative. Then since x ≥ y, x is positive and y is negative.
Then, since the powers are odd, x3 is positive and y3 is negative, so x3 ≥ y3 .
Case 2. Suppose x and y do not have opposite signs. Then x2 + x y + y2 ≥ 0 and
also x − y ≥ 0 because x ≥ y. Thus we have x3 − y3 = ( x − y)( x2 + x y + y2 ) ≥ 0. From
this we get x3 − y3 ≥ 0, so x3 ≥ y3 .
In either case we have x3 ≥ y3 .
25. For all a, b, c ∈ Z, if a | bc, then a | b or a | c.
This is false.
Disproof: Let a = 6, b = 3 and c = 4. Note that a | bc, but a | b and a | c.
27. The equation x2 = 2 x has three real solutions.
Proof. By inspection, the numbers x = 2 and x = 4 are two solutions of this
equation. But there is a third solution. Let m be the real number for which
m2m = 1 . Then negative number x = −2 m is a solution, as follows.
2
x2 = (−2 m)2 = 4 m2 = 4 m2 m
2m 2 =4 12
2
2m = 1
22m = 2−2m = 2 x . Therefore we have three solutions 2, 4 and m.
29. If x, y ∈ R and | x + y| = | x − y|, then y = 0.
This is false.
Disproof: Let x = 0 and y = 1. Then | x + y| = | x − y|, but y = 1. 251
31. No number appears in Pascal’s Triangle more than four times.
Disproof: The number 120 appears six times. Check that 130 =
16 = 120 = 120 = 120.
119
14
1 10
7 = 16
2 = 33. Suppose f ( x) = a 0 + a 1 x + a 2 x2 + · · · + a n x n is a polynomial of degree 1 or greater,
and for which each coeﬃcient a i is in N. Then there is an n ∈ N for which the
integer f (n) is not prime.
Proof. (Outline) Note that, because the coeﬃcients are all positive and the
degree is greater than 1, we have f (1) > 1. Let b = f (1) > 1. Now, the polynomial
f ( x) − b has a root 1, so f ( x) − b = ( x − 1) g( x) for some polynomial g. Then
f ( x) = ( x − 1) g( x) + b. Now note that f ( b + 1) = b g( b) + b = b( g( b) + 1). If we can
now show that g(b) + 1 is an integer, then we have a nontrivial factoring
f ( b + 1) = b( g( b) + 1), and f ( b + 1) is not prime. To complete the proof, use the
fact that f ( x) − b = ( x − 1) g( x) has integer coeﬃcients, and deduce that g( x) must
also have integer coeﬃcients. Chapter 10 Exercises
1. For every integer n ∈ N, it follows that 1 + 2 + 3 + 4 + · · · + n = n2 + n
.
2 Proof. We will prove this with mathematical induction.
2 +
(1) Observe that if n = 1, this statement is 1 = 1 2 1 , which is obviously true. (2) Consider any integer k ≥ 1. We must show that S k implies S k+1 . In other
2+
words, we must show that if 1 + 2 + 3 + 4 + · · · + k = k 2 k is true, then
1 + 2 + 3 + 4 + · · · + k + ( k + 1) = is also true. We use direct proof.
Suppose k ≥ 1 and 1 + 2 + 3 + 4 + · · · + k = k2 + k
2. 1 + 2 + 3 + 4 + · · · + k + ( k + 1) Observe that = (1 + 2 + 3 + 4 + · · · + k) + ( k + 1)
k2 + k
+ ( k + 1)
2 ( k + 1)2 + ( k + 1)
2 =
=
=
= k2 + k + 2( k + 1)
2
k2 + 2 k + 1 + k + 1
2
( k + 1)2 + ( k + 1)
.
2
2 +
Therefore we have shown that 1 + 2 + 3 + 4 + · · · + k + (k + 1) = (k+1) 2 (k+1) . The proof by induction is now complete. Solutions 252
3. For every integer n ∈ N, it follows that 13 + 23 + 33 + 43 + · · · + n3 = n2 ( n+1)2
.
4 Proof. We will prove this with mathematical induction.
2+2
(1) When n = 1 the statement is 13 = 1 (14 1) = 4 = 1, which is true.
4
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
2+2
13 + 23 + 33 + 43 + · · · + k3 = k (k4 1) . Observe that this implies the statement is
true for n = k + 1.
13 + 23 + 33 + 43 + · · · + k3 + ( k + 1)3 = 3 = 3 3 3 3 3 (1 + 2 + 3 + 4 + · · · + k ) + ( k + 1)
k2 ( k + 1)2
+ ( k + 1)3
4 k2 ( k + 1)2 4( k + 1)3
+
4
4
k2 ( k + 1)2 + 4( k + 1)3
4
( k + 1)2 ( k2 + 4( k + 1)1 )
4
( k + 1)2 ( k2 + 4 k + 4)
4
( k + 1)2 ( k + 2)2
4
( k + 1)2 (( k + 1) + 1)2
4 =
=
=
=
=
= 2 2 k
Therefore 13 + 23 + 33 + 43 + · · · + k3 + (k + 1)3 = (k+1) ((4+1)+1) , which means the
statement is true for n = k + 1.
This completes the proof by mathematical induction. 5. If n ∈ N, then 21 + 22 + 23 + · · · + 2n = 2n+1 − 2.
Proof. The proof is by mathematical induction.
(1) When n = 1, this statement is 21 = 21+1 − 2, or 2 = 4 − 2, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
21 + 22 + 23 + · · · + 2k = 2k+1 − 2. Observe this implies that the statement is true
for n = k + 1, as follows:
21 + 22 + 23 + · · · + 2k + 2k+1
k = 3 (2 + 2 + 2 + · · · + 2 ) + 2
2 k+1 −2+2 = k+1 = 2 · 2k+1 − 2
2k+2 − 2 = 2 k+1 = 1 2(k+1)+1 − 2 Thus we have 21 + 22 + 23 + · · · + 2k + 2k+1 = 2(k+1)+1 − 2, so the statement is true
for n = k + 1.
Thus the result follows by mathematical induction. 253
7. If n ∈ N, then 1 · 3 + 2 · 4 + 3 · 5 + 4 · 6 + · · · + n(n + 2) = n( n + 1)(2 n + 7)
.
6 Proof. The proof is by mathematical induction.
(1) When n = 1, we have 1 · 3 = 1(1+1)(2+7) , which is the true statement 3 = 18 .
6
6
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
1 · 3 + 2 · 4 + 3 · 5 + 4 · 6 + · · · + k( k + 2) = k(k+1)(2k+7) . Now observe that
6
1 · 3 + 2 · 4 + 3 · 5 + 4 · 6 + · · · + k( k + 2) + ( k + 1)(( k + 1) + 2) = (1 · 3 + 2 · 4 + 3 · 5 + 4 · 6 + · · · + k( k + 2)) + ( k + 1)(( k + 1) + 2)
k( k + 1)(2 k + 7)
+ ( k + 1)(( k + 1) + 2)
6
k( k + 1)(2 k + 7) 6( k + 1)( k + 3)
+
6
6
k( k + 1)(2 k + 7) + 6( k + 1)( k + 3)
6
( k + 1)( k(2 k + 7) + 6( k + 3))
6
( k + 1)(2 k2 + 13 k + 18)
6
( k + 1)( k + 2)(2 k + 9)
6
( k + 1)(( k + 1) + 1)(2( k + 1) + 7)
6 = Thus we have 1·3+2·4+3·5+4·6+· · ·+k(k+2)+(k+1)((k+1)+2) =
and this means the statement is true for n = k + 1.
Thus the result follows by mathematical induction. =
=
=
=
=
= ( k+1)(( k+1)+1)(2( k+1)+7)
,
6 9. For any integer n ≥ 0, it follows that 24 |(52n − 1).
Proof. The proof is by mathematical induction.
(1) For n = 0, the statement is 24 |(52·0 − 1). This is 24 | 0, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
24 |(52k − 1). This means 52k − 1 = 24a for some integer a, and from this we get
52k = 24a + 1. Now observe that
52(k+1) − 1 = 2 k+2 −1 = 2 2k −1 = 5 (24a + 1) − 1 = 25(24a + 1) − 1 = 25 · 24a + 25 − 1 = 5 55
2 24(25a + 1) This shows 52(k+1) − 1 = 24(25a + 1), which means 24 | 52(k+1) − 1.
This completes the proof by mathematical induction. Solutions 254
11. For any integer n ≥ 0, it follows that 3 |(n3 + 5n + 6). Proof. The proof is by mathematical induction.
(1) When n = 0, the statement is 3 |(03 + 5 · 0 + 6), or 3 | 6, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 0, that is assume
3 |( k3 + 5 k + 6). This means k3 + 5 k + 6 = 3a for some integer a. We need to show
that 3 |(( k + 1)3 + 5( k + 1) + 6). Observe that
( k + 1)3 + 5( k + 1) + 6 = k3 + 3 k2 + 3 k + 1 + 5 k + 5 + 6 = ( k3 + 5 k + 6) + 3 k2 + 3 k + 6 = 3a + 3 k2 + 3 k + 6 = 3(a + k2 + k + 2) Thus we have deduced (k + 1)3 − (k + 1) = 3(a + k2 + k + 2). Since a + k2 + k + 2 is
an integer, it follows that 3 |((k + 1)3 + 5( k + 1) + 6).
It follows by mathematical induction that 3 |(n3 + 5 n + 6) for every n ≥ 0.
13. For any integer n ≥ 0, it follows that 6 |(n3 − n).
Proof. The proof is by mathematical induction.
(1) When n = 0, the statement is 6 |(03 − 0), or 6 | 0, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 0, that is assume
6 |( k3 − k). This means k3 − k = 6a for some integer a. We need to show that
6 |(( k + 1)3 − ( k + 1)). Observe that
( k + 1)3 − ( k + 1) = k3 + 3 k2 + 3 k + 1 − k − 1 = ( k3 − k) + 3 k2 + 3 k = 6a + 3 k2 + 3 k = 6a + 3 k( k + 1) Thus we have deduced (k + 1)3 − (k + 1) = 6a + 3k( k + 1). Since one of k or (k + 1)
must be even, it follows that k(k + 1) is even, so k(k + 1) = 2 b for some integer
b. Consequently ( k + 1)3 − ( k + 1) = 6a + 3 k( k + 1) = 6a + 3(2 b) = 6(a + b). Since
( k + 1)3 − ( k + 1) = 6(a + b) it follows that 6 |(( k + 1)3 − ( k + 1)).
Thus the result follows by mathematical induction.
15. If n ∈ N, then 1
1·2 1
1
+ 213 + 314 + 415 + · · · + n(n+1) = 1 − n+1 .
·
·
· Proof. The proof is by mathematical induction.
1
(1) When n = 1, the statement is 1(11 1) = 1 − 1+1 , which simpliﬁes to 1 = 1 .
+
2
2
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
1
1
1
1
1
1
1·2 + 2·3 + 3·4 + 4·5 + · · · + k( k+1) = 1 − k+1 . Next we show that the statement for
n = k + 1 is true. Observe that 255 1
1
1
1
1
1
+
+
+
+···+
+
1·2 2·3 3·4 4·5
k( k + 1) ( k + 1)(( k + 1) + 1)
1
1
1
1
1
1
+
+
+
+···+
+
1·2 2·3 3·4 4·5
k( k + 1)
( k + 1)( k + 2)
1
1
1−
+
k+1
( k + 1)( k + 2)
1
1
+
1−
k + 1 ( k + 1)( k + 2)
k+2
1
1−
+
( k + 1)( k + 2) ( k + 1)( k + 2)
k+1
1−
( k + 1)( k + 2)
1
1−
k+2
1
1−
( k + 1) + 1 =
=
=
=
=
=
= This establishes 112 + 213 + 314 + 415 +· · ·+ (k+1)((1 +1)+1 = 1 − (k+1 +1 , which is to say
·
·
·
·
k
1)
that the statement is true for n = k + 1.
This completes the proof by mathematical induction.
17. Suppose A 1 , A 2 , . . . A n are sets in some universal set U , and n ≥ 2. Prove that
A1 ∩ A2 ∩ · · · ∩ A n = A1 ∪ A2 ∪ · · · ∪ A n .
Proof. The proof is by strong induction.
(1) When n = 2 the statement is A 1 ∩ A 2 = A 1 ∪ A 2 . This is not an entirely
obvious statement, so we have to prove it. Observe that
A1 ∩ A2 = { x : ( x ∈ U ) ∧ ( x ∉ A 1 ∩ A 2 )} (deﬁnition of complement) = { x : ( x ∈ U )∧ ∼ ( x ∈ A 1 ∩ A 2 )} = { x : ( x ∈ U )∧ ∼ (( x ∈ A 1 ) ∧ ( x ∈ A 2 ))} (deﬁnition of ∩) = { x : ( x ∈ U ) ∧ (∼ ( x ∈ A 1 )∨ ∼ ( x ∈ A 2 ))} (DeMorgan) = { x : ( x ∈ U ) ∧ (( x ∉ A 1 ) ∨ ( x ∉ A 2 ))} = { x : ( x ∈ U ) ∧ ( x ∉ A 1 ) ∨ ( x ∈ U ) ∧ ( x ∉ A 2 )} (distributive prop.) = { x : (( x ∈ U ) ∧ ( x ∉ A 1 ))} ∪ { x : (( x ∈ U ) ∧ ( x ∉ A 2 ))} (def. of ∪) = A 1 ∪ A 2 (deﬁnition of complement) (2) Let k ≥ 2. Assume the statement is true if it involves k or fewer sets. Then
A 1 ∩ A 2 ∩ · · · ∩ A k−1 ∩ A k ∩ A k+1 = A 1 ∩ A 2 ∩ · · · ∩ A k−1 ∩ ( A k ∩ A k+1 ) = A 1 ∪ A 2 ∪ · · · ∪ A k−1 ∪ A k ∩ A k+1 = A 1 ∪ A 2 ∪ · · · ∪ A k−1 ∪ A k ∪ A k+1 Solutions 256
Thus the statement is true when it involves k + 1 sets.
This completes the proof by strong induction.
19. Prove n
1/ k2
k=1 ≤ 2 − 1/ n for every n. Proof. This clearly holds for n = 1. Assume it holds for some n ≥ 1. Then
2−
n+1
1/ k2 ≤ 2 − 1/ n + 1/( n + 1)2 = 2 − (n+1) 1)2n ≤ 2 − 1/( n + 1). The proof is complete.
k=1
n( n+
21. If n ∈ N, then 1
1 1
1
+ 2 + 1 + · · · + 2n ≥ 1 + n .
3
2 Proof. If n = 1, the result is obvious.
Assume the proposition holds for some n > 1. Then
1
111
+ + + · · · + n+1
123
2 =
≥ Now, the sum 1
2n +1 111
1
1
1
1
1
+ + +···+ n + n
+
+
+ · · · + n+1
123
2
2 + 1 2n + 2 2n + 3
2
n
1
1
1
1
1+
+n
+n
+n
+ · · · + n+1
2
2 +1 2 +2 2 +3
2 1
+ 2n1 2 + 2n1 3 + · · · + 2n+1 on the right has 2n+1 − 2n = 2n terms,
+
+ all greater than or equal to
we get
1+ n+1
2. 1
1 1
,
2n+1 so the sum is greater than 2n 2n1+1 = 1 . Therefore
2 1
1
+ 1 + 3 + · · · + 2n+1 ≥ 1 + n +
2
2 1
2n +1 1
+ 2n1 2 + 2n1 3 + · · · + 2n+1 ≥ 1 + n + 1 =
+
+
2
2 This means the result is true for n + 1, so the theorem is proved. 23. Use induction to prove the Binomial Theorem ( x + y)n =
Proof. Notice that when n = 1, the formula is ( x + y)1 =
which is true. n
n
i =0 i
1
0 x n− i y i . x 1 y0 + 1
1 x0 y1 = x + y, Now assume the theorem is true for some n > 1. We will show that this implies
that it is true for the power n + 1. Just observe that
( x + y)n+1 = ( x + y)( x + y)n = ( x + y) n
i =0
n =
i =0
n =
i =0
n =
i =0
n+1 =
i =0 n n− i i
x
y
i n (n+1)− i i
x
y+
i
n
n
+
i
i−1 n
i =0 n n− i i+1
x
y
i x(n+1)− i y i n + 1 (n+1)− i i
x
y
i + + yn+1 n + 1 n+1
y
n+1 n + 1 (n+1)− i i
x
y.
i This shows that the formula is true for ( x + y)n+1 , so the theorem is proved. 257
25. Concerning the Fibonacci Sequence, prove that F1 + F2 + F3 + F4 + . . . + F n = F n+2 − 1.
Proof. The proof is by induction.
(1) When n = 1 the statement is F1 = F1+2 − 1 = F3 − 1 = 2 − 1 = 1, which is true.
Also when n = 2 the statement is F1 + F2 = F2+2 − 1 = F4 − 1 = 3 − 1 = 2, which is
true, as F1 + F2 = 1 + 1 = 2.
(2) Now assume k ≥ 1 and F1 + F2 + F3 + F4 + . . . + F k = F k+2 − 1. We need to show
F1 + F2 + F3 + F4 + . . . + F k + F k+1 = F k+3 − 1. Observe that
F1 + F2 + F3 + F4 + . . . + F k + F k+1 = (F1 + F2 + F3 + F4 + . . . + F k ) + F k+1 = F k+2 − 1 + +F k+1 = (F k+1 + F k+2 ) − 1 = F k+3 − 1. This completes the proof by induction.
27. Concerning the Fibonacci Sequence, prove that F1 + F3 + · · · + F2n−1 = F2n .
Proof. If n = 1, the result is immediate. Assume for some n > 1 we have
n
n+1
n
i =1 F2 i −1 = F2 n . Then i =1 F2 i −1 = F2 n+1 + i =1 F2 i −1 = F2 n+1 + F2 n = F2 n+2 = F2( n+1)
as desired.
n
0 n−1
1 n−2
2 n−3
3 1
n−1 + 0
n = F n+1 . Proof. (Strong Induction) For n = 1 this is
the assertion is true when n = 1. 1
0 + 0
1 29. Prove that + + + +···+ k
Now ﬁx n and assume that 0 + k−1 + k−2 +
1
2
k < n. In what follows we use the identity
a
b = 0 whenever it is untrue that 0 ≤ b ≤ a. = 1 + 0 = 1 = F2 = F1+1 . Thus k−3
1
0
3 + · · · + k−1 + k = F k+1
n
n−1
n−1
k = k−1 + k . We also whenever
often use n−1
n−2
1
0
n
+
+
+
+···+
0
1
2
n−1
n
= n
n−1
n−2
1
+
+
+···+
0
1
2
n−1 = n−1
n−1
n−2
n−2
n−3
n−3
0
0
+
+
+
+
+
+···+
+
−1
0
0
1
1
2
n−1
n = n−1
n−2
n−2
n−3
n−3
0
0
+
+
+
+
+···+
+
0
0
1
1
2
n−1
n =
= n−1
n−2
0
+
+···+
0
1
n−1
F n + F n−1 = F n This completes the proof. + n−2
n−3
0
+
+···+
0
1
n−2 Solutions 258 +1
31. Prove that n=0 k = n+1 , where r ∈ N.
r
r
k
Hint: Use induction on the integer n. After doing the basis step, break up the
−1
expression k as k = k−1 + k−1 . Then regroup, use the induction hypothesis,
r
r
r
r
and recombine using the above identity. 33. Suppose that n inﬁnitely long straight lines lie on the plane in such a way that
no two are parallel, and no three intersect at a single point. Show that this
2
arrangement divides the plane into n +2n+2 regions.
Proof. The proof is by induction. For the basis step, suppose n = 1. Then there
is one line, and it clearly divides the plane into 2 regions, one on either side of
21
2
the line. As 2 = 1 +2 +2 = n +2n+2 , the formula is correct when n = 1.
Now suppose there are n + 1 lines on the plane, and that the formula is correct
for when there are n lines on the plane. Single out one of the n + 1 lines on the
plane, and call it . Remove line , so that there are now n lines on the plane.
By the induction hypothesis, these
2n
n lines divide the plane into n +2 +2
regions. Now add line back. Doing this adds an additional n + 1 regions. (The diagram illustrates the
case where n + 1 = 5. Without , there
are n = 4 lines. Adding back produces n + 1 = 5 new regions.) 5
4
3
2
1 Thus, with n + 1 lines there are all together (n + 1) +
( n + 1) + n2 + n+2
2 regions. Observe n2 + n + 2 2 n + 2 + n2 + n + 2 ( n + 1)2 + ( n + 1) + 2
=
=
.
2
2
2
2 (
Thus, with n + 1 lines, we have (n+1) +2 n+1)+2 regions, which means that the
formula is true for when there are n + 1 lines. We have shown that if the
formula is true for n lines, it is also true for n + 1 lines. This completes the
proof by induction. 35. If n, k ∈ N, and n is even and k is odd, then n
k is even. Proof. Notice that if k is not a value between 0 and n, then n = 0 is even; thus
k
from here on we can assume that 0 < k < n. We will use strong induction.
For the basis case, notice that the assertion is true for the even values n = 2
and n = 4: 2 = 2; 4 = 4; 4 = 4 (even in each case).
1
1
3
Now ﬁx and even n assume that m is even whenever m is even, k is odd, and
k
−1
m < n. Using the identity n = n−1 + n−1 three times, we get
k
k
k 259 n
k = n−1
n−1
+
k−1
k = n−2
n−2
n−2
n−2
+
+
+
k−2
k−1
k−1
k = n−2
n−2
n−2
+2
+
.
k−2
k−1
k Now, n − 2 is even, and k and k − 2 are odd. By the inductive hypothesis, the
outer terms of the above expression are even, and the middle is clearly even;
thus we have expressed n as the sum of three even integers, so it is even.
k
Chapter 11 Exercises
Section 11.0 Exercises
1. Let A = {0, 1, 2, 3, 4, 5}. Write out the relation R that expresses > on A . Then
illustrate it with a diagram.
2 R = (5, 4), (5, 3), (5, 3), (5, 3), (5, 1), (5, 0), (4, 3), (4, 2), (4, 1),
(4, 0), (3, 2), (3, 1), (3, 0), (2, 1), (2, 0), (1, 0) 1 3 0
4 5 3. Let A = {0, 1, 2, 3, 4, 5}. Write out the relation R that expresses ≥ on A . Then
illustrate it with a diagram.
2 R = (5, 5), (5, 4), (5, 3), (5, 2), (5, 1), (5, 0),
(4, 4), (4, 3), (4, 2), (4, 1), (4, 0),
(3, 3), (3, 2), (3, 1), (3, 0),
(2, 2), (2, 1), (2, 0), (1, 1), (1, 0), (0, 0) 1 3 0
4 5 5. The following diagram represents a relation R on a set A . Write the sets A
and R . Answer: A = {0, 1, 2, 3, 4, 5}; R = {(3, 3), (4, 3), (4, 2), (1, 2), (2, 5), (5, 0)}
7. Write the relation < on the set A = Z as a subset R of Z × Z. This is an inﬁnite
set, so you will have to use set-builder notation.
Answer: R = {( x, y) ∈ Z × Z : y − x ∈ N}
9. How many diﬀerent relations are there on the set A = 1, 2, 3, 4, 5, 6 ?
Consider forming a relation R ⊆ A × A on A . For each ordered pair ( x, y) ∈ A × A ,
we have two choices: we can either include ( x, y) in R or not include it. There
are 6 · 6 = 36 ordered pairs in A × A . By the Multiplication Principle, there are
thus 236 diﬀerent subsets R and hence also this many relations on A . Solutions 260
Section 11.1 Exercises 1. Consider the relation R = {(a, a), (b, b), ( c, c), (d , d ), (a, b), ( b, a)} on the set A = {a, b, c, d }.
Which of the properties reﬂexive, symmetric and transitive does R possess and
why? If a property does not hold, say why.
This is reﬂexive because ( x, x) ∈ R (i.e. xRx )for every x ∈ A .
It is symmetric because it is impossible to ﬁnd an ( x, y) ∈ R for which ( y, x) ∉ R .
It is transitive because ( xR y ∧ yR z) ⇒ xR z always holds.
3. Consider the relation R = {(a, b), (a, c), ( c, b), (b, c)} on the set A = {a, b, c}. Which
of the properties reﬂexive, symmetric and transitive does R possess and why?
If a property does not hold, say why.
This is not reﬂexive because (a, a) ∉ R (for example).
It is not symmetric because (a, b) ∈ R but (b, a) ∉ R .
It is transitive because ( xR y ∧ yR z) ⇒ xR z always holds. For example (aRb ∧
bRa) ⇒ aRa is true, etc.
5. Consider the relation R = (0, 0), ( 2, 0), (0, 2), ( 2, 2) on R. Say whether this
relation is reﬂexive, symmetric and transitive. If a property does not hold, say
why.
This is not reﬂexive because (1, 1) ∉ R (for example).
It is symmetric because it is impossible to ﬁnd an ( x, y) ∈ R for which ( y, x) ∉ R .
It is transitive because ( xR y ∧ yR z) ⇒ xR z always holds.
7. There are 16 possible diﬀerent relations R on the set A = {a, b}. Describe all of
them. (A picture for each one will suﬃce, but don’t forget to label the nodes.)
a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b 9. Deﬁne a relation on Z by declaring xR y if and only if x and y have the same
parity. Say whether this relation is reﬂexive, symmetric and transitive. If a
property does not hold, say why. What familiar relation is this?
This is reﬂexive because xRx since x always has the same parity as x.
It is symmetric because if x and y have the same parity, then y and x must
have the same parity (that is, xR y ⇒ yRx).
It is transitive because if x and y have the same parity and y and z have the
same parity, then x and z must have the same parity. (That is ( xR y ∧ yR z) ⇒ xR z
always holds.)
11. Suppose A = {a, b, c, d } and R = {(a, a), (b, b), ( c, c), (d , d )}. Say whether this relation
is reﬂexive, symmetric and transitive. If a property does not hold, say why. 261
This is reﬂexive because ( x, x) ∈ R for every x ∈ A .
It is symmetric because it is impossible to ﬁnd an ( x, y) ∈ R for which ( y, x) ∉ R .
It is transitive because ( xR y ∧ yR z) ⇒ xR z always holds.
(For example (aRa ∧ aRa) ⇒ aRa is true, etc.)
13. Consider the relation R = {( x, y) ∈ R × R : x − y ∈ Z} on R. Prove that this relation
is reﬂexive and symmetric, and transitive.
Proof. In this relation, xR y means x − y ∈ Z.
To see that R is reﬂexive, take any x ∈ R and observe that x − x = 0 ∈ Z, so xRx.
Therefore R is reﬂexive.
To see that R is symmetric, we need to prove xR y ⇒ yRx for all x, y ∈ R. We
use direct proof. Suppose xR y. This means x − y ∈ Z. Then it follows that
−( x − y) = y − x is also in Z. But y − x ∈ Z means yRx. We’ve shown xR y implies
yRx, so R is symmetric.
To see that R is transitive, we need to prove ( xR y ∧ yR z) ⇒ xR z is always
true. We prove this conditional statement with direct proof. Suppose xR y and
yR z. Since xR y, we know x − y ∈ Z. Since yR z, we know y − z ∈ Z. Thus x − y
and y − z are both integers; by adding these integers we get another integer
( x − y) + ( y − z) = x − z. Thus x − z ∈ Z, and this means xR z. We’ve now shown that
if xR y and yR z, then xR z. Therefore R is transitive.
15. Prove or disprove: If a relation is symmetric and transitive, then it is also
reﬂexive.
This is false. For a counterexample, consider the relation R = {(a, a), (a, b), (b, a), (b, b)}
on the set A = {a, b, c}. This is symmetric and transitive but it is not reﬂexive.
Section 11.2 Exercises
1. Let A = {1, 2, 3, 4, 5, 6}, and consider the following equivalence relation on A : R =
{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 3), (3, 2), (4, 5), (5, 4), (4, 6), (6, 4), (5, 6), (6, 5)}. List
the equivalence classes of R .
The equivalence classes are: [1] = {1}; [2] = [3] = {2, 3}; [4] = [5] = [6] = {4, 5, 6}. 3. Let A = {a, b, c, d , e}. Suppose R is an equivalence relation on A . Suppose R has
three equivalence classes. Also aRd and bR c. Write out R as a set.
Answer: R = {(a, a), (b, b), ( c, c), (d , d ), ( e, e), (a, d ), ( d , a), (b, c), ( c, b)}
5. There are two diﬀerent equivalence relations on the set A = {a, b}. Describe
them all. Diagrams will suﬃce.
Answer: R = {(a, a), (b, b)} and R = {(a, a), (b, b), (a, b), (b, a)}
7. Deﬁne a relation R on Z as xR y if and only if 3 x − 5 y is even. Prove R is an
equivalence relation. Describe its equivalence classes.
To prove that R is an equivalence relation, we must show it’s reﬂexive, symmetric and transitive. Solutions 262 The relation R is reﬂexive for the following reason. If x ∈ Z, then 3 x − 5 x = −2 x
is even. But then since 3 x − 5 x is even, we have xRx. Thus R is reﬂexive.
To see that R is symmetric, suppose xR y. We must show yRx. Since xR y, we
know 3 x − 5 y is even, so 3 x − 5 y = 2a for some integer a. Now reason as follows.
3x − 5 y = 2a 3x − 5 y + 8 y − 8x = 2a + 8 y − 8 x 3 y − 5x = 2(a + 4 y − 4 x) From this it follows that 3 y − 5 x is even, so yRx. We’ve now shown xR y implies
yRx, so R is symmetric.
To prove that R is transitive, assume that xR y and yR z. (We will show that this
implies xR z.) Since xR y and yR z, it follows that 3 x − 5 y and 3 y − 5 z are both even,
so 3 x −5 y = 2a and 3 y−5 z = 2 b for some integers a and b. Adding these equations,
we get (3 x − 5 y) + (3 y − 5 z) = 2a + 2b, and this simpliﬁes to 3 x − 5 z = 2(a + b + y).
Therefore 3 x − 5 z is even, so xR z. We’ve now shown that if xR y and yR z, then
xR z, so R is transitive.
We’ve now shown that R is reﬂexive, symmetric and transitive, so it is an
equivalence relation.
The completes the ﬁrst part of the problem. Now we move on the second part.
To ﬁnd the equivalence classes, ﬁrst note that
[0] = { x ∈ Z : xR 0} = { x ∈ Z : 3 x − 5 · 0 is even} = { x ∈ Z : 3 x is even} = { x ∈ Z : x is even} . Thus the equivalence class [0] consists of all even integers. Next, note that
[1] = { x ∈ Z : xR 1} = { x ∈ Z : 3 x − 5 · 1 is even} = { x ∈ Z : 3 x − 5 is even} = x ∈ Z : x is odd . Thus the equivalence class [1] consists of all odd integers.
Consequently there are just two equivalence classes {. . . , −4, −2, 0, 2, 4, . . .} and
{. . . , −3, −1, 1, 3, 5, . . .} 9. Deﬁne a relation R on Z as xR y if and only if 4 |( x +3 y). Prove R is an equivalence
relation. Describe its equivalence classes.
This is reﬂexive, because for any x ∈ Z we have 4 |( x + 3 x), so xRx.
To prove that R is symmetric, suppose xR y. Then 4 |( x + 3 y), so x + 3 y = 4a
for some integer a. Multiplying by 3, we get 3 x + 9 y = 12a, which becomes
y + 3 x = 12a − 8 y. Then y + 3 x = 4(3a − 2 y), so 4 |( y + 3 x), hence yRx. Thus we’ve
shown xR y implies yRx, so R is symmetric.
To prove transitivity, suppose xR y and yR z. Then 4|( x + 3 y) and 4|( y + 3 z), so
x + 3 y = 4a and y + 3 z = 4 b for some integers a and b. Adding these two equations
produces x + 4 y + 3 z = 4a + 4b, or x + 3 z = 4a + 4b − 4 y = 4(a + b − y). Consequently
4|( x + 3 z), so xR z, and R is transitive. 263
As R is reﬂexive, symmetric and transitive, it is an equivalence relation.
Now let’s compute its equivalence classes.
[0] = { x ∈ Z : xR 0} = { x ∈ Z : 4 |( x + 3 · 0)} = { x ∈ Z : 4 | x} =
{. . . − 4, 0, 4, 8, 12, 16 . . .}
[1] = { x ∈ Z : xR 1} = { x ∈ Z : 4 |( x + 3 · 1)} = { x ∈ Z : 4 |( x + 3)} = {. . . − 3, 1, 5, 9, 13, 17 . . .}
[2] = { x ∈ Z : xR 2} = { x ∈ Z : 4 |( x + 3 · 2)} = { x ∈ Z : 4 |( x + 6)} = {. . . − 2, 2, 6, 10, 14, 18 . . .}
[3] = { x ∈ Z : xR 3} = { x ∈ Z : 4 |( x + 3 · 3)} = { x ∈ Z : 4 |( x + 9)} = {. . . − 1, 3, 7, 11, 15, 19 . . .} 11. Prove or disprove: If R is an equivalence relation on an inﬁnite set A , then R
has inﬁnitely many equivalence classes.
This is False. The equivalence relation in Exercise 7 above is a counterexample.
It is a relation on the inﬁnite set Z, but it has only two equivalence classes. Section 11.3 Exercises
1. List all the partitions of the set A = {a, b}. Compare your answer to the answer
to Exercise 5 of Section 11.2.
There are just two partitions {{a} , {b}} and {{a, b}}. These correspond to the two
equivalence relations R1 = {(a, a), (b, b)} and R2 = {(a, a), (a, b), ( b, a), (b, b)}, respectively, on A .
3. Describe the partition of Z resulting from the equivalence relation ≡ (mod 4).
Answer: The partition is {[0], [1], [2], [3]} =
{. . . , −4, 0, 4, 8, 12, . . .} , {. . . , −3, 1, 5, 9, 13, . . .} , {. . . , −2, 2, 4, 6, 10, 14, . . .} , {. . . , −1, 3, 7, 11, 15, . . .} Section 11.4 Exercises
1. Write the addition and multiplication tables for Z2 .
+ [0] [1] · [0] [1] [0] [0] [1] [0] [0] [0] [1] [1] [0] [1] [0] [1] 3. Write the addition and multiplication tables for Z4 .
+ [0] [1] [2] [3] · [0] [1] [2] [3] [0] [0] [1] [2] [3] [0] [0] [0] [0] [0] [1] [1] [2] [3] [0] [1] [0] [1] [2] [3] [2] [2] [3] [0] [1] [2] [0] [2] [0] [2] [3] [3] [0] [1] [2] [3] [0] [3] [2] [1] 264 Solutions 5. Suppose [a], [ b] ∈ Z5 and [a] · [b] = [0]. Is it necessarily true that either [a] = [0]
or [ b] = [0]?
The multiplication table for xZ5 is shown in Section 11.4. In the body of that
table, the only place that [0] occurs is in the ﬁrst row or the ﬁrst column. That
row and column are both headed by [0]. It follows that if [a] · [b] = [0], then
either [a] or [b] must be [0].
7. Do the following calculations in Z9 , in each case expressing your answer as [a]
with 0 ≤ a ≤ 8.
(a) [8] + [8] = [7]
(b) [24] + [11] = [8]
(c) [21] · [15] = [0]
(d) [8] · [8] = [1] Chapter 12 Exercises
Section 12.1 Exercises
1. Suppose A = {0, 1, 2, 3, 4}, B = {2, 3, 4, 5} and f = {(0, 3), (1, 3), (2, 4), (3, 2), (4, 2)}. State
the domain and range of f . Find f (2) and f (1).
Domain is A ; Range is {2, 3, 4}; f (2) = 4; f (1) = 3.
3. There are four diﬀerent functions f : {a, b} → {0, 1}. List them all. Diagrams will
suﬃce.
f 1 = {(a, 0), ( b, 0)} f 2 = {(a, 1), ( b, 0)} , f 3 = {(a, 0), ( b, 1)} f 4 = {(a, 1), ( b, 1)} 5. Give an example of a relation from {a, b, c, d } to {d , e} that is not a function.
One example is {(a, d ), (a, e), ( b, d ), ( c, d ), (d , d )}.
7. Consider the set f = {( x, y) ∈ Z × Z : 3 x + y = 4}. Is this a function from Z to Z?
Explain.
Yes, since 3 x + y = 4 if and only if y = 4 − 3 x, this is the function f : Z → Z deﬁned
as f ( x) = 4 − 3 x.
9. Consider the set f = ( x2 , x) : x ∈ R . Is this a function from R to R? Explain.
No. This is not a function. Observe that f contains the ordered pairs (4, 2) and
(4, −2). Thus the real number 4 occurs as the ﬁrst coordinate of more than one
element of f .
11. Is the set θ = {( X , | X |) : X ⊆ Z5 } a function? If so, what is its domain and range?
Yes, this is a function. The domain is P (Z5 ). The range is {0, 1, 2, 3, 4, 5}.
Section 12.2 Exercises
1. Let A = {1, 2, 3, 4} and B = {a, b, c}. Give an example of a function f : A → B that
is neither injective nor surjective.
Consider f = {(1, a), (2, a), (3, a), (4, a)}.
Then f is not injective because f (1) = f (2).
Also f is not surjective because it sends no element of A to the element c ∈ B. 265
3. Consider the cosine function cos : R → R. Decide whether this function is injective
and whether it is surjective. What if it had been deﬁned as cos : R → [−1, 1]?
The function cos : R → R is not injective because, for example, cos(0) = cos(2π). It
is not surjective because if b = 5 ∈ R (for example), there is no real number for
which cos( x) = b. The function cos : R → [−1, 1] is surjective. but not injective.
5. A function f : Z → Z is deﬁned as f ( n) = 2n + 1. Verify whether this function is
injective and whether it is surjective.
This function is injective. To see this, suppose m, n ∈ Z and f (m) = f (n).
This means 2m + 1 = 2n + 1, from which we get 2 m = 2n, and then m = n.
Thus f is injective.
This function is not surjective. To see this notice that f (n) is odd for all
n ∈ Z.
So given the (even) number 2 in the codomain Z, there is no n with f (n) = 2.
7. A function f : Z × Z → Z is deﬁned as f ((m, n)) = 2n − 4m. Verify whether this
function is injective and whether it is surjective.
This is not injective because (0, 2) = (−1, 0), yet f ((0, 2)) = f ((−1, 0)) = 4. This is
not surjective because f ((m, n)) = 2n − 4m = 2( n − 2m) is always even. If b ∈ Z
is odd, then f ((m, n)) = b, for all (m, n) ∈ Z × Z.
+
9. Prove that the function f : R − {2} → R − {5} deﬁned by f ( x) = 5xx−21 is bijective. Proof. First, let’s check that f is injective. Suppose f ( x) = f ( y). Then
5x + 1
x−2
(5 x + 1)( y − 2) = 5y + 1
y−2
(5 y + 1)( x − 2) 5 x y − 10 x + y − 2 = 5 yx − 10 y + x − 2 −10 x + y = −10 y + x 11 y = 11 x y = x = Since f ( x) = f ( y) implies x = y, it follows that f is injective.
Next, let’s check that f is surjective. For this, take an arbitrary element
b ∈ R − {5}. We want to see if there is an x ∈ R − {2} for which f ( x) = b, or 5xx+1 = b.
−2
Solving this for x, we get: 5x + 1 = b( x − 2) 5x + 1 = bx − 2 b 5 x − xb = −2 b − 1 x(5 − b) = −2 b − 1 Solutions 266 Since we have assumed b ∈ R − {5}, the term (5 − b) is not zero, and we can
−2 b − 1 . This is an x for which f ( x) = b, so f is
divide with impunity to get x =
5−b
surjective.
Since f is both injective and surjective, it is bijective.
11. Consider the function θ : {0, 1} × N → Z deﬁned as θ (a, b) = (−1)a b. Is θ injective?
Is it surjective? Explain.
First we show that θ is injective. Suppose θ (a, b) = θ ( c, d ). Then (−1)a b = (−1) c d .
Since b and d are both in N, they are both positive. Therefore since (−1)a b =
(−1) c d it follows that (−1)a and (−1)b have the same sign. Since each of (−1)a
and (−1)b equals ±1, we have (−1)a = (−1)b , so then (−1)a b = (−1) c d implies b = d .
But also (−1)a = (−1)b means a and b have the same parity, and since a, b ∈ {0, 1}
if that follows a = b. Thus (a, b) = ( c, d ), so θ is injective.
Next note that θ is not surjective because θ (a, b) = (−1)a b is either positive or
negative, but never zero. Therefore there exist no element (a, b) ∈ {0, 1} × N for
which θ (a, b) = 0 ∈ Z.
13. Consider the function f : R2 → R2 deﬁned by the formula f ( x, y) = ( x y, x3 ). Is f
injective? Is it surjective?
Notice that f (1, 0) = (0, 0) and f (0, 0) = (0, 0), so f is not injective. To show that f
is also not surjective, we will show that it’s impossible to ﬁnd an ordered pair
( x, y) with f ( x, y) = (1, 0). If there were such a pair, then f ( x, y) = ( x y, x3 ) = (1, 0),
which yields x y = 1 and x3 = 0. From x3 = 0 we get x = 0, so x y = 0, a contradiction.
15. This question concerns functions f : { A , B, C , D , E , F , G } → {1, 2, 3, 4, 5, 6, 7}. How
many such functions are there? How many of these functions are injective?
How many are surjective? How many are bijective?
Function f can described as a list ( f ( A ), f (B), f (C ), f (D ), f (E ), f (F ), f (G )), where
there are seven choices for each entry. By the multiplication principle, the total
number of functions f is 77 = 823543.
If f is injective, then this list can’t have any repetition, so there are 7! = 5040
injective functions. Since any injective function sends the seven elements of the
domain to seven distinct elements of the co-domain, all of the injective functions
are surjective, and vice versa. Thus there are 5040 surjective functions and
5040 bijective functions.
17. This question concerns functions f : { A , B, C , D , E , F , G } → {1, 2}. How many such
functions are there? How many of these functions are injective? How many
are surjective? How many are bijective?
Function f can described as a list ( f ( A ), f (B), f (C ), f (D ), f (E ), f (F ), f (G )), where
there are two choices for each entry. Therefore the total number of functions
is 27 = 128. It is impossible for any function to send all seven elements of
{ A , B, C , D , E , F , G } to seven distinct elements of {1, 2}, so none of these 128
functions is injective, hence none are bijective.
How many are surjective? Only two of the 128 functions are not surjective, and
they are the “constant” functions {( A , 1), (B, 1), (C , 1), (D , 1), (E , 1), (F , 1), (G , 1)} and
{( A , 2), (B, 2), (C , 2), (D , 2), (E , 2), (F , 2), (G , 2)}. So there are 126 surjective functions. 267
Section 12.3 Exercises
1. If 6 integers are chosen at random, at least two will have the same remainder
when divided by 5.
Proof. Write Z as follows: Z = 4=0 {5k + j : k ∈ Z}. This is a partition of Z into 5
j
sets. If six integers are picked at random, by the pigeonhole principle, at least
two will be in the same set. However, each set corresponds to the remainder
of a number after being divided by 5 (for example, {5k + 1 : k ∈ Z} are all those
integers that leave a remainder of 1 after being divided by 5).
3. Given any six positive integers, there are two for which their sum or diﬀerence
is divisible by 9.
Proof. If for two of the integers n, m we had n ≡ m (mod 9), then n− m ≡ 0 (mod 9),
and we would be done. Thus assume this is not the case. Observe that the
only two element subsets of positive integers that sum to 9 are {1, 8}, {2, 7}, {3, 6},
and {4, 5}. However, since at least ﬁve of the six integers must have distinct
remainders from 1, 2, ..., 8 it follows from the pigeonhole principle that two
integers n, m are in the same set. Hence n + m ≡ 0 (mod 9) as desired.
5. Prove that any set of seven distinct natural numbers contains a pair of numbers
whose sum or diﬀerence is divisible by 10.
Proof. Let S = {a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 } be any set of 7 natural numbers. Let’s say
that a 1 < a 2 < a 3 < · · · < a 7 . Consider the set
A = {a 1 − a 2 , a 1 − a 3 , a 1 − a 4 , a 1 − a 5 , a 1 − a 6 , a 1 − a 7 ,
a1 + a2 , a1 + a3 , a1 + a4 , a1 + a5 , a1 + a6 , a1 + a7 } Thus | A | = 12. Now let B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, so |B| = 10. Let f : A → B be the
function for which f (n) equals the last digit of n. (That is f (97) = 7, f (12) = 2,
f (230) = 0, etc.) Then, since | A | > |B|, the pigeonhole principle guarantees that
f is not injective. Thus A contains elements a 1 ± a i and a 1 ± a j for which
f (a 1 ± a i ) = f (a 1 ± a j ). This means the last digit of a 1 ± a i is the same as the last
digit of a 1 ± a j . Thus the last digit of the diﬀerence (a 1 ± a i ) − (a 1 ± a j ) = ±a i ± a j
is 0. Hence ±a i ± a j is a sum or diﬀerence of elements of S that is divisible by
10.
Section 12.4 Exercises
1. Suppose A = {5, 6, 8}, B = {0, 1}, C = {1, 2, 3}. Let f : A → B be the function f =
{(5, 1), (6, 0), (8, 1)}, and g : B → C be g = {(0, 1), (1, 1)}. Find g ◦ f .
g ◦ f = {(5, 1), (6, 1), (8, 1)} 3. Suppose A = {1, 2, 3}. Let f : A → A be the function f = {(1, 2), (2, 2), (3, 1)}, and let
g : A → A be the function g = {(1, 3), (2, 1), (3, 2)}. Find g ◦ f and f ◦ g.
g ◦ f = {(1, 1), (2, 1), (3, 3)}; f ◦ g = {(1, 1), (2, 2), (3, 2)}. Solutions 268
5. Consider the functions f , g : R → R deﬁned as f ( x) =
the formulas for g ◦ f and f ◦ g.
3
g ◦ f ( x) = x + 1; f ◦ g ( x) = x3 + 1 3 x + 1 and g( x) = x3 . Find 7. Consider the functions f , g : Z × Z → Z × Z deﬁned as f (m, n) = (mn, m2 ) and
g( m, n) = ( m + 1, m + n). Find the formulas for g ◦ f and f ◦ g.
Note g ◦ f (m, n) = g( f (m, n)) = g(mn, m2 ) = (mn + 1, mn + m2 ).
Thus g ◦ f (m, n) = ( mn + 1, mn + m2 )
Note f ◦ g (m, n) = f ( g(m, n)) = f (m + 1, m + n) = (( m + 1)( m + n), (m + 1)2 ).
Thus f ◦ g (m, n) = ( m2 + mn + m + n, m2 + 2 m + 1)
9. Consider the functions f : Z × Z → Z deﬁned as f ( m, n) = m + n and g : Z → Z × Z
deﬁned as g(m) = ( m, m). Find the formulas for g ◦ f and f ◦ g.
g ◦ f ( m, n) = ( m + n, m + n)
f ◦ g ( m) = 2 m Section 12.5 Exercises
1. Check that the function f : Z → Z deﬁned by f (n) = 6 − n is bijective. Then
compute f −1 .
It is injective as follows. Suppose f (m) = f (n). Then 6 − m = 6 − n, which reduces
to m = n.
It is surjective as follows. If b ∈ Z, then f (6 − b) = 6 − (6 − b) = b.
Inverse: f −1 (n) = 6 − n.
3. Let B = {2n : n ∈ Z} = . . . , 1 , 1 , 1, 2, 4, 8, . . . . Show that the function f : Z → B deﬁned
42
as f (n) = 2n is bijective. Then ﬁnd f −1 .
It is injective as follows. Suppose f (m) = f (n), which means 2m = 2n . Taking
log2 of both sides gives log2 (2m ) = log2 (2n ), which simpliﬁes to m = n.
The function f is surjective as follows. Suppose b ∈ B. By deﬁnition of B this
means b = 2n for some n ∈ Z. Then f (n) = 2n = b.
Inverse: f −1 (n) = log2 (n).
5. The function f : R → R deﬁned as f ( x) = π x − e is bijective. Find its inverse.
x+e
Inverse: f −1 ( x) =
.
π 7. Show that the function f : R2 → R2 deﬁned by the formula f (( x, y) = (( x2 + 1) y, x3 )
is bijective. Then ﬁnd its inverse.
First we prove the function is injective. Assume f ( x1 , y1 ) = f ( x2 , y2 ). Then
2
2
3
3
( x1 + 1) y1 = ( x2 + 1) y2 and x1 = x2 . Since the real-valued function f ( x) = x3 is one2
to-one, it follows that x1 = x2 . Since x1 = x2 , and x1 + 1 > 0 we may divide both
2
2
2
sides of ( x1 + 1) y1 = ( x1 + 1) y2 by ( x1 + 1) to get y1 = y2 . Hence ( x1 , y1 ) = ( x2 , y2 ).
Now we prove the function is surjective. Let (a, b) ∈ R2 . Set x = b1/3 and y =
a
a/( b2/3 + 1). Then f ( x, y) = (( b2/3 + 1) b2/3 +1 , ( b1/3 )3 ) = (a, b). It now follows that f is
bijective.
Finally, we compute the inverse. Write f ( x, y) = (u, v). Interchange variables to
get ( x, y) = f ( u, v) = ((u2 + 1)v, u3 ). Thus x = (u2 + 1)v and y = u3 . Hence u = y1/3 and
x
x
v = y2/3 +1 . Therefore f −1 ( x, y) = ( u, v) = y1/3 , y2/3 +1 . 269
9. Consider the function f : R × N → N × R deﬁned as f ( x, y) = ( y, 3 x y). Check that
this is bijective; ﬁnd its inverse.
To see that this is injective, suppose f (a, b) = f ( c, d ). This means (b, 3ab) =
( d , 3 cd ). Since the ﬁrst coordinates must be equal, we get b = d . As the second
coordinates are equal, we get 3ab = 3dc, which becomes 3ab = 3 bc. Note that,
from the deﬁnition of f , b ∈ N, so b = 0. Thus we can divide both sides of
3ab = 3 bc by the non-zero quantity 3 b to get a = c. Now we have a = c and b = d ,
so (a, b) = ( c, d ). It follows that f is injective.
Next we check that f is surjective. Given any (b, c) in the codomain N × R, notice
that ( 3cb , b) belongs to the domain R × N, and f ( 3cb , b) = (b, c). This proves f is
surjective. As it is both injective and surjective, it is bijective; thus the inverse
exists.
To ﬁnd the inverse, recall that we obtained f ( 3cb , b) = (b, c). Then f −1 f ( 3cb , b) =
f −1 ( b, c), which reduces to ( 3cb , b) = f −1 ( b, c). Replacing b and c with x and y,
respectively, we get f −1 ( x, y) = ( 3yx , x).
Section 12.6 Exercises
1. Consider the function f : R → R deﬁned as f ( x) = x2 + 3. Find f ([−3, 5]) and
f −1 ([12, 19]). Answers: f ([−3, 5]) = [3, 28]; f −1 ([12, 19]) = [−4, −3] ∪ [3, 4].
3. This problem concerns functions f : {1, 2, 3, 4, 5, 6, 7} → {0, 1, 2, 3, 4}. How many
such functions have the property that | f −1 ({3})| = 3? Answer: 44 7 .
3
5. Consider a function f : A → B and a subset X ⊆ A . We observed in Section 12.6
that f −1 ( f ( X )) = X in general. However X ⊆ f −1 ( f ( X )) is always true. Prove this.
Proof. Suppose a ∈ X . Thus f (a) ∈ { f ( x) : x ∈ X } = f ( X ), that is f (a) ∈ f ( X ). Now,
by deﬁnition of preimage, we have f −1 ( f ( X )) = { x ∈ A : f ( x) ∈ f ( X )}. Since a ∈ A
and f (a) ∈ f ( X ), it follows that a ∈ f −1 ( f ( X )). This proves X ⊆ f −1 ( f ( X )).
7. Given a function f : A → B and subsets W , X ⊆ A , prove f (W ∩ X ) ⊆ f (W ) ∩ f ( X ).
Proof. Suppose b ∈ f (W ∩ X ). This means b ∈ { f ( x) : x ∈ W ∩ X }, that is b = f (a)
for some a ∈ W ∩ X . Since a ∈ W we have b = f (a) ∈ { f ( x) : x ∈ W } = f (W ). Since
a ∈ X we have b = f (a) ∈ { f ( x) : x ∈ X } = f ( X ). Thus b is in both f (W ) and f ( X ), so
b ∈ f (W ) ∩ f ( X ). This completes the proof that f (W ∩ X ) ⊆ f (W ) ∩ f ( X ).
9. Given a function f : A → B and subsets W , X ⊆ A , prove f (W ∪ X ) = f (W ) ∪ f ( X ).
Proof. First we will show f (W ∪ X ) ⊆ f (W ) ∪ f ( X ). Suppose b ∈ f (W ∪ X ). This
means b ∈ { f ( x) : x ∈ W ∪ X }, that is, b = f (a) for some a ∈ W ∪ X . Thus a ∈ W
or a ∈ X . If a ∈ W , then b = f (a) ∈ { f ( x) : x ∈ W } = f (W ). If a ∈ X , then b = f (a) ∈
{ f ( x) : x ∈ X } = f ( X ). Thus b is in f (W ) or f ( X ), so b ∈ f (W ) ∪ f ( X ). This completes
the proof that f (W ∪ X ) ⊆ f (W ) ∪ f ( X ).
Next we will show f (W ) ∪ f ( X ) ⊆ f (W ∪ X ). Suppose b ∈ f (W ) ∪ f ( X ). This means
b ∈ f (W ) or b ∈ f ( X ). If b ∈ f (W ), then b = f (a) for some a ∈ W . If b ∈ f ( X ), then 270 Solutions b = f (a) for some a ∈ X . Either way, b = f (a) for some a that is in W or X . That
is, b = f (a) for some a ∈ W ∪ X . But this means b ∈ f (W ∪ X ). This completes the
proof that f (W ) ∪ f ( X ) ⊆ f (W ∪ X ).
The previous two paragraphs show f (W ∪ X ) = f (W ) ∪ f ( X ). 11. Given f : A → B and subsets Y , Z ⊆ B, prove f −1 (Y ∪ Z ) = f −1 (Y ) ∪ f −1 ( Z ).
Proof. First we will show f −1 (Y ∪ Z ) ⊆ f −1 (Y ) ∪ f −1 ( Z ). Suppose a ∈ f −1 (Y ∪ Z ).
By Deﬁnition 12.9, this means f (a) ∈ Y ∪ Z . Thus, f (a) ∈ Y or f (a) ∈ Z . If
f (a) ∈ Y , then a ∈ f −1 (Y ), by Deﬁnition 12.9. Similarly, if f (a) ∈ Z , then a ∈
f −1 ( Z ). Hence a ∈ f −1 (Y ) or a ∈ f −1 ( Z ), so a ∈ f −1 (Y ) ∪ f −1 ( Z ). Consequently
f −1 (Y ∪ Z ) ⊆ f −1 (Y ) ∪ f −1 ( Z ).
Next we show f −1 (Y ) ∪ f −1 ( Z ) ⊆ f −1 (Y ∪ Z ). Suppose a ∈ f −1 (Y ) ∪ f −1 ( Z ). This
means a ∈ f −1 (Y ) or a ∈ f −1 ( Z ). Hence, by Deﬁnition 12.9, f (a) ∈ Y or f (a) ∈ Z ,
which means f (a) ∈ Y ∪ Z . But by Deﬁnition 12.9, f (a) ∈ Y ∪ Z means a ∈ f −1 (Y ∪ Z ).
Consequently f −1 (Y ) ∪ f −1 ( Z ) ⊆ f −1 (Y ∪ Z ).
The previous two paragraphs show f −1 (Y ∪ Z ) = f −1 (Y ) ∪ f −1 ( Z ). Chapter 13 Exercises
Section 13.1 Exercises
1. R and (0, ∞)
Observe that the function f ( x) = e x sends R to (0, ∞). It is injective because
f ( x) = f ( y) implies e x = e y , and taking ln of both sides gives x = y. It is surjective
because if b ∈ (0, ∞), then f (ln(b)) = b. Therefore, because of the bijection
f : R → (0, ∞), it follows that |R| = |(0, ∞)|.
3. R and (0, 1)
1
Observe that the function π f ( x) = cot−1 ( x) sends R to (0, 1). It is injective and
surjective by elementary trigonometry. Therefore, because of the bijection
f : R → (0, 1), it follows that |R| = |(0, 1)|.
5. A = {3k : k ∈ Z} and B = {7k : k ∈ Z}
Observe that the function f ( x) = 7 x sends A to B. It is injective because
3
f ( x) = f ( y) implies 7 x = 7 y, and multiplying both sides by 3 gives x = y. It is
3
3
7
surjective because if b ∈ B, then b = 7 k for some integer k. Then 3k ∈ A , and
f (3 k) = 7 k = b. Therefore, because of the bijection f : A → B, it follows that
| A | = | B |.
1
7. Z and S = . . . , 1 , 1 , 2 , 1, 2, 4, 8, 16, . . .
84
Observe that the function f : Z → S deﬁned as f (n) = 2n is bijective: It is injective
because f (m) = f ( n) implies 2m = 2n , and taking log2 of both sides produces m = n.
It is surjective because any element b of S has form b = 2n for some integer n,
and therefore f ( n) = 2n = b. Because of the bijection f : Z → S , it follows that
| Z| = | S | . 271
9. {0, 1} × N and N
Consider the function f : {0, 1} × N → N deﬁned as f (a, n) = 2n − a. This is injective
because if f (a, n) = f (b, m), then 2 n − a = 2m − b. Now if a were unequal to b, one
of a or b would be 0 and the other would be 1, and one side of 2n − a = 2m − b
would be odd and the other even, a contradiction. Therefore a = b. Then
2 n − a = 2 m − b becomes 2 n − a = 2 m − a; add a to both sides and divide by 2 to
get m = n. Thus we have a = b and m = n, so (a, n) = (b, m), so f is injective.
To see that f is surjective, take any b ∈ N. If b is even, then b = 2n for some
integer n, and f (0, n) = 2n − 0 = b. If b is odd, then b = 2 n + 1 for some integer n.
Then f (1, n + 1) = 2(n + 1) − 1 = 2n + 1 = b. Therefore f is surjective. Then f is a
bijection, so |{0, 1} × N| = |N|.
Section 13.2 Exercises
1. Prove that the set A = {ln( n) : n ∈ N} ⊆ R is countably inﬁnite.
Just note that its elements can be written in inﬁnite list form as ln(1), ln(2), ln(3), · · · .
Thus A is countably inﬁnite.
3. Prove that the set A = {(5 n, −3n) : n ∈ Z} is countably inﬁnite.
Consider the function f : Z → A deﬁned as f (n) = (5n, −3n). This is clearly
surjective, and it is injective because f (n) = f (m) gives (5 n, −3n) = (5 m, −3m), so
5 n = 5 m, hence m = n. Thus, because f is surjective, |Z| = | A |, and | A | = |Z| = ℵ0 .
Therefore A is countably inﬁnite.
5. Prove or disprove: There exists a countably inﬁnite subset of the set of irrational
numbers.
This is true. Just consider the set consisting of the irrational numbers
ππππ
1 , 2 , 3 , 4 ,···.
7. Prove or disprove: The set Q100 is countably inﬁnite.
This is true. Note Q100 = Q × Q × · · · × Q (100 times), and since Q is countably
inﬁnite, it follows from the corollary of Theorem 13.5 that this product is
countably inﬁnite.
9. Prove or disprove: The set {0, 1} × N is countably inﬁnite.
This is true. Note that {0, 1} × N can be written in inﬁnite list form as
(0, 1), (1, 1), (0, 2), (1, 2), (0, 3), (1, 3), (0, 4), (1, 4), · · · . Thus the set is countably inﬁnite.
11. Partition N into 8 countably inﬁnite sets.
For each i ∈ {1, 2, 3, 4, 5, 6, 7, 8}, let X i be those natural numbers that are congruent
to i modulo 8, that is
X1 = {1, 9, 17, 25, 33, . . .} X2 = {2, 10, 18, 26, 34, . . .} X3 = {3, 11, 19, 27, 35, . . .} X4 = {4, 12, 20, 28, 36, . . .} X5 = {5, 13, 21, 29, 37, . . .} Solutions 272
X6 = {6, 14, 22, 30, 38, . . .} X7 = {7, 15, 13, 31, 39, . . .} X8 = {8, 16, 24, 32, 40, . . .} 13. If A = { X ⊂ N : X is ﬁnite }. Then | A | = ℵ0 .
Proof. This is true. To show this we will describe how to arrange the items of
A in an inﬁnite list X 1 , X 2 , X 3 , X 4 , . . ..
For each natural number n, let p n be the nth prime number, that is p 1 =
2, p 2 = 3, p 3 = 5, p 4 = 7, and so on. Now consider any element X ∈ A , so X =
{ n 1 , n 2 , n 3 , ..., n k }, where k = | X | and n i ∈ N for each 1 ≤ i ≤ k. Deﬁne a function
f : A → N as follows: f ( X ) = p n1 p n2 · · · p n k . That is, we treat X ∈ A as an “index"
for the prime sequence and just map the entire set to the product of all the
primes with corresponding index. For example, take the set X = {1, 2, 3}. Then
f ( X ) = f ({1, 2, 3}) = p 1 p 2 p 3 = 2 · 3 · 5 = 30. Note that f is an injection from A to N. Assume f ( X ) = f (Y ). Then, by deﬁnition
of the function, f ( X ) = p n1 p n2 · · · p n k . Similarly, f (Y ) = p m1 p m2 · · · p m s . By the
Fundamental Theorem of Arithmetic, these are the prime decompositions of
each f ( X ) and f (Y ). Furthermore, the fundamental theorem guarantees that
these decompositions are unique. Hence {n1 , n2 , ..., n k } = {m 1 , m 2 , ..., m s } or X = Y .
This means each ﬁnite set X ⊆ N is associated with a unique natural number
f ( X ). Thus we can list the elements in X in A in increasing order of the
numbers f ( X ). The ﬁrst several terms of this list would be
{1}, {2}, {3}, {1, 2}, {4}, {1, 3}, {5}, {6}, {1, 4}, {2, 3}, {7}, . . . It follows that A is countably inﬁnite.
Section 13.3 Exercises
1. Suppose B is an uncountable set and A is a set. Given that there is a surjective
function f : A → B, what can be said about the cardinality of A ?
The set A must be uncountable, as follows. For each b ∈ B, let a b be an element
of A for which f (a b ) = b. (Such an element must exist because f is surjective.
Now form the set U = {a b : b ∈ B}. Then the function f : U → B is bijective,
by construction. Then since B is uncountable, so is U . Therefore U is an
uncountable subset of A , so A is uncountable by Theorem 13.9.
3. Prove or disprove: If A is uncountable, then | A | = |R|.
This is false. Let A = P (R). Then A is uncountable, and by Theorem 13.7,
|R| < |P (R)| = | A |.
5. Prove or disprove: The set {0, 1} × R is uncountable.
This is true. To see why, ﬁrst note that the function f : R → {0} × R deﬁned as
f ( x) = (0, x) is a bijection. Thus |R| = |{0} × R|, and since R is uncountable, so is 273
{0} × R. Then {0} × R is an uncountable subset of the set {0, 1} × R, so {0, 1} × R is
uncountable by Theorem 13.9. 7. Prove or disprove: If A ⊆ B and A is countably inﬁnite and B is uncountable,
then B − A is uncountable.
This is true. To see why, suppose to the contrary that B − A is countably inﬁnite.
Then B = A ∪ (B − A ) is a union of countably inﬁnite sets, and thus countable,
by Theorem 13.6. This contradicts the fact that B is uncountable. Index C ( n, k), 68 Absolute Value, 6
Addition Principle, 76
And, 33
Biconditional Statement, 39
Bijection, see Bijective
Bijective, 186
Byte, 58
Cantor, Georg, 204
Cardinality, 4, 202
Cartesian plane, 9
Cartesian power, 10
Cartesian product, 8
Closed Interval, 6
Codomain of a function, 183
Complement of a set, 19
Composition of functions, 193
Conditional Statement, 36
Conjecture, 135
Contrapositive, 94
Converse of a statement, 39, 94
Corollary, 82
Countable set, 206
Counterexample, 137
Counting, 57
Deﬁnition, 81
DeMorgan’s Laws, 45, 52
Diﬀerence of sets, 17
Disproof, 134
Divides, 84
Divisor, 84
Domain of a function, 183
Element of a set, 3 Empty Set, 4
Entries of a list, 57
Equality of functions, 185
Equality of lists, 58
Equality of sets, 3
Equivalence class, 171
Equivalence Relation, 169
Equivalent Statement, 115
Euclid, 106, 128
Euler, Leonhard, 99, 130
Existence Theorem, 117
Existential quantiﬁer, 46
Existential Statement, 117
Expression, 29
Factorial, 64
False, 29
Fermat’s Last Theorem, 31
Fermat, Pierre de, 31
Fibonacci Sequence, 153
Function, 182
range of, 183
bijective, 202
codomain of, 183
composition of, 193
domain of, 183
equality, 185
injective, 186
inverse, 196
notation, 184
surjective, 186
Function notation, 184
Fundamental Theorem of Calculus, viii
Gamma Function, 67
Geometric sequence, 155
Goldbach Conjecture, 31, 50
Goldbach, Christian, 31 275
Golden ratio, 155
Graph, 150
cycle, 150
edges, 150
vertices, 150 Logical Inference, 55
Mean Value Theorem, 50
Mersenne prime, 132
Multiple, 84
Multiplication Principle, 59 Half-open Interval, 6
If-and-only-if Theorem, 113
Image, 199
Inclusion-Exclusion formula, 75
Index Set, 25
Indexed Set, 24
Induction, 142
strong, 148
Inﬁnite Interval, 6
Injection, see Injective
Injective function, 186
Integers, 3, 4
congruence, 97, 167
modulo n, 177
Intersection of sets, 17
Interval, 6
Closed, 6
Half-open, 6
Inﬁnite, 6
Open, 6
Inverse of a function, 196
Inverse relation, 196
Irrational, 105
Lemma, 82
Length, 57
List, 57
empty, 58
entries, 57
equal, 58
non-repetitive, 60
order, 57
repetition, 60
Logic, 28
contradiction, 103
equivalence, 44
inference, 55
quantiﬁer, 46
existential, 46
universal, 46
symbols, 41, 46
Logical Equivalence, 44 Natural numbers, 4
Necessary condition, 38
Negation of a statement, 35
One-to-one, 186
Onto, 186
Open Interval, 6
Open sentence, 30, 48
Or, 34
Ordered pair, 8
Ordered triple, 9
Parity, 83
Partition, 174
Pascal’s Triangle, 73
Pascal, Blaise, 73
Perfect number, 127, 130
Pigeonhole principle, 191
Pisano, Leonardo, 153
Power Set, 14
Power, Cartesian, 10
Preimage, 199
Prime number, 31, 84
Proof
by cases, 90
by contradiction, 103
by Induction, 142
strong, 148
by smallest counterexample, 152
Contrapositive, 94
Direct, 81, 85
existence, 116
involving sets, 119
within-a-proof, 109
Proposition, 82, 85
Putin, Vladimir, 19
Pythagorean theorem, 31
Quadratic formula, 31
Quantiﬁer, 46
Range of a function, 183
Rational numbers, 6 Index 276
Real numbers, 4
Reﬂexive property of a relation, 165
Relations, 161
between sets, 180
equivalence, 169
class, 171
inverse, 196
reﬂexive, 165
symmetric, 165
transitive, 165
Remainder, 97
Set(s)
builder-notation, 5, 119
cardinalities of
comparison of, 211
equal, 202
unequal, 202
cardinality of, 3, 202
complement, 19
countable, 206
element of, 3
empty, 4
equal, 3
partition of, 174
subset of, 11
uncountable, 206
Sigma notation, 24
Size, see Cardinality
Statement, 29
Biconditional, 39
Conditional, 36
necessary, 38
suﬃcient, 38
converse, 39
equivalent, 115
existential, 117
negation, 52
Stirling’s Formula, 67
Strong Induction, 148
Subset, 11
Suﬃcient condition, 38
Surjection, see Surjective
Surjective function, 186
Symmetric property of a relation, 165
Theorem, 81
existence, 117
if-and-only-if, 113 Three-dimensional space, 10
Transitive property of a relation, 165
Tree, 150
Triple, ordered, 9
True, 29
Truth
table, 33
value, 33
Uncountable set, 206
Union of sets, 17
Unit circle, 13, 19
Universal quantiﬁer, 46
Universal Set, 19
Variable, 30
Vector space, 119
Venn diagram, 21
Wiles, Andrew, 31
WLOG, 92 ...

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