Unformatted text preview: Reading, Discovering and Writing Proofs
Version 0.1.7
Steven Furino
February 1, 2012 Contents
1 In the beginning
1.1 What Makes a Mathematician a Mathematician? . . . . . . . . . . . .
1.2 How The Course Works . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Why do we reason formally? . . . . . . . . . . . . . . . . . . . . . . . .
2 The
2.1
2.2
2.3
2.4 First Time
Objectives . . .
The Language .
Implications . .
Our First Proof 7
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24 4 I Swear to Tell The Whole Truth
4.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Truth Tables as Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Truth Tables to Evaluate Logical Expressions . . . . . . . . . . . . . . 26
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28 5 The
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. 3 Set
3.1
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3.3 It Up
Objectives . . . . . . .
Describing a Set . . .
Set Operations . . . .
3.3.1 Venn Diagrams
3.4 Comparing Sets . . . .
3.5 Working With Sets . . .
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. Undiscovered Country
Objectives . . . . . . . . .
Discovering a Proof . . . .
Reading A Proof . . . . .
The Division Algorithm . .
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41 7 The Greatest Common Divisor
7.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Greatest Common Divisor . . . . . . . . . . . . . . . . . . . . . . . . . 43
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43 6 To Be or Not To Be
6.1 Objectives . . . . . . .
6.2 Quantiﬁers . . . . . .
6.3 The Object Method .
6.4 The Construct Method
6.5 The Select Method . .
6.6 A NonProof . . . . . .
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. 2 Section 0.0 CONTENTS
7.3
7.4 3
Certiﬁcate of Correctess . . . . . . . . . . . . . . . . . . . . . . . . . .
The Extended Euclidean Algorithm (EEA) . . . . . . . . . . . . . . . 47
49 8 Properties Of GCDs
8.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Some Useful Propositions . . . . . . . . . . . . . . . . . . . . . . . . . 51
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51 9 Linear Diophantine Equations
9.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 The Select Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Linear Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . 57
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59 10 Nested Quantiﬁers
10.1 Objectives . . . . . . . . . . . . .
10.2 Onto . . . . . . . . . . . . . . . .
10.2.1 Deﬁnition . . . . . . . . .
10.2.2 Reading . . . . . . . . . .
10.2.3 Discovering . . . . . . . .
10.3 Limits . . . . . . . . . . . . . . .
10.3.1 Deﬁnition . . . . . . . . .
10.3.2 Reading A Limit Proof .
10.3.3 Discovering a Limit Proof .
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74 11 Practice, Practice, Practice: Quantiﬁers and Sets
11.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
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76 12 Congruence
12.1 Objectives . . . . . . . . . . . . .
12.2 Congruences . . . . . . . . . . . .
12.2.1 Deﬁnition of Congruences
12.3 Elementary Properties . . . . . . .
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99 13 Modular Arithmetic
13.1 Objectives . . . . . . . . .
13.2 Modular Arithmetic . . .
13.2.1 [0] ∈ Zm . . . . . .
13.2.2 [1] ∈ Zm . . . . . .
13.2.3 Subtraction in Zm
13.2.4 Division in Zm . .
13.3 Extending Equivalencies .
13.4 Fermat’s Little Theorem .
14 Linear Congruences
14.1 Objectives . . . . . . . .
14.2 The Problem . . . . . .
14.3 Extending Equivalencies
14.4 Examples . . . . . . . . .
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. 15 Chinese Remainder Theorem
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15.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
15.2 An Old Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4 Chapter 0 CONTENTS 15.3 Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . 102
16 Practice, Practice, Practice: Congruences
16.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 Linear and Polynomial Congruences . . . . . . . . . . . . . . . . . . .
16.3 Preparing for RSA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
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110 17 The
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116 RSA Scheme
Objectives . . . . . . . . . . . . .
Why Public Key Cryptography?
Implementing RSA . . . . . . . .
17.3.1 Setting up RSA . . . . . .
17.3.2 Sending a Message . . . .
17.3.4 Example . . . . . . . . . .
17.3.3 Receiving a Message . . .
17.4 Does M = R? . . . . . . . . . . .
17.5 How Secure Is RSA? . . . . . . . .
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. 18 Just Say No
18.1 Objectives . . . . . . . . . . . . . . . .
18.2 Negating Statements . . . . . . . . . .
18.3 Negating Statements with Quantiﬁers
18.3.1 Counterexamples . . . . . . . . .
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120 19 Contradiction
19.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . .
19.2 How To Use Contradiction . . . . . . . . . . . . . . . . . .
19.2.1 When To Use Contradiction . . . . . . . . . . . . .
19.2.2 Reading a Proof by Contradiction . . . . . . . . .
19.2.3 Discovering and Writing a Proof by Contradiction .
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. 20 Contrapositive
20.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20.2 The Contrapositive . . . . . . . . . . . . . . . . . . . . . . . . . . .
20.2.1 When To Use The Contrapositive . . . . . . . . . . . . . . .
20.3 Reading a Proof That Uses the Contrapositive . . . . . . . . . . .
20.3.1 Discovering and Writing a Proof Using The Contrapositive
21 Uniqueness
21.1 Objectives . . . . . . . .
21.2 Introduction . . . . . . .
21.3 Showing X = Y . . . . .
21.4 Finding a Contradiction
21.5 The Division Algorithm .
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. 22 Induction
22.1 Objectives . . . . . . . . . . . . . . . . . .
22.2 Introduction . . . . . . . . . . . . . . . . .
22.3 Principle of Mathematical Induction . . .
22.3.1 Why Does Induction Work? . . . .
22.3.2 Two Examples of Simple Induction
22.3.3 A Diﬀerent Starting Point . . . . . .
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. Section 0.0 CONTENTS 5 22.4 Strong Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
22.4.1 Interesting Example . . . . . . . . . . . . . . . . . . . . . . . . 143
23 Introduction to Primes
23.1 Objectives . . . . . . . . . . . . . . .
23.2 Introduction to Primes . . . . . . . .
23.3 Induction . . . . . . . . . . . . . . .
23.4 Fundamental Theorem of Arithmetic
23.5 Finding a Prime Factor . . . . . . .
23.6 Working With Prime Factorizations .
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151 24 Introduction to Fermat’s Last Theorem
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24.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
24.2 History of Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . 153
24.3 Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
25 Characterization of Pythagorean Triples
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25.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
25.2 Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
26 Fermat’s Theorem for n = 4
26.1 Objectives . . . . . . . . .
26.2 n = 4 . . . . . . . . . . .
26.3 Reducing the Problem . .
26.4 History . . . . . . . . . . .
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165 27 Problems Related to FLT
166
27.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
27.2 x4 − y 4 = z 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
28 Practice, Practice, Practice: Prime Numbers
169
28.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
28.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
29 Complex Numbers
29.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29.2 Diﬀerent Equations Require Diﬀerent Number Systems . . . . . . . . .
29.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
170
170
171 30 Properties Of
30.1 Objectives
30.2 Conjugate
30.3 Modulus . 174
174
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175 Complex Numbers
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.................................. 31 Graphical Representations of Complex Numbers
31.1 Objectives . . . . . . . . . . . . . . . . . . . . . . .
31.2 The Complex Plane . . . . . . . . . . . . . . . . .
31.2.1 (x, y ) . . . . . . . . . . . . . . . . . . . . . .
31.2.2 Modulus . . . . . . . . . . . . . . . . . . . .
31.3 Polar Representation . . . . . . . . . . . . . . . . .
31.4 Converting Between Representations . . . . . . . . .
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179 6 Chapter 0 CONTENTS 32 De Moivre’s Theorem
32.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.2 De Movre’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.3 Complex Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
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184 33 Roots of Complex Numbers
33.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33.2 Complex nth Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33.3 More Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
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188 34 An Introduction to Polynomials
189
34.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
34.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
34.3 Operations on Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 190
35 Factoring Polynomials
193
35.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
35.2 Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
36 The
36.1
36.2
36.3
36.4
36.5 Shortest Path
Objectives . . .
The Problem .
Abstraction . .
Algorithm . . .
Extensions . . . Problem
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200 37 Paths, Walks, Cycles and Trees
201
37.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
37.2 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
37.3 Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
38 Trees
207
38.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
38.2 Properties of Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
39 Dijkstra’s Algorithm
211
39.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
39.2 Dijkstra’s Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
39.3 Certiﬁcate of Optimality . . . . . . . . . . . . . . . . . . . . . . . . . . 216
40 Certiﬁcate of Optimality  Path
40.1 Objectives . . . . . . . . . . . .
40.2 Certiﬁcate of Optimality . . . .
40.3 Weighted Graphs . . . . . . . .
40.4 Certiﬁcate of Optimality  Tree
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226 Chapter 1 In the beginning
1.1 What Makes a Mathematician a Mathematician? Welcome to MATH 135!
Let me begin with a question. What makes a mathematician a mathematician?
Many people would answer that someone who works with numbers is a mathematician.
But bookkeepers for small businesses work with numbers and we don’t normally
consider a bookkeeper as a mathematician. Others might think of geometry and
answer that someone who works with shapes is a mathematician. But architects work
with shapes and we don’t normally consider architects as mathematicians. Still others
might answer that people who use formulas are mathematicians. But engineers work
with formulas and we don’t normally consider engineers as mathematicians. A more
insightful answer would be that people who ﬁnd patterns and provide descriptions
and evidence for those patterns are mathematicians. But scientists search for and
document patterns and we don’t normally consider scientists as mathematicians.
The answer is proof  a rigorous, formal argument that establishes the truth of a
statement. This has been the deﬁning characteristic of mathematics since ancient
Greece.
MATH 135 is about reading, writing and discovering proofs. If you have never done
this before, do not worry. This course will provide you with techniques that will help,
and we will practice those techniques in the context of some very interesting algebra. 1.2 How The Course Works
He who seeks for methods without having a deﬁnite problem in mind seeks
for the most part in vain.
David Hilbert Let me describe how the course works.
Throughout the course, we will work on three problems all of which illustrate the
need for proof. The ﬁrst problem resolves a very important practical commercial
problem. The second problem begins work on one of the most notorious problems in
7 8 Chapter 1 In the beginning all of mathematics. The last problem yields a surprising and beautiful result. Here
are the three problems.
How do we secure internet commerce? Have you ever bought a song or movie
over iTunes? Have you ever done your banking over the web? How do you make
sure that your credit card number and personal information are not intercepted
by bad guys? Number theory allows us to enable secure web transactions. And
that theory is backed by proof.
How many solutions are there to xn + y n = z n where x, y and z must be positive integers and n is an integer greater than or equal to three? This is one of
the most famous problems in the history of mathematics and it took over 350
years to solve. It was ﬁrst conjectured by the French mathematician Pierre de
Fermat in 1637 and was only solved in 1995 by Andrew Wiles.
Why does eiπ + 1 = 0 ? e is a very unusual number. Of all the real numbers a,
ax
there is exactly one where ddx = ax . And that number is e. i is a very unusual
number with its deﬁning property of i2 = −1. π is a very unusual number even
if it is common. It is the unique ratio of the circumference of a circle to its
diameter. Why should that ratio be unique? One is the basis of the natural
numbers, hence the integers, hence the rationals. Zero is a diﬃcult number
and was only accepted into the mathematics of western Europe because of the
inﬂuence of Hindu and Islamic scholars. Why should all of these numbers be
connected in so simple and elegant a form?
To work with these problems we will need to learn about congruences, modular arithmetic, primes and complex numbers. And to work with these topics, we must learn
about proof techniques. The proof techniques will be introduced as we need them.
There is a separate reading for each proof technique. After you do the assigned reading, there will be a short online quiz for you to take to reinforce the material you have
studied. 1.3 Why do we reason formally? Since many people dislike proofs, and think that humans already know enough mathematics, let me deal with the question: “Why bother with proofs?” There are quite
a few reasons.
To prevent silliness. In solving quadratic equations with nonreal roots, some of
you will have encountered the number i which has the special property that
i2 = −1. But then,
√
√√
√
−1 = i2 = i × i = −1 −1 = −1 × −1 = 1 = 1
Clearly, something is amiss. Section 1.3 Why do we reason formally? 9 To understand better. How would most of us answer the question “What’s a real
number?” We would probably say that any number written as a decimal expansion is a real number and any two diﬀerent expansions represent diﬀerent
numbers. But then what about this?
Let x = 0.9 = 0.999 . . . .
Multiplying by 10 and subtracting gives
10x = 9.9
−
x = 0 .9
9x = 9
which implies x = 1, not x = 0.9.
Or suppose we wanted to evaluate the inﬁnite sum
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ...
If we pair up the ﬁrst two terms we get zero and every successive pair of terms
also gives us 0 so the sum is zero.
1 − 1+1 − 1+1 − 1+1 − 1+...
On the other hand, if we pair up the second and third term we get 0 and all
successive pairs of terms give 0 so the sum is 1.
1 −1 + 1 −1 + 1 −1 + 1 −1 + 1 + . . .
Or suppose we wanted to resolve Zeno’s paradox. Zeno was a famous ancient
Greek philosopher who posed the following problem. Suppose the Greek hero
Achilles was going to race against a tortoise and suppose, in recognition of the
slowness of the tortoise, that the tortoise gets a 100m head start. By the time
Achilles has run half the distance between he and the tortoise, the tortoise has
moved ahead. And now again, by the time Achilles has run half the remaining
distance between he and the tortoise, the tortoise has moved ahead. No matter
how fast Achilles runs, the tortoise will always be ahead! You might object that
your eyes see Achilles pass the tortoise, but what is logically wrong with Zeno’s
argument?
To make better commercial decisions. Building pipelines is expensive. And lots
of pipelines will be built in the next few decades. Pipelines will ship oil, natural gas, water and sewage. Finding the shortest route given physical constraints (mountains, rivers, lakes, cities), environmental constraints (protection
of the water table, no access through national or state parks), and supply chain
constraints (access to concrete and steel) is very important. How do pipeline
builders prove that the route they have chosen for the pipeline is the shortest
possible route given the constraints?
To discover solutions. Formal reasoning provides a set of tools that allow us to
think rationally and carefully about problems in mathematics, computing, engineering, science, economics and any discipline in which we create models.
Poor reasoning can be very expensive. Inaccurate application of ﬁnancial models
led to losses of hundreds of billions of dollars during the ﬁnancial crisis of 2008.
To experience joy. Mathematics can be beautiful, just as poetry can be beautiful.
But to hear the poetry of mathematics, one must ﬁrst understand the language. Chapter 2 The First Time
2.1 Objectives The technique objectives are:
1. Deﬁne statement, hypothesis, conclusion and implication.
2. Learn how to structure the analysis of a proof.
3. Carry out the analysis of a proof.
The content objectives are:
1. Deﬁne divisibility.
2. State and prove the Transitivity of Divisibility. 2.2 The Language
Mathematics is the language of mathematicians, and a proof is a method
of communicating a mathematical truth to another person who speaks the
“language”.
(Solow, How to Read and Do Proofs) Mathematics is an unusual language. It is extraordinarily precise. When a proof is
fully and correctly presented, there is no ambiguity and no doubt about its correctness.
However, understanding a proof requires understanding the language. This course will
help you with the basic grammar of the language of mathematics and is applicable
to all proofs. Just as in learning any new language, you will need lots of practice to
become ﬂuent.
With respect to learning proof techniques, the broad objectives of the course are
simple. 10 Section 2.2 The Language 11 1. Explain and categorize proof techniques that can be used in any proof. This
course will teach not only how a technique works, but when it is most likely to
be used and why it works.
2. Learn how to read a proof. This will require you to identify the techniques of
the ﬁrst objective.
3. Discover your own proofs. Knowledge of technique is essential but inadequate.
Or, as we would say in the language of mathematics, technique is “necessary but
not suﬃcient”. Discovering your own proof requires not only technique but also
understanding, creativity, intuition and experience. This course will help with
the technique and experience. Understanding, creativity, and intuition come
with time. Talent helps of course.
4. Write your own proofs. Having discovered a proof, you must distill your discovery into mathematical prose that is targeted at a speciﬁc audience.
Hopefully, in the previous lecture, I convinced you of why we need to prove things.
Now what is it that mathematicians prove? Mathematicians prove statements. Deﬁnition A statement is a sentence which is either true or false. Statement Example 1 Here are some examples of statements.
1. 2 + 2 = 4. (A true statement.)
2. 2 + 2 = 5. (A false statement.)
3. x2 − 1 = 0 has two distinct real roots. (A true statement.)
4. There exists an angle θ such that sin(θ) > 1. (A false statement.) Example 2 Now consider the following sentences.
1. x > 0.
2. ABC is congruent to P QR. These are statements only if we have an appropriate value for x in the ﬁrst sentence
and appropriate instances of ABC and P QR in the second sentence. For example,
if x is the number 5, then the sentence “5 > 0” is a statement since the sentence is
true. If x is the number −5, then the sentence “−5 > 0” is also a statement since
the sentence is false. The key point is that a statement is a sentence which must be
true or false. If x is the English word algebra, then the sentence “algebra > 0” is not
a statement since the sentence is neither true nor false. Sentences like the two above
are called open sentences. 12 Chapter 2 Deﬁnition
Open
Sentence The First Time An open sentence is a sentence that
• contains one or more variables, where
• each variable has values that come from a designated set called the domain of
the variable, and
• where the sentence is either true or false whenever values from the respective
domains of the variables are substituted for the variables. For example, if the domain of x is the set of real numbers, then for any real number
chosen and substituted for x, the sentence “x > 0” is a statement.
In this course, we will treat all open sentences as statements under the assumption
that the values of the variables always come from a suitable domain. Self Check 1 For each of the following, choose one of the following possible answers.
(a) This is not a statement.
(b) This is a true statement.
(c) This is a false statement.
(d) This is an open sentence.
1. {1, 3, 5, 7, 9}
2. sin2 θ + cos2 θ
3. For all real values θ, sin2 θ + cos2 θ = 1
4. Therefore, x = π/2.
5. If x is a positive real number, then log10 x > 0.
6. x2 + 1 = 0 has two real roots. Section 2.3 Implications 2.3
Deﬁnition
Implication 13 Implications The most common type of statement we will prove is an implication. Implications
have the form
If A is true , then B is true
where A and B are themselves statements. An implication is more commonly read as
If A, then B
or
A implies B
or
A⇒B Deﬁnition
Compound
Statement An implication is a compound statement, that is, it is made up of more than one
statement. In the statement “A implies B ”, A is a statement which may be true or false. B is a
statement which may be true or false. “A implies B ” is also a statement which may
be true or false. Deﬁnition The statement A is called the hypothesis. The statement B is called the conclusion. Hypothesis,
Conclusion REMARK
To prove the implication “A implies B ”, you assume that A is true and you use this
assumption to show that B is true. A is what you start with. B is where you must
end up.
To use the implication “A implies B ”, you must ﬁrst establish that A is true. After
you have established that A is true, then you can invoke B .
It is crucial that you are able to identify
1. the hypothesis
2. the conclusion
3. whether you are using or proving an implication Here are some examples of implications. Example 3 If x is a positive real number, then log10 x > 0. 14 Chapter 2 The First Time Hypothesis: x is a positive real number.
Conclusion: log10 x > 0. Example 4 Let f (x) = x sin(x). Then f (x) = x for some real number x with 0 ≤ x ≤ 2π .
Hypothesis: f (x) = x sin(x).
Conclusion: f (x) = x for some real number x with 0 ≤ x ≤ 2π . Example 5 In plane geometry, ∠ABC = ∠XY Z whenever
Hypothesis: All ﬁgures are in the plane. ABC is similar to
ABC ∼ XY Z. XY Z. Conclusion: ∠ABC = ∠XY Z . Self Check 2 Identify the hypothesis and the conclusion in each of the following statements.
1. If a, b and c are real numbers, and b2 − 4ac > 0, then ax2 + bx + c = 0 has two
distinct, real roots.
(a) a, b and c are real numbers.
(b) b2 − 4ac > 0.
(c) a, b and c are real numbers and b2 − 4ac > 0.
(d) ax2 + bx + c = 0.
(e) ax2 + bx + c = 0 has two distinct, real roots.
2. If the line segment AB intersects the line segment P Q at O, then ∠AOQ =
∠P OB .
(a) The line segment AB
(b) The line segment AB intersects the line segment P Q.
(c) The line segment AB intersects the line segment P Q at O.
(d) ∠AOQ = ∠P OB .
(e) None of the above.
3. y = ax2 − 1 has no real root if a < 0.
(a) y = ax2 − 1.
(b) y = ax2 − 1 has no real roots.
(c) a < 0.
(d) None of the above. Section 2.4 Our First Proof 15 4. When x is an integer, the maximum value of y = −x4 + 4x2 + 0.5 is 4.
(a) x is an integer.
(b) y = −x4 + 4x2 + 0.5
(c) x is a maximum value.
(d) The maximum value of y = −x4 + 4x2 + 0.5 is 4.
(e) None of the above. 2.4 Our First Proof Let us read our ﬁrst proof. We begin with a deﬁnition. Deﬁnition
Divisibility An integer m divides an integer n, and we write m  n, if there exists an integer k
so that n = km. Example 6
• 3  6 since we can ﬁnd an integer k , 2 in this case, so that 6 = 2 × 3.
• 5 6 since no integer k exists so that 6 = k × 5.
• For all integers a, a  0 since 0 = 0 × a. This is true for a = 0 as well.
• For all nonzero integers a, 0 a since there is no integer k so that k × 0 = a. Some comments about deﬁnitions are in order. If mathematics is thought of as a
language, then deﬁnitions are the vocabulary and our prior mathematical knowledge
indicates our experience and versatility with the language.
Mathematics and the English language both share the use of deﬁnitions as extremely
practical abbreviations. Instead of saying “a domesticated carnivorous mammal
known scientiﬁcally as Canis familiaris” we would say “dog.” Instead of writing
down “there exists an integer k so that n = km”, we write “m  n.”
However, mathematics diﬀers greatly from English in precision and emotional content.
Mathematical deﬁnitions do not allow ambiguity or sentiment. Deﬁnition
Proposition Proposition 1 A proposition is a true statement that you are trying to prove. Consider the following proposition and proof. (Transitivity of Divisibility (TD))
Let a, b and c be integers. If a  b and b  c, then a  c. 16 Chapter 2 The First Time Proof: Since a  b, there exists an integer r so that ra = b. Since b  c, there exists
an integer s so that sb = c. Substituting ra for b in the previous equation, we get
(sr)a = c. Since sr is an integer, a  c.
Though this is a simple proof, other proofs can be diﬃcult to read because of the habits
of writing for professional audiences. Many proofs share the following properties which
can be frustrating for students.
1. Proofs are economical. That is, a proof includes what is needed to verify the
truth of a proposition but nothing more.
2. Proofs do not usually identify the hypothesis and the conclusion.
3. Proofs sometimes omit or combine steps.
4. Proofs do not always explicitly justify steps.
5. Proofs do not reﬂect the process by which the proof was discovered.
The reader of the proof must be conscious of the hypothesis and conclusion, ﬁll in
the omitted parts and justify each step.
Let’s analyze the proof of the Transitivity of Divisibility in detail because it will give
us some sense of how to analyze proofs in general. First, observe that “If a  b and
b  c, then a  c.” is an open sentence, and that the domains for the variables a, b and
c are speciﬁed in the ﬁrst sentence, “Let a, b and c be integers.” REMARK
When you are reading a proof of an implication, do the following.
1. Explicitly identify the hypothesis and the conclusion. If the hypothesis contains
no statements write “No explicit hypothesis”. At the end of the proof, you
should be able to identify where each part of the hypothesis has been used.
2. Explicitly identify the core proof technique. When reading a proof, the reader
usually works forward from the hypothesis until the conclusion is reached. Speciﬁc techniques will be covered later in the course.
3. Record any preliminary material needed, usually deﬁnitions or propositions that
have already been proved. Judgement is needed here about how much to include.
4. Justify each step with reference to the deﬁnitions, previously proved propositions
or techniques used.
5. Add missing steps where necessary and justify these steps with reference to the
deﬁnitions, previously proved propositions or techniques used. Professional mathematicians do all of these things implicitly but for the ﬁrst part of
this course, we will do these things explicitly.
We will do a line by line analysis, so to make our work easier, we will write each
sentence on a separate line. Section 2.4 Our First Proof 17 Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Since a  b, there exists an integer r so that ra = b.
2. Since b  c, there exists an integer s so that sb = c.
3. Substituting ra for b in the previous equation, we get (sr)a = c.
4. Since sr is an integer, a  c. Let’s analyze the proof. What we do now will seem like overkill but it serves two
purposes. It gives practice at justifying every line of a proof, and it gives us a structure that we can use for other proofs. Lastly, recall that the author is proving an
implication. The author assumes that the hypothesis is true, and uses the hypothesis
to demonstrate that the conclusion is true. Here goes.
Analysis of Proof We begin by explicitly identifying the hypothesis and the conclusion.
Hypothesis: a, b and c are integers. a  b and b  c.
Conclusion: a  c.
Core Proof Technique: Work forwards from the hypothesis.
Preliminary Material: The deﬁnition of divides. An integer m divides an
integer n, and we write m  n, if there exists an integer k so that n = km.
Sentence 1 Since a  b, there exists an integer r so that ra = b.
In this sentence, the author of the proof uses the hypothesis a  b and the
deﬁnition of divides.
Sentence 2 Since b  c, there exists an integer s so that sb = c.
In this sentence, the author uses the hypothesis b  c and the deﬁnition of
divides.
Sentence 3 Substituting ra for b in the previous equation, we get (sr)a = c.
Here, the author works forward using arithmetic. The actual work is:
sb = c and ra = b implies s(ra) = c which implies (sr)a = c.
Sentence 4 Since sr is an integer, a  c.
Lastly, the author uses the deﬁnition of divides. In this case, the m, k and n
of the deﬁnition apply to the a, sr and c of the proof. It is important to note
that sr is an integer, otherwise the deﬁnition of divides does not apply.
At the end of each proof, you should be able to identify where each part of the
hypothesis was used. It is obvious where a  b and b  c were used. The hypothesis “a,
b and c are integers” was needed to allow the author to use the deﬁnition of divides.
This completes the analysis of our ﬁrst proof. Between the readings, lectures, quizzes,
assignments and tests, you will work your way through roughly one hundred proofs. Chapter 3 Set It Up
3.1 Objectives The technique objectives are:
1. Deﬁne and gain experience with set, element, setbuilder notation, deﬁning property, subset, superset, equality of sets, empty set, universal set, complement,
cardinality, union, intersection and diﬀerence.
2. Be able to read and use Venn diagrams. 3.2 Describing a Set Sets are foundational in mathematics and literally appear everywhere. Deﬁnition
Set,
Element A set is a collection of objects. The objects that make up a set are called its elements
(or members). Sets can contain any type of objects. Since this is a math course, we frequently
use sets of numbers. But sets could contain letters, the letters of the alphabet for
example, or books, such as those in a library collection.
It is customary to use uppercase letters (A, B, C . . .) to represent sets and lowercase
letters (a, b, c, . . .) to represent elements. If a is an element of the set A, we write
a ∈ A. If a is not an element of the set A, we write a ∈ A.
Small sets can be explicitly listed. For example, the set of primes less than 10 is
{2, 3, 5, 7}
Many sets are either too large to be listed (the set of all primes less than 10,000) or
are deﬁned by a rule. In these cases, we employ setbuilder notation which makes use
of a deﬁning property of the set. For example, the set of all real numbers between 1
and 2 could be written as
{x ∈ R  1 ≤ x ≤ 2}
18 Section 3.2 Describing a Set 19 The part of the description following the bar () is the deﬁning property of the set.
Some authors use a colon (:) instead of a bar and write
{x ∈ R : 1 ≤ x ≤ 2}
Some letters have become associated with speciﬁc sets.
N
Z
Q
Q
R
C Deﬁnition
Subset natural numbers, 1, 2, 3, . . .
integers
rational numbers, { p  p, q ∈ Z, q = 0}
q
irrational numbers
real numbers
complex numbers {x + yi  x, y ∈ R} A set A is called a subset of a set B , and is written A ⊆ B , if every element of A
belongs to B . Symbolically, we write
A ⊆ B means x ∈ A ⇒ x ∈ B
or equivalently
A ⊆ B means “For all x ∈ A, x ∈ B
We sometimes say that A is contained in B . Example 7
{1, 2, 3} ⊆ {1, 2, 3, 4} Deﬁnition
Proper
Subset A set A is called a proper subset of a set B , and written A ⊂ B , if every element
of A belongs to B and there exists an element in B which does not belong to A. In the previous example, it is also the case that Example 8
{1, 2, 3} ⊂ {1, 2, 3, 4} Deﬁnition
Superset A set A is called a superset of a set B , and written A ⊇ B , if every element of B
belongs to A. A ⊇ B is equivalent to B ⊆ A. Example 9
{1, 2, 3, 4} ⊇ {1, 2, 3} Deﬁnition
Proper
Subset As before, a set A is called a proper superset of a set B , and written A ⊃ B , if
every element of B belongs to A and there exists an element in A which does not
belong to B . 20 Chapter 3 Set It Up Example 10
{1, 2, 3, 4} ⊃ {1, 2, 3} Deﬁnition
Set Equality Saying that two sets A and B are equal, and writing A = B , means that A and B
have exactly the same elements. Equivalently, and the more usual way of showing
A = B , is to show mutual inclusion, that is, show A is contained in B and B is
contained in A. Symbolically, we write
A = B means A ⊆ B AND B ⊆ A Deﬁnition
Empty Set Deﬁnition
Universal
Set Deﬁnition
Set Complement There is a special set, called the empty set and denoted by ∅, which contains no
elements. The empty set is a subset of every set. When we discuss sets, we are often concerned with subsets of some implicit or speciﬁed
set U , called the universal set. In our work on divisibility and greatest common
divisors, we will be concerned with integers as the universal set, even when we don’t
explicitly say so. Relative to a universal set U , the complement of a subset A of U , written A, is the
set of all elements in U but not in A. Symbolically, we write
A = {x  x ∈ U AND x ∈ A} Deﬁnition Lastly, the cardinality of a set A, written A, is the number of elements in the set. Cardinality Example 11 For example, if A = {1, 2, 3, 4}, then A = 4. Here’s a pair of mindblowing questions.
What is the cardinality of N? How much larger is Q than N? Example 12 Let S = {x ∈ R  x2 = 2} and T = {x ∈ Q  x2 = 2}.
1. Describe the set S by listing its elements. What is the cardinality of S ?
2. Describe the set T by listing its elements. What is the cardinality of T ?
3. List all of the subsets of S .
Solution:
√
√
1. S = { 2, − 2}. S  = 2.
2. T = ∅. T  = 0.
√
√
3. ∅, { 2}, {− 2}, S Section 3.3 Set Operations Example 13 21 Let the universal set for this question be U , the set of natural numbers less than
twenty. Let T be the set of integers divisible by three and F be the set of integers
divisible by ﬁve.
1. Describe T by explicitly listing the set and by using setbuilder notation in at
least two ways.
2. Find a subset of T of cardinality three.
3. Find an element which belongs to both T and F .
4. Find an element which belongs to neither T nor F .
5. Explicitly list the set T .
Solution:
1. Explicitly listing the set gives T = {3, 6, 9, 12, 15, 18}. Two setbuilder descriptions of the set are T = {n ∈ N : 3  n, n ≤ 20} and T = {3k  k ∈ N, 3k ≤ 20}
2. {3, 6, 9}. There are several choices possible.
3. 15. Notice that this is an element, not a set.
4. 1. There are several choices possible.
5. {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19} 3.3
Deﬁnition
Union Set Operations The union of two set A and B , written A ∪ B , is the set of all elements belonging to
either set A or set B . Symbolically we write
A ∪ B = {x  x ∈ A OR x ∈ B } Note that when we say “set A or set B ” we mean the mathematical use of or. That
is, the element can belong to A, B or both A and B . Deﬁnition
Intersection The intersection of two set A and B , written A ∩ B , is the set of all elements
belonging to both set A and set B . Symbolically we write
A ∩ B = {x  x ∈ A AND x ∈ B } 22 Chapter 3 Deﬁnition
Diﬀerence Set It Up The diﬀerence of two set A and B , written A − B (or A \ B ), is the set of all elements
belonging to A but not B . Symbolically we write
A − B = {x  x ∈ A AND x ∈ B }
If U is the universal set and A ⊂ U then A = U − A. Example 14 Let the universal set for this question be U , the set of natural numbers less than or
equal to twelve. Let T be the set of integers divisible by three, F be the set of integers
divisible by ﬁve and P the set of primes. Determine each of the following.
1. T ∪ F
2. T ∩ F
3. P
4. P ∩ (T ∪ F )
5. T ∪ F
6. (T ∪ F ) − P
Solution:
1. T ∪ F = {3, 5, 6, 9, 10, 12}
2. T ∩ F = ∅
3. P = {1, 4, 6, 8, 9, 10, 12}
4. P ∩ (T ∪ F ) = {3, 5}
5. T ∪ F = {1, 2, 4, 5, 7, 8, 10, 11}
6. (T ∪ F ) − P = {6, 9, 10, 12} Section 3.4 Comparing Sets 3.3.1 23 Venn Diagrams Venn diagrams can serve as useful illustrations of set relationships. In Figure 3.3.1
below, the universal set is U = {a, b, c, d, e, w}, the set A = {a, b, c, d} and the set
B = {d, e}. The element d lies in the intersection of sets A and B . Since d is the only
such element, A ∩ B = {d}. The element w does not lie in either set A or B . w
A B b
d
a c e Figure 3.3.1: Venn Diagram Add schematic Venn diagrams for intersection, union, disjoint, subset,
superset, complement 3.4 Comparing Sets One common use of sets is to describe values which are solutions to an equation, but
care in expression is required here. The following two sentences mean diﬀerent things.
1. Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. The solutions to the quadratic equation
ax2 + bx + c = 0
are
x= −b ± √
b2 − 4ac
2a 2. Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. Then
√
−b ± b2 − 4ac
x=
2a
are solutions to the quadratic equation
ax2 + bx + c = 0
The ﬁrst sentence asserts that a complete description of all solutions is given. The
√
−b ± b2 − 4ac
second sentence only asserts that x =
are solutions, not that they
2a
are the complete solution. In the √
language of sets, if S is the complete solution to
−b ± b2 − 4ac
ax2 + bx + c = 0, and T = {
}, Sentence 1 asserts that S = T (which
2a
implies S ⊆ T and T ⊆ S ) but Sentence 2 only asserts that T ⊆ S . 24 Chapter 3 Set It Up This point can be confusing. Statements about solutions are often implicitly divided
into two sets: the set S of all solutions and a set T of proposed solutions. One must be
careful to determine whether the statement is equivalent to S = T or T ⊆ S . Phrases
like the solution or complete solution or all solutions indicate S = T . Phrases like a
solution or are solutions indicate T ⊆ S .
Similar confusion arises when showing that sets have more than one representation.
For example, a circle centred at the origin O is often deﬁned geometrically as the set
of points equidistant from O. Others deﬁne a circle algebraically in the Cartesian
plane as the set of points satisfying x2 + y 2 = r2 . To show that the two deﬁnitions
describe the same object, one must show that the two sets of points are equal. 3.5 Working With Sets Given a set S and a set T , there are two very frequent tasks one must perform: one
must show S ⊆ T or S = T . In fact, the second task is just two instances of the ﬁrst
task: to show S = T one must show S ⊆ T and T ⊆ S .
So, the important message here is that mathematicians must become skilled at demonstrating that S ⊆ T . The plan in all cases is the same: choose a generic element of S
and show that it belongs to T . Symbolically
S ⊆ T means x ∈ S ⇒ x ∈ T
or equivalently
S ⊆ T means For all x ∈ S, x ∈ T
The element chosen must be completely generic and could, if forced, be instantiated
as any element of the set S . Showing that a speciﬁc element of S belongs to T is
inadequate. Example 1 Consider the statement:
Integer multiples of π are roots of f (x) = (x2 − 1) sin x.
1. Explicitly identify two sets used in this statement.
2. Are the two sets equal?
3. Is the statement true?
Solution:
1. Let S be the set of all roots of f (x) = (x2 − 1) sin x. (We could write S more
symbolically as S = {x ∈ R  f (x) = 0}.) Let T be the set of integer multiples
of π . (We could also write T more symbolically as T = {nπ  n ∈ Z}). Section 3.5 Working With Sets 25 2. To show that S = T we must show T ⊆ S and S ⊆ T . Since sin(nπ ) = 0 for
all integers n, we know that f (nπ ) = 0. Now, the deﬁning property of S is that
a real number x belongs to S if f (x) = 0. Since f (nπ ) = 0, nπ ∈ S . This is
equivalent to: if nπ ∈ T then nπ ∈ S , or equivalently, T ⊆ S . Now consider
x = 1. The value x = 1 is a solution to (x2 − 1) sin x = 0 and so belongs to S ,
but it is not an integer multiple of π , so it does not belong to T . That is, S ⊆ T
and so the two sets are not equal.
3. The statement is true. The statement only claims that T ⊆ S , not S = T . Chapter 4 I Swear to Tell The Whole Truth
4.1 Objectives The technique objectives are:
1. Deﬁne and gain experience with truth tables.
2. Deﬁne not, and and or.
3. Use truth tables to establish the equivalence of DeMorgan’s Laws. 4.2
Deﬁnition
Compound,
Component Truth Tables as Deﬁnitions All of the statements we need to prove will be compound statements, that is, statements composed of several individual statements called component statements. For example, the compound statement
If a  b and b  c, then a  c.
is composed of the three statements
a  b,
b  c, and
ac
Suppose we let X be the statement a  b and Y be the statement b  c and Z be the
statement a  c. Then our original statement
If a  b and b  c, then a  c.
becomes 26 Section 4.3 Truth Tables as Deﬁnitions 27 X and Y imply Z .
If we knew the truth values of X , Y and Z , then we would be able to determine the
truth value of the compound statement “X and Y imply Z ”. And that is where truth
tables come in. Truth tables contain all possible values of the component statements
and determine the truth value of the compound statement.
Truth tables can be used to deﬁne the truth value of a statement or evaluate the
truth value of a statement. For logical operations like not, and, or and implies, truth
tables are used to deﬁne the truth value of the compound statement.
For example, the simplest deﬁnition is that of “NOT A”, written ¬A.
A ¬A
TF
FT
In English, if the statement A is true, then the statement “NOT A” is false. If the
statement A is false, then the statement “NOT A” is true.
Two very important and common logical connectives are AND and OR. Note that
these do not always coincide with our use of the words and and or in the English
language!
The deﬁnition of “A AND B ”, written A ∧ B , is
A
T
T
F
F B A∧B
T
T
F
F
T
F
F
F The deﬁnition of “A OR B ”, written A ∨ B , is
A
T
T
F
F B A∨B
T
T
F
T
T
T
F
F This is an opportune moment to highlight the diﬀerence between mathematical language and the English language. If you are visiting a friend and your friend oﬀers
you “coﬀee or tea”, we interpret that to mean that you may have coﬀee or tea but
not both. However, the logical A ∨ B results in a true statement when A is true, B
is true or both are true.
The deﬁnition of “A implies B ”, written A ⇒ B , often seems strange.
A
T
T
F
F B A⇒B
T
T
F
F
T
T
F
T The ﬁrst two rows in the table make sense. The last two make less sense. How can a
false hypothesis result in a true statement? The basic idea is that if one is allowed to
assume an hypothesis which is false, any conclusion can be derived. 28 Chapter 4 4.3 I Swear to Tell The Whole Truth Truth Tables to Evaluate Logical Expressions We can construct truth tables for compound statements by evaluating parts of the
compound statement separately and then evaluating the larger statement. Consider
the following truth table which shows the truth values of ¬(A ∨ B ) for all possible
combinations of truth values of the component statements A and B . Example 2 Construct a truth table for ¬(A ∨ B ).
A
T
T
F
F B A ∨ B ¬(A ∨ B )
T
T
F
F
T
F
T
T
F
F
F
T Here is another example. Example 3 Construct a truth table for A ⇒ (B ∨ C ).
A
T
T
T
T
F
F
F
F Deﬁnition
Logically
equivalent Example 4 B
T
T
F
F
T
T
F
F C B ∨ C A ⇒ (B ∨ C )
T
T
T
F
T
T
T
T
T
F
F
F
T
T
T
F
T
T
T
T
T
F
F
T Two compound statements are logically equivalent if they have the same truth
values for all combinations of their component statements. We write S1 ≡ S2 to
mean S1 is logically equivalent to S2 . Construct a single truth table for ¬(A ∨ B ) and (¬A) ∧ (¬B ). Are these statements
logically equivalent?
A
T
T
F
F B A∨B
T
T
F
T
T
T
F
F ¬(A ∨ B )
F
F
F
T ¬A ¬B
F
F
F
T
T
F
T
T (¬A) ∧ (¬B )
F
F
F
T Since the columns representing ¬(A ∨ B ) and (¬A) ∧ (¬B ) are identical,
we can conclude that ¬(A ∨ B ) ≡ (¬A) ∧ (¬B ). Section 4.3 Truth Tables to Evaluate Logical Expressions 29 The preceding example and your assignments demonstrate DeMorgan’s Laws. Proposition 1 (De Morgan’s Law’s (DML))
If A and B are statements, then
1. ¬(A ∨ B ) ≡ (¬A) ∧ (¬B )
2. ¬(A ∧ B ) ≡ (¬A) ∨ (¬B ) Exercise 1 Use truth tables to show that for statements A, B and C , the Associativity Laws
hold. That is
1. A ∨ (B ∨ C ) ≡ (A ∨ B ) ∨ C
2. A ∧ (B ∧ C ) ≡ (A ∧ B ) ∧ C Exercise 2 Use truth tables to determine which of the following statements is equivalent to A ∨
(B ∧ C ).
1. A ∨ B ∧ C
2. (A ∧ B ) ∨ (A ∧ C )
3. (A ∨ B ) ∧ (A ∨ C ) Exercise 3 What logical statement is equivalent to ¬(A ⇒ B )? Provide evidence in the form of
a truth table. Chapter 5 The Undiscovered Country
5.1 Objectives The technique objectives are:
1. Discover a proof using the Direct Proof technique.
2. Write a condensed proof.
3. Read a proof.
The content objectives are:
1. Prove the Divisibility of Integer Combinations.
2. Prove the Bounds By Divisibility.
3. State the Division Algorithm. 5.2 Discovering a Proof Discovering a proof of a statement is generally hard. There is no recipe for this, but
there are some tips that may be useful, and as we go on through the course, you will
learn speciﬁc techniques. Consider the following proposition. Proposition 2 (Divisibility of Integer Combinations (DIC))
Let a, b and c be integers.
If a  b and a  c, then a  (bx + cy ) for any integers x and y . The very ﬁrst thing to do is explicitly identify the hypothesis and the conclusion.
Hypothesis: a, b, c ∈ Z, a  b and a  c. x, y ∈ Z
Conclusion: a  (bx + cy )
30 Section 5.2 Discovering a Proof 31 Since we are proving a statement, not using a statement, we assume that the hypothesis is true, and then demonstrate that the conclusion is true. This straightforward
approach is called Direct Proof. However, in actually discovering a proof we do not
need to work only forwards from hypothesis. We can work backwards from the conclusion and meet somewhere in the middle. When writing the proof we must ensure
that we begin with the hypothesis and end with the conclusion.
Whether working forwards or backwards, I ﬁnd it best to proceed by asking questions.
When working backwards, I ask
What mathematical fact would allow me to deduce the conclusion?
For example, in the proposition under consideration I would ask
What mathematical fact would allow me to deduce that a  (bx + cy )?
The answer tells me what to look for or gives me another statement I can work
backwards from. In this case the answer would be
1. If there exists an integer k so that bx + cy = ak , then a  (bx + cy ).
Note that the answer makes use of the deﬁnition of divides. Now I could ask the
question
How can I ﬁnd such a k ?
The answer is not obvious so let’s turn to working forwards from the hypothesis. In
this case my standard two questions are
Have I seen something like this before?
What mathematical fact can I deduce from what I already know?
I have seen a  b in an hypothesis before, in the proof of the Transitivity of Divisibility.
I can use the deﬁnition of divisibility to assert that
2. There exists an integer r such that b = ra.
I also know that a  c so I can again use the deﬁnition of divisibility to assert that
3. There exists an integer s such that c = sa.
Hmmm, what now? Let’s look again at Sentence 1.
1. If there exists an integer k so that bx + cy = ak , then a  (bx + cy ). 32 Chapter 5 The Undiscovered Country There is a bx + cy in Sentence 1 and an algebraic expression for b and c in Sentences
2 and 3. Substituting gives
bx + cy = (ra)x + (sa)y
and factoring out the a gives
bx + cy = (ra)x + (sa)y = a(rx + sy )
If we let k = rx + sy then k is an integer, since adding integers gives integers and
multiplying integers gives integers, and so there exists an integer k so that bx+cy = ak .
Hence, a  (bx + cy ).
We are done. Almost. We have discovered a proof but this is rough work. We must
now write a condensed proof. Just like any other writing, the amount of detail needed
in expressing your thoughts depends upon the audience. A proof of a statement
targeted at an audience of professional specialists in algebra will not look the same as
a proof targeted at a high school audience. When you approach a proof, you should
ﬁrst make a judgement about the audience. I suggest that you write for your peers.
That is, you write your proof so that you could hand it to a classmate and expect
that they would understand the proof.
Proof: Since a  b, there exists an integer r such that b = ra. Since a  c, there
exists an integer s such that c = sa. Let x and y be any integers. Now bx + cy =
(ra)x + (sa)y = a(rx + sy ). Since rx + sy is an integer, it follows from the deﬁnition
of divisibility that a  (bx + cy ).
Note that this proof does not reﬂect the discovery process, and it is a Direct Proof.
It begins with the hypothesis and ends with the conclusion. 5.3 Reading A Proof Here is another proposition and condensed proof. Proposition 3 (Bounds By Divisibility (BBD))
Let a and b be integers. If a  b and b = 0 then a ≤ b. Proof: Since a  b, there exists an integer q so that b = qa. Since b = 0, q = 0. But
if q = 0, q  ≥ 1. Hence, b = qa = q a ≥ a.
Let’s analyze this proof. First, we will rewrite the proof line by line.
Proof: (For reference purposes, each sentence of the proof is written on a separate
line.)
1. Since a  b, there exists an integer q so that b = qa.
2. Since b = 0, q = 0. Section 5.4 The Division Algorithm 33 3. But if q = 0, q  ≥ 1.
4. Hence, b = qa = q a ≥ a. Now the analysis.
Analysis of Proof As usual, we begin by explicitly identifying the hypothesis and
the conclusion.
Hypothesis: a and b are integers. a  b and b = 0.
Conclusion: a ≤ b.
Core Proof Technique: Direct Proof.
Preliminary Material: The deﬁnition of divides.
Now we justify every sentence in the proof.
Sentence 1 Since a  b, there exists an integer q so that b = qa.
In this sentence, the author of the proof uses the hypothesis a  b and the
deﬁnition of divides.
Sentence 2 Since b = 0, q = 0.
If q were zero, then b = qa would imply that b is zero. Since b is not zero, q
cannot be zero.
Sentence 3 But if q = 0, q  ≥ 1.
Since q is an integer from Sentence 1, and q is not zero from Sentence 2, q ≥ 1
or q ≤ −1. In either case, q  ≥ 1.
Sentence 4 Hence, b = qa = q a ≥ a.
Sentence 1 tells us that b = qa. Taking the absolute value of both sides gives
b = qa and using the properties of absolute values we get qa = q a. From
Sentence 3, q  ≥ 1 so q a ≥ a. Exercise 4 Prove the following statement. Let a, b, c and d be integers.
If a  c and b  d, then ab  cd. Exercise 5 Prove the following statement. Let x be an integer. If 2  (x2 − 1), then 4  (x2 − 1). 5.4 The Division Algorithm As you have known since grade school, not all integers are divided evenly by other
integers. There is usually a remainder. We record this as the Division Algorithm. 34 Chapter 5 Proposition 4 The Undiscovered Country (Division Algorithm (DA))
If a and b are integers, and b > 0, then there exist unique integers q and r such that
a = qb + r where 0 ≤ r < b. We will not prove this statement now. You will see a proof of the uniqueness part
later on and a complete proof is available in the appendix. Add to appendix. Let’s
see some examples before a few remarks. Example 5
a=q×b+r
20 = 2 × 7 + 6
21 = 3 × 7 + 0
−20 = −3 × 7 + 1 REMARK
• The integer q is called the quotient.
• The integer r is called the remainder.
• The integer r is always strictly less than b.
• The integer r is always positive or zero.
• Observe that b  a if and only if the remainder is 0.
• Though the proposition is commonly known as the Division Algorithm, it is not
really an algorithm since it doesn’t provide a ﬁnite sequence of steps that will
construct q and r. It turns out that the Division Algorithm is remarkably useful. To see how, we must
ﬁrst deﬁne the greatest common divisor which we do soon. Chapter 6 To Be or Not To Be
6.1 Objectives The technique objectives are:
1. Learn the basic structure of quantiﬁers.
2. Use the Object, Construct and Select Methods. 6.2 Quantiﬁers Not all mathematical statements are in the form “If A, then B ”. You will encounter
statements of the form there is, there are, there exists or for all, for each, for every, for
any. The ﬁrst three are all examples of the existential quantiﬁer there is and the
ﬁnal four are all examples of the universal quantiﬁer for all. The word existence is
used to make it clear that we are looking for or looking at a particular mathematical
object. The word universal is used to make it clear that we are looking for or looking
at a set of objects all of which share some desired behaviour. REMARK
All statements which use quantiﬁers look basically like one of the following two open
sentences, though some elements of the sentence may be implicit or appear in a
diﬀerent order.
There exists an x in the set S such that P (x) is true.
For every x in the set S , P (x) is true.
where P (x) is an open sentence that uses the variable x.
Some mathematicians prefer a more symbolic approach. The symbol ∃ stands for the
English expression “there exists”. The symbol ∀ stands for the English expression
“for all”. Symbolically, the two quantiﬁed sentences above are written as:
∃x ∈ S, P (x)
∀x ∈ S, P (x)
35 36 Chapter 6 To Be or Not To Be REMARK
All statements which use quantiﬁers share a basic structure.
1. a quantiﬁer which will be either an existential or universal quantiﬁer,
2. a variable which can be any mathematical object,
3. a set which is the domain of the variable, often implicit, and
4. an open sentence which involves the variable,
It is crucial that you be able to identify the four parts in the structure of quantiﬁed
statements. Here are some examples. Let’s begin with something we have already seen. Example 6
1. There exists an integer k so that n = km
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
k
Z
n = km Our next example could come from any of several branches of mathematics.
2. There exists a real number x such that f (x) = 0.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
x
R
f (x) = 0 This is a good point to illustrate the inﬂuence of the domain. Suppose in this
example we are interested in the speciﬁc function f (x) = x2 − 2. Then the
statement
There exists a real number x such that x2 − 2 = 0.
is true since we can ﬁnd an x, √ 2, so that x2 − 2 = 0. But if we change the domain to integers, the statement
There exists an integer x such that x2 − 2 = 0.
√
√
is false because neither of the two real roots, 2 or − 2 are integers. So
changing the domain can change the truth value of the statement. In practice,
the domain is often not explicitly stated and is inferred from context.
3. For every integer n > 5, 2n > n2 . Section 6.3 The Object Method 37 Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
n
Z, n > 5
2n > n2 The sentence might appear as “2n > n2 for all integers n > 5”. The order is
diﬀerent but the meaning is the same.
4. There exists an angle θ such that sin(θ) = 1.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
an angle θ
R, inferred from the context
sin(θ) = 1 Note that in this example, the universe of discourse is implicit. Note also that
there can be many objects, many angles θ, which satisfy the statement.
5. For every angle θ, sin2 (θ) + cos2 (θ) = 1.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
θ
R, inferred from context
sin2 (θ) + cos2 (θ) = 1 6. If f is continuous on [a, b] and diﬀerentiable on (a, b) and f (a) = f (b), then
there exists a real number c ∈ (a, b) such that f (c) = 0.
This conclusion of this implication uses an existential quantiﬁer. The hypothesis
and the conclusion are:
Hypothesis: f is continuous on [a, b] and diﬀerentiable on (a, b) and f (a) =
f (b).
Conclusion: There exists a real number c ∈ (a, b) such that f (c) = 0.
For the conclusion, the parts of the quantiﬁed statement are given below.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
c
(a, b) ⊂ R
f (c) = 0 It takes practice to become ﬂuent in reading and writing statements that use quantiﬁers. 6.3 The Object Method 38 Chapter 6 To Be or Not To Be REMARK
We use the Object Method when an existential quantiﬁer occurs in the hypothesis.
Suppose that we must prove “A implies B ” and A uses an existential quantiﬁer. That
is, A looks like
There exists an x in the set S such that P (x) is true.
We proceed exactly as the English language interpretation would suggest  we assume
that the object x exists. We should:
1. Identify the four parts of the quantiﬁed statement.
2. Assume that a mathematical object x exists within the set S so that the statement P (x) is true.
3. Make use of this information to generate another statement. For example, let’s look at the proof of the Transitivity of Divisibility again. Proposition 5 (Transitivity of Divisibility (TD))
Let a, b and c be integers.
If a  b and b  c, then a  c. Proof: Since a  b, there exists an integer r so that ra = b. Since b  c, there exists
an integer s so that sb = c. Substituting ra for b in the previous equation, we get
(sr)a = c. Since sr is an integer, a  c.
You might ﬁrst ask “Where is the existential quantiﬁer?”. It isn’t obvious – yet. But
recall the deﬁnition of divisibility.
An integer m divides an integer n, and we write m  n, if there exists an
integer k so that n = km.
The sentence “there exists an integer k so that n = km” uses the existential quantiﬁer.
It is very common in mathematics that sentences contain implicit quantiﬁers and you
should be alert for them. Returning to divisibility, we have already identiﬁed the four
parts of the quantiﬁed sentence.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
k
Z
n = km Section 6.4 The Construct Method 39 How would the Object Method work? Consider the statement a  b. It uses an
implicit existential quantiﬁer. Since a  b occurs in the hypothesis, we assume the
existence of an integer, say r, so that ra = b. And if you return to examine our proof
of Transitivity of Divisibility, this is precisely what appears in the ﬁrst sentence of
the proof. Similarly, the Object Method can be used with b  c to assert that there
exists an integer s so that sb = c. Together, the ﬁrst two sentences allow us to derive
the third sentence. 6.4 The Construct Method REMARK
We use the Construct Method when an existential quantiﬁer occurs in the conclusion.
Suppose that we must prove “A implies B ” and B uses an existential quantiﬁer. That
is, B looks like
There exists an x in the set S such that P (x) is true.
We proceed exactly as the English language interpretation would suggest  we show
that the object x exists, that x is in the set S , and that P (x) is true. We should:
1. Identify the four parts of the quantiﬁed statement.
2. Construct a mathematical object x.
3. Show that x ∈ S .
4. Show that P (x) is true. For example, let us discover a proof of the following proposition. Proposition 6 If n is of the form 4 + 1 for some positive integer , then 8  (n2 − 1). As usual, let us begin by explicitly identifying the hypothesis, the conclusion and the
core proof technique.
Hypothesis: n is of the form 4 + 1 for some integer .
Conclusion: 8  (n2 − 1).
Core Proof Technique: Since the deﬁnition of divisibility contains an existential
quantiﬁer, and
8  (n2 − 1) occurs in the conclusion, we will use the Construct Method. 40 Chapter 6 To Be or Not To Be What, precisely should we construct? Again, thinking of the deﬁnition of divisibility
and the requirement of the Construct Method, we should construct a k and then show
that k is an integer and that 8k = n2 − 1. Where is this k going to come from? Let’s
start with the hypothesis, n is of the form 4 + 1 for some integer . Substituting
n = 4 + 1 into n2 − 1 gives
n2 − 1 = (4 + 1)2 − 1 = 16 2 + 8 + 1 − 1 = 16 2 + 8 = 8(2 2 +) It seems that a suitable choice for k would be 2 2 + . Since is an integer and the
product of integers is an integer and the sum of integers is an integer, k is an integer.
It is also clear from the equation above that 8k = n2 − 1.
A proof might look like the following.
Proof: Substituting n = 4 + 1 into n2 − 1 gives
n2 − 1 = (4 + 1)2 − 1 = 16
Since 2 2 2 + 8 + 1 − 1 = 16 2 + 8 = 8(2 2 +) + is an integer, 8  (n2 − 1). Note that the proof does not explicitly name the Construct Method. Exercise 6 Where was the Construct Method used in the proof of the Transitivity of Divisibility? 6.5 The Select Method REMARK
We use the Select Method whenever a universal quantiﬁer occurs. Suppose a statement looks like
For every x in the set S , P (x) is true.
Observe that this statement is equivalent to
If x is in the set S , then P (x) is true.
We proceed exactly as the English language interpretation would suggest  we show
that whenever an object x in the set S exists, P (x) is true. We should:
1. Identify the four parts of the quantiﬁed statement.
2. Select a representative mathematical object x ∈ S . This cannot be a speciﬁc
object. It has to be a placeholder so that our argument would work for any
speciﬁc member of S . Note that if the the set S is empty, we proceed no
further. The statement is vacuously true.
3. Show that P (x) is true. Section 6.6 A NonProof 41 For example, let us discover a proof of the following proposition. Proposition 7 For every odd integer n, 4  (n2 + 4n + 3). Let’s begin by identifying the four parts of the quantiﬁed statement.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
n
odd integers
4  (n2 + 4n + 3) Now we select a representative mathematical object from the set. Let’s call the
odd integer that we selected n0 . We could certainly call it n. I am using n0 to
emphasize that we have selected a representative element. Now we must show that
4  (n2 + 4n0 + 3). Since n0 is odd, we can write it as n0 = 2m + 1. Substituting into
0
n2 + 4n0 + 3 gives
0
n2 + 4n0 + 3 = 4m2 + 4m + 1 + 8m + 4 + 3 = 4m2 + 12m + 8 = 4(m2 + 3m + 2)
0
which implies 4  (n2 + 4n0 + 3).
0
A proof might look like the following.
Proof: Let n0 be a positive, odd integer. We can write n0 as 2m + 1 for some integer
m. Substituting n0 = 2m + 1 into n2 + 4n0 + 3 gives
0
n2 + 4n0 + 3 = 4m2 + 4m + 1 + 8m + 4 + 3 = 4m2 + 12m + 8 = 4(m2 + 3m + 2)
0
Since m2 + 3m + 2 is an integer, 4  (n2 + 4n0 + 3).
0
The same proof would work if we converted the universal statement into an “If ...
then ” form. The equivalent statement would be Proposition 8 If n is a positive, odd integer, then 4  (n2 + 4n + 3). 6.6 A NonProof Making mistakes is easy. Let’s take a look at a “proof” which is not a proof. Let’s
ﬁnd out why it fails. Proposition 9 1 If r is a positive real number with r = 1, then there is an integer n such that 2 n < r. Proof: (For reference purposes, each sentence of the proof is written on a separate
line.) 42 Chapter 6 1. Let n be any integer with n >
2. It then follows that To Be or Not To Be 1
.
log2 (r) 1
< log2 (r).
n 1 3. Hence 2 n < 2log2 (r) = r. Analysis of Proof An interpretation of sentences 1 through 3 follows.
Sentence 1 Let n be any integer with n > 1/ log2 (r).
Since an existential quantiﬁer occurs in the conclusion, the author is using the
Construct Method. The four parts of the quantiﬁer are:
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
n
Z
1
2n < r In the ﬁrst sentence of the proof, the author constructs an integer n. Later in
the proof, the author intends to show that n satisﬁes the open sentence of the
quantiﬁer. Since r is a real number (not equal to 1), 1/ log2 (r) evaluates to a
real number and we can certainly ﬁnd an integer greater than any given real
number.
Sentence 2 It then follows that 1
n < log2 (r). Here the author takes the reciprocal of n > 1/ log2 (r).
1 Sentence 3 Hence 2 n < 2log2 (r) = r.
1
Use the left and right sides of n < log2 (r) as exponents of 2 and recall that the
function 2x always increases as x increases. Even the analysis looks good. What went wrong? Let’s look again at Sentence 2.
Here we used the statement Statement 10 If a, b ∈ R, neither equal to 0, and a < b, then 1/b < 1/a. A proof seems pretty straightforward – divide both sides of a < b by ab. Except that
1
1
the statement is false. Consider the case a = −2 and b = 4. −2 < 4 but 4
−2 . Our
proposition really should be Statement 11 If a, b ∈ R, and 0 < a < b, then 1/b < 1/a. Now we can ﬁnd the problem in our proof. Choose r so that 0 < r < 1, say r = 1/2.
That will make log2 (r) negative and hence 1/ log2 (r) negative. Choose n = 1. Now
Sentence 1 is satisﬁed but Sentence 2 fails.
Can you think of any way to correct the proposition or the proof? Chapter 7 The Greatest Common Divisor
7.1 Objectives The content objectives are:
1. To discover a proof of the proposition GCD With Remainders.
2. State the Euclidean Algorithm.
3. Compute gcds using the Euclidean Algorithm.
4. Prove the GCD Characterization Theorem.
5. Compute gcds and certiﬁcates using the Extended Euclidean Algorithm. 7.2
Deﬁnition
Greatest
Common
Divisor Greatest Common Divisor Let a and b be integers, not both zero. An integer d > 0 is the greatest common
divisor of a and b, written gcd(a, b), if and only if
1. d  a and d  b (this captures the common part of the deﬁnition), and
2. if c  a and c  b then c ≤ d (this captures the greatest part of the deﬁnition). Example 7
• gcd(24, 30) = 6
• gcd(17, 25) = 1
• gcd(−12, 0) = 12
• gcd(−12, −12) = 12
• gcd(0, 0) =?? 43 44 Chapter 7 Deﬁnition
gcd(0, 0) The Greatest Common Divisor For a = 0, the deﬁnition implies that gcd(a, 0) = a and gcd(a, a) = a. We deﬁne
gcd(0, 0) as 0. This may sound counterintuitive, since all integers are divisors of 0,
but it is consistent with gcd(a, 0) = a and gcd(a, a) = a. Let’s prove a seemingly unusual proposition about gcds. Proposition 12 (GCD With Remainders (GCD WR))
If a and b are integers not both zero, and q and r are integers such that a = qb + r,
then gcd(a, b) = gcd(b, r). How would we discover a proof for this proposition? Let’s try the usual approach:
identify the hypothesis and conclusion, and begin asking questions.
Hypothesis: a, b, q and r are integers such that a = qb + r.
Conclusion: gcd(a, b) = gcd(b, r)
My ﬁrst question typically starts with the conclusion and works backward. What is
a suitable ﬁrst question? How about “How do we show that two integers are equal?”
There are lots of possible answers: show that their diﬀerence is zero, their ratio is
one, each is less than or equal the other. However, here we are working with gcds
rather than generic integers so perhaps a better question would be “How do we show
that a number is a gcd?” The broad answer is relatively easy. Use the deﬁnition of
gcd. After all, right now it is the only thing we have! A speciﬁc answer is less easy.
Do we want to focus on gcd(a, b) or gcd(b, r)? Here is an easy way to do both. Let
d = gcd(a, b). Then show that d = gcd(b, r). That gets us two statement in our proof.
Proof in Progress
1. Let d = gcd(a, b).
2. To be completed.
3. Hence d = gcd(b, r).
But how do we show that d = gcd(b, r)? Use the deﬁnition. Our proof can expand to
Proof in Progress
1. Let d = gcd(a, b).
2. We will show
(a) d  b and d  r, and
(b) if c  b and c  r then c ≤ d.
3. To be completed.
4. Hence d = gcd(b, r). Section 7.2 Greatest Common Divisor 45 For the ﬁrst part of the deﬁnition, we ask “How do we show that one number divides
another number?” Interestingly enough, there are two diﬀerent answers  one for b
and one for r, though that is not obvious. For b there is already a connection between
d and b in the ﬁrst sentence. Since d = gcd(a, b), we know from the deﬁnition of gcd
that d  b.
What about r? Using the deﬁnition of divisibility seems problematic. What propositions could we use? Transitivity of Divisibility (Proposition 1) doesn’t seem to apply.
How about using the Divisibility of Integer Combinations (Proposition 2)? Proposition 13 (Divisibility of Integer Combinations)
Let a, b and c be integers. If a  b and a  c, then a  (bx + cy ) for any x, y ∈ Z. Observe that r = a − qb. Since d  a and d  b, d divides any integer combination of a
and b by the Divisibility of Integer Combinations. That is, d  (a(1) + b(−q )) so d  r.
Let’s extend our proof in progress.
Proof in Progress
1. Let d = gcd(a, b).
2. We will show
(a) d  b and d  r, and
(b) if c  b and c  r then c ≤ d.
3. Since d = gcd(a, b), we know from the deﬁnition of gcd that d  b.
4. Observe that r = a − qb. Since d  a and d  b, d  (a(1) + b(−q )) by the
Divisibility of Integer Combinations, so d  r.
5. To be completed.
6. Hence d = gcd(b, r).
That leaves us with the greatest part of greatest common divisor. This second part
of the deﬁnition is itself an implication, so we assume that c  b and c  r and we
must show c ≤ d. How do we show one number is less than or equal to another
number? There doesn’t seem to be anything obvious but ask “Have I seen this
anywhere before?”. Yes, we have. In the second part of the deﬁnition of gcd. But
then you might ask “Isn’t that assuming what we have to prove?” Let’s be precise
about what we are saying. We can use d for one inequality.
Since d = gcd(a, b), for any c where c  a and c  b, c ≤ d.
What we need to show is: if c  b and c  r then c ≤ d.
These two statements are close, but not the same. If we assume that c  b and
c  r, then c  (b(q ) + r(1)) by the Divisibility of Integer Combinations (again). Since
a = qb + r, c  a. And now, since d = gcd(a, b) and c  a and c  b, c ≤ d as needed.
Let’s add that to our proof in progress.
Proof in Progress 46 Chapter 7 The Greatest Common Divisor 1. Let d = gcd(a, b).
2. We will show
(a) d  b and d  r, and
(b) if c  b and c  r then c ≤ d.
3. Since d = gcd(a, b), we know from the deﬁnition of gcd that d  b.
4. Observe that r = a − qb. Since d  a and d  b, d  (a(1) + b(−q )) by the
Divisibility of Integer Combinations, so d  r.
5. Let c  b and c  r. Then c  (b(q ) + r(1)) by the Divisibility of Integer Combinations. Since a = qb + r, c  a. And now, since d = gcd(a, b) and c  a and c  b,
c ≤ d by the second part of the deﬁnition of gcd.
6. Hence d = gcd(b, r).
Having discovered a proof, we should now write the proof. Whenever you write, you
should have an audience in mind. You actually have two audiences to keep in mind:
your peers with whom you collaborate, and the markers. You do not need to specify
each proof technique, since your peers and markers know all of them. It does help to
provide an overall plan if you can. Also, proofs tend to work much more forwards than
backwards because that helps to emphasize the notion of starting with hypotheses
and ending with the conclusion. Here is one possible proof.
Proof: Let d = gcd(a, b). We will use the deﬁnition of gcd to show that d = gcd(b, r).
Since d = gcd(a, b), d  b. Observe that r = a − qb. Since d  a and d  b, d  (a − qb)
by the Divisibility of Integer Combinations. Hence d  r, and d is a common divisor
of b and r.
Let c be a divisor of b and r. Since c  b and c  r, c  (qb + r) by the Divisibility of
Integer Combinations. Now a = qb + r, so c  a. Because d = gcd(a, b) and c  a and
c  b, c ≤ d. REMARK
1. If a = b = 0 this proposition is also true since the only possible choices for b
and r are b = r = 0.
2. In general, there are many ways to work forwards and backwards.
3. The proof may records steps in a diﬀerent order than their appearance in the
discovery process.
4. Proofs are short and usually omit the discovery process.
5. Be sure that you can identify where each of your hypotheses was used in the
proof. Section 7.3 Certiﬁcate of Correctess 7.3 47 Certiﬁcate of Correctess Suppose we wanted to compute gcd(1386, 322). We could factor both numbers, ﬁnd
their common factors and select the greatest. In general, this is very slow.
Repeated use of GCD With Remainders allows us to eﬃciently compute gcds. For
example, let’s compute gcd(1386, 322). Example 8
Since
Since
Since
Since 1386 = 4 × 322 + 98,
322 = 3 × 98 + 28,
98 = 3 × 28 + 14,
28 = 2 × 14 + 0, gcd(1386, 322) = gcd(322, 98).
gcd(322, 98) = gcd(98, 28).
gcd(98, 28) = gcd(28, 14).
gcd(28, 14) = gcd(14, 0). Since gcd(14, 0) = 14, the chain of equalities from the column on the right gives us
gcd(1386, 322) = gcd(322, 98) = gcd(98, 28) = gcd(28, 14) = gcd(14, 0) = 14. This process is known as the Euclidean Algorithm. Exercise 7 With the person beside you, randomly pick two positive integers and compute their
gcd using the Euclidean Algorithm. How do you know that you have the correct
answer? Keep your work. You’ll need it soon. Because mistakes happen when performing arithmetic by hand, and mistakes happen
when programming computers, it would be very useful if there were a way to certify
that an answer is correct. Think of a certiﬁcate of correctness this way. You are a
manager. You ask one of your staﬀ to solve a problem. The staﬀ member comes
back with the proposed solution and a certiﬁcate of correctness that can be used to
verify that the proposed solution is, in fact, correct. The certiﬁcate has two parts: a
theorem which you have already proved and which relates to the problem in general,
and data which relates to this speciﬁc problem.
For example, here’s a proposition that allows us to produce a certiﬁcate for gcd(a, b). Proposition 14 (GCD Characterization Theorem (GCD CT))
If d is a positive common divisor of the integers a and b, and there exist integers x
and y so that ax + by = d, then d = gcd(a, b). Our certiﬁcate would consist of this theorem along with integers x and y . If our
proposed solution was d and d  a, d  b and ax + by = d, then we could conclude
without doubt that d = gcd(a, b).
In the example 8 above, the proposed gcd of gcd(1386, 322) is 14. Our certiﬁcate of
correctness consists of the GCD Characterization Theorem and the integers d = 14,
x = 10 and y = −43. Note that 14  1386 and 14  322 and 1386×10+322×(−43) = 14,
so we can conclude that 14 = gcd(1386, 322).
Here is a proof of the GCD Characterization Theorem. 48 Chapter 7 The Greatest Common Divisor Proof: (For reference, each sentence of the proof is written on a separate line.)
1. We will show that d satisﬁes the deﬁnition of gcd(a, b).
2. From the hypotheses, d  a and d  b.
3. Now let c  a and c  b.
4. By the Divisibility of Integer Combinations (Proposition 2), c  (ax + by ) so
c  d.
5. By the Bounds by Divisibility (Proposition 3), c ≤ d, and so d = gcd(a, b). Let’s do an analysis of the proof.
Analysis of Proof As usual, we will begin by explicitly identifying the hypothesis
and the conclusion.
Hypothesis: d is a positive common divisor of the integers a and b. There
exist integers x and y so that ax + by = d.
Conclusion: d = gcd(a, b)
Core Proof Technique: Work forwards recognizing an existential quantiﬁer
in the hypothesis.
Preliminary Material: Deﬁnition of gcd. An integer d > 0 is the gcd(a, b) if
and only if
1. d  a and d  b, and
2. if c  a and c  b then c ≤ d.
Sentence 1 We will show that d satisﬁes the deﬁnition of gcd(a, b).
The author states the plan  always a good idea. The author is actually answering the question “How do I show that one number is the gcd of two other
numbers?”
Sentence 2 From the hypotheses, d  a and d  b.
The author is working forwards from the hypothesis. This handles the ﬁrst part
of the deﬁnition of gcd.
Sentence 3 Now let c  a and c  b.
The second part of the deﬁnition of gcd is an implication with hypothesis c  a
and c  b. The author must show c ≤ d.
Sentence 4 By the Divisibility of Integer Combinations, c  (ax + by ) so c  d.
This is where the author uses an existential quantiﬁer in the hypothesis. The
author assumes the existence of two integers x and y such that ax + by = d.
The author does not state this explicitly.
Having made this assumption, the author can use Sentence 3 to satisfy the
hypotheses of Divisibility of Integer Combinations and so invoke the conclusion,
that is, c  (ax + by ). Section 7.4 The Extended Euclidean Algorithm (EEA) 49 Sentence 5 By the Bounds By Divisibility, c ≤ d, and so d = gcd(a, b).
Having determined that c ≤ d, both parts of the deﬁnition of gcd are satisﬁed
and so the author can conclude that d = gcd(a, b).
Now the obvious questions is: “How do we ﬁnd x and y ?” 7.4 The Extended Euclidean Algorithm (EEA) Given two positive integers, a and b, the EEA is an eﬃcient way to compute not only
d = gcd(a, b) but the data x and y for the certiﬁcate. Here is the algorithm and an
example.
Algorithm 1 Extended Euclidean Algorithm
Require: a > b > 0 are integers.
Ensure: The following conditions hold at the end of the algorithm.
rn+1 = 0.
rn = gcd(a, b).
ri−2 = qi ri−1 + ri where 0 ≤ ri < ri−1 .
In every row, axi + byi = ri .
x = xn , y = yn is a solution to ax + by = gcd(a, b).
{Initialize}
Construct a table with four columns so that
The columns are labelled x, y , r and q .
The ﬁrst row in the table is (1, 0, a, 0).
The second row in the table is (0, 1, b, 0).
{To produce the remaining rows (i ≥ 3)}
repeat
ri−2
qi ← ri−1
Rowi ← Rowi−2 − qi Rowi−1
until ri = 0
This may be easier to understand with an example. Let’s compute gcd(1386, 322)
using the EEA. Since 1386 > 322 > 0 we can, in fact, legitimately use the EEA.
Initially we get
xy
r
q
1 0 1386 0
0 1 322 0
To generate the third row we must ﬁrst compute q3 . Using the formula
qi ← ri−2
ri−1 we get
q3 = r1
1386
=
=4
r2
322 Now we use the formula
Rowi ← Rowi−2 − qi Rowi−1 50 Chapter 7 The Greatest Common Divisor when i = 3 to get
Row3 ← Row1 − q3 Row2
With q3 = 4 we get
Row3 ← Row1 − 4 × Row2
Representing this in the table gives Row1
−4 × Row2
= Row3 xy
r
q
1 0 1386 0
01
322 0
1 −4 98 4 In a similar fashion we get the fourth row. Row2
−3 × Row3
= Row4 x
y
r
q
1
0 1386 0
0
1
322 0
1 −4 98 4
−3 13
28 3 The completely worked out example follows.
x
y
r
q
1
0
1386 0
0
1
322 0
1
−4
98 4
−3
13
28 3
10 −43 14 3
−23 99
0
2
We stop when the remainder is 0. The second last row provides the desired d, x and
y . The gcd d is the entry in the r column, x is the entry in the x column and y is the
entry in the y column. Hence, d = 14 (as before), and we can check the conditions of
the GCD Characterization Theorem to certify correctness. Since 141386 and 14322
and 1386 × 10 + 322 × (−43) = 14, we can conclude that 14 = gcd(1386, 322).
If a or b is negative, apply the EEA to gcd(a, b) and then change the signs of x
and y after the EEA is complete. If a < b, simply swap their places in the algorithm.
This works because gcd(a, b) = gcd(b, a).
We treat the EEA as a proposition where the preconditions are the hypotheses and
the postconditions are the conclusions. A proof of the correctness of the EEA is
available in the appendix. Add to appendix. Exercise 8 A few minutes ago you computed the gcd of two numbers. Repeat that exercise using
the EEA and verify that you can produce a certiﬁcate of correctness for your proposed
gcd. Chapter 8 Properties Of GCDs
8.1 Objectives The technique objectives are:
1. To practice working with existential quantiﬁers.
The content objectives are:
1. Discover a proof of Coprimeness and Divisibility.
2. Discover a proof of GCD Of One
3. Exercise: Discover a proof of Division by the GCD.
4. Exercise: Discover a proof of Primes and Divisibility. 8.2 Some Useful Propositions We begin with a proposition on coprimeness and divisibility. Deﬁnition Two integers a and b are coprime if gcd(a, b) = 1. Coprime Proposition 15 (Coprimeness and Divisibility (CAD))
If a, b and c are integers and c  ab and gcd(a, c) = 1, then c  b. This proposition has two implicit existential quantiﬁers, one in the hypothesis and one
in the conclusion. You might object and ask “Where?” It’s hidden  in the deﬁnition
of divides. Recall the deﬁnition. An integer m divides an integer n if there exists an
integer k so that n = km.
We treat an existential quantiﬁer in the hypothesis diﬀerently from an existential
quantiﬁer in the conclusion. Recall the following remarks from the chapter on quantiﬁers.
51 52 Chapter 8 Properties Of GCDs REMARK
When proving that “A implies B ” and A uses an existential quantiﬁer, use the
Object Method.
1. Identify the four parts of the quantiﬁed statement “there exists an x in the set
S such that P (x) is true.”
2. Assume that a mathematical object x exists within the domain S so that the
statement P (x) is true.
3. Make use of this information to generate another statement.
When proving that “A implies B ” and B uses an existential quantiﬁer, use the
Construct Method.
1. Identify the four parts of the quantiﬁed statement. “there exists an x in the set
S such that P (x) is true.”
2. Construct a mathematical object x.
3. Show that x ∈ S .
4. Show that P (x) is true. Let’s be clear about what “there exists an integer k so that b = kc”, the existential
statement in the conclusion, means.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
k
Z
b = kc With all of this in mind, how do we go about discovering a proof for Coprimeness
and Divisibility? As usual, we will begin by explicitly identifying the hypothesis, the
conclusion, the core proof technique and any preliminary material we think we might
need.
Hypothesis: a, b and c are integers and c  ab and gcd(a, c) = 1.
Conclusion: c  b.
Core Proof Technique: We use the Object Method because of the existential quantiﬁer in the hypothesis, and the Construct Method because of the existential
quantiﬁer in the conclusion.
Preliminary Material: Deﬁnition of divides and gcd.
When discovering proofs I prefer to start by working backwards from the conclusion.
In this case, I would begin by asking “How do we show that one integer divides
another?” We can answer with the deﬁnition of divisibility. We must construct an
integer k so that b = ck . We will record this as follows. Section 8.2 Some Useful Propositions 53 Proof in Progress
1. To be completed.
2. Since b = kc, c  b.
The problem is that it is not at all clear what k should be. Let’s work forwards from
the hypothesis.
Somehow we need an equation with a b alone on one side of the equality sign. We
can’t start there but we can get an equation with a b. Since gcd(a, c) = 1, the EEA
guarantees that we can ﬁnd integers x and y so that ax + cy = 1. We could multiply
this equation by b. Let’s record these forward statements.
Proof in Progress
1. Since gcd(a, c) = 1, the EEA guarantees that we can ﬁnd integers x and y so
that ax + cy = 1 (1).
2. Multiplying (1) by b gives abx + cby = b (2).
3. To be completed.
4. Since b = kc, c  b.
If we could factor the left hand side of (2), we’d be able to get a c and other stuﬀ
that we could treat as our k . But the ﬁrst term has no c. Or maybe it does. Since
c  ab there exists an integer h so that ch = ab. Substituting ch for ab in (2) gives
chx + cby = b (3). We record this as
Proof in Progress
1. Since gcd(a, c) = 1, the EEA guarantees that we can ﬁnd integers x and y so
that ax + cy = 1 (1).
2. Multiplying (1) by b gives abx + cby = b (2).
3. Since c  ab there exists an integer h so that ch = ab. Substituting ch for ab in
(2) gives chx + cby = b (3).
4. To be completed.
5. Since b = kc, c  b.
Now factor.
Proof in Progress
1. Since gcd(a, c) = 1, the EEA guarantees that we can ﬁnd integers x and y so
that ax + cy = 1 (1).
2. Multiplying (1) by b gives abx + cby = b (2).
3. Since c  ab there exists an integer h so that ch = ab. Substituting ch for ab in
(2) gives chx + cby = b (3). 54 Chapter 8 Properties Of GCDs 4. This gives c(hx + by ) = b.
5. But then if we let k = hx + by we have an integer k so that ck = b.
6. Since b = kc, c  b.
Here is a proof.
Proof: By the Extended Euclidean Algorithm and the hypothesis gcd(a, c) = 1, there
exist integers x and y so that ax + cy = 1. Multiplying by b gives abx + cby = b.
Since c  ab there exists an integer h so that ch = ab. Substituting ch for ab gives
chx + cby = b. Lastly, factoring produces (hx + by )c = b. Since hx + by is an integer,
c  b.
Let us consider more properties of the greatest common divisor. Proposition 16 (GCD Of One (GCD OO))
Let a and b be integers. Then gcd(a, b) = 1 if and only if there are integers x and y
with ax + by = 1. This proposition has similar elements to the one we just proved, so it won’t be a
surprise if we use similar reasoning. REMARK
The important diﬀerence is that this statement is an “if and only if” statement. To
prove A if and only if B we must prove two statements:
1. If A, then B .
2. If B , then A. We can restate the proposition as Proposition 17 (GCD Of One)
Let a and b be integers.
1. If gcd(a, b) = 1, then there are integers x and y with ax + by = 1.
2. If there are integers x and y with ax + by = 1, then gcd(a, b) = 1. In statement (1), there is an existential quantiﬁer in the conclusion, so we would
expect to use the Construction Method. The problem is “Where do we get x and y ?”
In the previous proof, we used the EEA and it makes sense to use it here as well. Section 8.2 Some Useful Propositions 55 By the EEA and the hypothesis gcd(a, b) = 1, there exist integers x and y so that
ax + by = 1.
In statement (2), an existential quantiﬁer occurs in the hypothesis so we can assume
the existence of integers x and y so that ax + by = 1. Also, 1  a and 1  b. These are
exactly the hypotheses of the GCD Characterization Theorem, so we can conclude
that gcd(a, b) = 1.
Here is a proof of the GCD Of One proposition.
Proof: Since gcd(a, b) = 1, the EEA assures the existence of integers x and y so that
ax + by = 1. Statement 1 is proved.
Now, 1  a and 1  b. Also, by the hypothesis of Statement 2, there exist integers x and
y so that ax + by = 1. These are exactly the hypotheses of the GCD Characterization
Theorem, so we can conclude that gcd(a, b) = 1 and Statement 2 is proved. REMARK
This proof illustrates the connection between the GCD Characterization Theorem
and the Extended Euclidean Algorithm. Both assume integers a and b. The GCD
Characterization Theorem starts with an integer d where d  a, d  b and integers x
and y so that ax + by = d and concludes that d = gcd(a, b). The Extended Euclidean
Algorithm computes a d so that d = gcd(a, b), hence it produces a d so that d  a and
d  b, and also computes integers x and y so that ax + by = d.
So, if we encounter a gcd in the conclusion, we can try the GCD Characterization
Theorem. If we encounter a gcd in the hypothesis, we can try the Extended Euclidean
Algorithm. Exercise 9 Proposition 18 Prove the following proposition. Compare your proof with the proof that follows. (Division by the GCD (DB GCD))
Let a and b be integers. If gcd(a, b) = d = 0, then gcd Proof: First, observe that gcd ab
,
dd ab
d, d = 1. is meaningful. Since d  a and d  b, both a
d b
are integers.
d
We will use the GCD Characterization Theorem. Since gcd(a, b) = d, the EEA assures
the existence of integers x and y so that ax + by = d. Dividing by d gives
and a
b
x+ y =1
d
d
Since 1 divides both
gcd ab
,
dd = 1. a
b
and , the GCD Characterization Theorem implies that
d
d 56 Chapter 8 Exercise 10 Properties Of GCDs This exercise illustrates the use of Proof by Elimination and proves a very useful
proposition that follows from Coprimeness and Divisibility.
1. Prove that
A ⇒ B ∨ C ≡ (A ⇒ B ) ∨ (A ⇒ C ) ≡ ¬(A ⇒ B ) ∧ (A ⇒ C )
2. A true statement of the form A ⇒ B ∨ C has two cases: A ⇒ B OR A ⇒ C .
One way to prove a statement of the form A ⇒ B ∨ C is to show that one of
the two cases must hold. This is often done by proving the statement
¬(A ⇒ B ) ∧ (A ⇒ C ) which is equivalent to proving “If the ﬁrst case is not
true, then the second case must be true.” This technique is called Proof By
Elimination since one of the two cases is eliminated. With this in mind prove
the proposition Primes and Divisibility below. Begin your proof with “Suppose
p a. We must show” Proposition 19 (Primes and Divisibility (PAD))
If p is a prime and p  ab, then p  a or p  b. Chapter 9 Linear Diophantine Equations
9.1 Objectives The technique objectives are:
1. To practice working with universal quantiﬁers.
2. To practice working with subsets.
The content objectives are:
1. Prove Coprimeness and Divisibility paying attention to the universal quantiﬁer.
2. Deﬁne Diophantine equations.
3. Prove the Linear Diophantine Equation Theorem (Part 1)
4. Discover a proof to the Linear Diophantine Equation Theorem (Part 2).
5. Examples of the Linear Diophantine Equation Theorem. 9.2 The Select Method We have already proved Proposition 20 (Coprimeness and Divisibility (CAD))
If a, b and c are integers and c  ab and gcd(a, c) = 1, then c  b. Let’s restate the proposition using a universal quantiﬁer. Proposition 21 (Coprimeness and Divisibility)
For all integers a, b and c where c  ab and gcd(a, c) = 1, c  b. 57 58 Chapter 9 Linear Diophantine Equations REMARK
Whenever we encounter a universal quantiﬁer, we use the Select Method. To prove a
statement of the form
For every x in the set S , P (x) is true.
we should
1. Identify the four parts of the quantiﬁed statement.
2. Select a representative mathematical object x ∈ S . This cannot be a speciﬁc
object. It has to be a placeholder so that our argument would work for any
speciﬁc member of S . Note that if the the set S is empty, we proceed no
further. The statement is vacuously true.
3. Show that P (x) is true. Let’s see how the Select Method is used in the following proof, most of which you
have seen before.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let a, b and c be integers where c  ab and gcd(a, c) = 1.
2. By the Extended Euclidean Algorithm and gcd(a, c) = 1, there exist integers x
and y so that ax + cy = 1.
3. Multiplying by b gives abx + cby = b.
4. Since c  ab there exists an integer h so that ch = ab.
5. Substituting ch for ab gives chx + cby = b.
6. Lastly, factoring produces c(hx + by ) = b.
7. Since hx + by is an integer, c  b. Here is an analysis.
Analysis of Proof The statement begins with a quantiﬁer so we should ﬁrst identify
the four parts of the quantiﬁed statement.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
a, b, c
Z
If c  ab and gcd(a, c) = 1, then c  b. Section 9.3 Linear Diophantine Equations 59 The open sentence is an implication so let’s identify the hypothesis and conclusion as well as the core proof technique and preliminary material.
Hypothesis: c  ab and gcd(a, c) = 1.
Conclusion: c  b.
Core Proof Technique: We begin with the Select Method. We also use the
Object Method because of the implicit existential quantiﬁer in the hypothesis (c  ab) and the Construct Method because of the implicit existential
quantiﬁer in teh conclusion (c  b).
Preliminary Material: Deﬁnitions of divide and gcd.
Sentence 1 Let a, b and c be integers where c  ab and gcd(a, c) = 1.
This follows exactly the plan of the Select Method. We begin by selecting
representative objects in the domain. We end by showing that, for the chosen
objects, the open sentence is true. This part appears in Sentence 7.
Sentences 2 – 6 These appear just as they did in the original proof of Coprimeness
and Divisibility (Proposition 15)
Sentence 7 Since hx + by is an integer, c  b.
The important part of this sentence is the demonstration that the open sentence
is true.
There are some important remarks to make here.
• Just as in English, there is often more than one way to say the same thing.
• Quantiﬁers are often implicit or hidden.
• Condensed proofs typically do not illustrate the discovery process or explicitly
identify techniques. 9.3 Linear Diophantine Equations In high school, you looked at linear equations that involved real numbers. We will
look at linear equations involving only integers. Deﬁnition
Diophantine
Equations Equations with integer coeﬃcients for which integer solutions are sought, are called
Diophantine equations after the Greek mathematician, Diophantus of Alexandria,
who studied such equations. Diophantine equations are called linear if each term in
the equation is a constant or a constant times a single variable of degree 1. The simplest linear Diophantine equation is
ax = b
To emphasize, a, b ∈ Z and we want an x ∈ Z that solves ax = b. From the deﬁnition
of divisibility, we know that this equation has an integer solution x if and only if a  b. 60 Chapter 9 Linear Diophantine Equations What about linear Diophantine equations with two variables?
ax + by = c Theorem 22 (Linear Diophantine Equation Theorem, Part 1 (LDET 1))
Let gcd(a, b) = d. The linear Diophantine equation
ax + by = c
has a solution if and only if d  c. Before we study a proof of this theorem, let’s see how it works in practice. Example 9 Which of the following linear Diophantine equations has a solution?
1. 33x + 18y = 10
2. 33x + 18y = 15
Solution:
1. Since gcd(33, 18) = 3, and 3 does not divide 10, the ﬁrst equation has no integer
solutions.
2. Since gcd(33, 18) = 3, and 3 does divide 15, the second equation does have an
integer solution. But how do we ﬁnd a solution? Here are two simple steps that will allow us to ﬁnd
a solution.
1. Use the Extended Euclidean Algorithm to ﬁnd d = gcd(a, b) and x1 and y1
where ax1 + by1 = d.
c
2. Multiply by k =
to get akx1 + bky1 = kd = c. A solution is x = kx1 and
d
y = ky1 .
Returning to the exercise, the Extended Euclidean Algorithm gives
x
y
rq
1
0
33 0
0
1
18 0
1
−1 15 1
−1
2
31
6 −11 0 5
hence
33 × −1 + 18 × 2 = 3 Section 9.3 Linear Diophantine Equations 61 Multiplying by k = c/d = 15/3 = 5 gives
33 × −5 + 18 × 10 = 15
so one particular solution is x = −5 and y = 10.
But are there more solutions? That’s where Part 2 of the Linear Diophantine Equation
Theorem comes in and we will cover it later.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. First, suppose that the linear Diophantine equation ax + by = c has an integer
solution x = x0 , y = y0 . That is, ax0 + by0 = c.
2. Since d = gcd(a, b), d  a and d  b.
3. But then, by the Divisibility of Integer Combinations, d  (ax0 + by0 ). That is
d  c.
4. Conversely, suppose that d  c.
5. Then there exists an integer k such that c = kd.
6. Now, by the Extended Euclidean Algorithm, there exist integers x1 and y1 so
that
ax1 + by1 = d.
7. Multiplying this equation by k gives
akx1 + bky1 = kd = c
which, in turn, implies that x = kx1 and y = ky1 is a solution to ax + by = c. Let’s perform an analysis of this proof.
Analysis of Proof This is an “if and only if” statement so we must prove two
statements.
1. If the linear Diophantine equation ax + by = c has a solution, then d  c.
2. If d  c, then the linear Diophantine equation ax + by = c has a solution.
Core Proof Technique: Both statements contain an existential quantiﬁer in
the hypothesis, so each will start with the Object Method. Though both
statements also contain an existential quantiﬁer in the conclusion, only
one uses the Construction Method. The other uses a proposition we have
already proved.
Sentence 1 First, suppose that the linear Diophantine equation ax + by = c has an
integer solution x = x0 , y = y0 . That is, ax0 + by0 = c.
The author does not explicitly rephrase the “if and only if” as two statements.
Rather, Sentence 1 indicates which of the two implicit statements will be proved
by stating the hypothesis of Statement 1. Moreover, the ﬁrst statement uses an
existential quantiﬁer in the hypothesis. The hypothesis of the ﬁrst statement
could be restated as 62 Chapter 9 Linear Diophantine Equations there exists an integer solution to the linear Diophantine equation
The four parts are
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
x0 , y0
Z
ax0 + by0 = c. Since the existential quantiﬁer occurs in the hypothesis, the author uses the
Object Method. The author assumes the existence of the corresponding objects
(x0 , y0 ) in a suitable domain (Z) and that these objects satisfy the related open
sentence (ax0 + by0 = c).
Sentence 2 Since d = gcd(a, b), d  a and d  b.
This follows from the deﬁnition of gcd.
Sentence 3 But then, by the Divisibility of Integer Combinations, d  (ax0 + by0 ).
That is d  c.
Since hypotheses of DIC (a, b and d are integers, and d  a and d  b) are
satisﬁed, the author can invoke the conclusion of DIC (d  (ax0 + by0 )). And
from Sentence 1, ax0 + by0 = c so d  c.
Sentence 4 Conversely, suppose that d  c.
The conversely indicates that the author is about to prove Statement 2. Recall
that an “if and only if” always consists of a statement and its converse. The
hypothesis of the converse is d  c. The deﬁnition of divides contains an existential quantiﬁer and so, in Sentence 5, the authors uses the Object Method.
The conclusion of Statement 2 contains an existential quantiﬁer (there exists
an integer solution to the linear Diophantine equation), so the author uses the
Construction Method and builds a suitable solution. Here are the parts of the
existential quantiﬁer in the conclusion.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∃
x, y
Z
ax + by = c. Sentence 5 Then there exists an integer k such that c = kd.
This is the Object Method and follows from the deﬁnition of divisibility.
Sentence 6 Now, by the Extended Euclidean Algorithm, there exist integers x1 and
y1 so that
ax1 + by1 = d.
This is prior knowledge.
Sentence 7 Multiplying this equation by k gives
akx1 + bky1 = kd = c
which, in turn, implies that x = kx1 and y = ky1 is a solution to ax + by = c. Section 9.3 Linear Diophantine Equations 63 This is where the solution is constructed, x = kx1 and y = ky1 , and where the
open sentence is veriﬁed. The author does not explicitly check that kx1 and
kx2 are integers, though we must when we analyse the proof.
LDET 1 tells us when solutions exist and how to construct a solution. It does not
ﬁnd all of the solutions. That happens next. Theorem 23 (Linear Diophantine Equation Theorem, Part 2, (LDET 2))
Let gcd(a, b) = d = 0. If x = x0 and y = y0 is one particular integer solution to the
linear Diophantine equation ax + by = c, then the complete integer solution is
b
a
x = x0 + n, y = y0 − n, ∀n ∈ Z.
d
d Before we discover a proof, let’s make sure we understand the statement. Example 10 Find all solutions to 33x + 18y = 15.
Solution: Since gcd(33, 18) = 3, and 3 does divide 15, this equation does have
integer solutions by the Linear Diophantine Equation Theorem, Part 1. If we can
ﬁnd one solution, we can use the Linear Diophantine Equation Theorem, Part 2 to
ﬁnd all solutions. Since we earlier found the solution x = −5 and y = 10 the complete
solution is
{(x, y )  x = −5 + 6n, y = 10 − 11n, n ∈ Z}
You we can check that these are solutions by substitution.
Check:
33x + 18y = 33(−5 + 6n) + 18(10 − 11n) = −165 + 198n + 180 − 198n = 15
This check does not verify that we have found all solutions. It veriﬁes that all of the
pairs f integers we have fund are solutions. The expression “complete integer solution” in the statement of LDET 2 hides the use
of sets. Let’s be explicit about those sets and what we need to do with them. There
are, in fact, two sets in the conclusion, the set of solutions and the set of x and y
pairs. We deﬁne them formally as follows.
Complete solution Let S = {(x, y )  x, y ∈ Z, ax + by = c}
b
Proposed solution Let T = {(x, y )  x = x0 + d n, y = y0 − a n, ∀n ∈ Z}
d The conclusion of LDET 2 is S = T .
How do we show that two sets are equal? Two sets S and T are equal if and only if
S ⊆ T and T ⊆ S . That is, at the risk of being repetitive, to establish that S = T
we must show two things. 64 Chapter 9 Linear Diophantine Equations 1. S ⊆ T and
2. T ⊆ S
Normally one of the two is easy and the other is harder.
Suppose we want to show S ⊆ T . How do universal quantiﬁers ﬁgure in? Showing
that S ⊆ T is equivalent to the following statement. Proposition 24 S ⊆ T if and only if, for every member s ∈ S , s ∈ T .
If you prefer symbolic notation you could write ∀s ∈ S, s ∈ T or s ∈ S ⇒ s ∈ T .
What are the components of the universal quantiﬁer in Proposition 24?
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
s
S
s∈T The Select Method works perfectly in these situations.
As frequently as sets are used, they are usually implicit and our ﬁrst task is to discern
what sets exist and how they are used. Let’s return to the proof of LDET 2 where
our sets are:
Complete solution Let S = {(x, y )  x, y ∈ Z, ax + by = c}
b
Proposed solution Let T = {(x, y )  x = x0 + d n, y = y0 − a n, ∀n ∈ Z}
d Let us discover a proof. We must keep in mind that we have two things to prove
1. S ⊆ T and
2. T ⊆ S
In this case, item 2 is easier so we will do it ﬁrst. How do we show that T ⊆ S ? We
must show that x ∈ T ⇒ x ∈ S . We certainly don’t want to individually check every
element of T so we choose a representative element of T , one that could be replaced
by any element of T and the subsequent argument would hold. This is just the Select
Method and it provides our ﬁrst statement.
b
Let n0 ∈ Z. Then (x0 + d n0 , y0 − a n0 ) ∈ T .
d To show that this element is in S we must show that the element satisﬁes the deﬁning
property of S , that is, the element is a solution.
b
ax + by = a x0 + n0
d a
+ b y0 − n 0
d
ab
ab
= ax0 + by0 + n0 − n0
d
d
= ax0 + by0
=c And now we can conclude by hypothesis, x = x0 and y = y0 is an integer solution Section 9.3 Linear Diophantine Equations 65 b
(x0 + d n0 , y0 − a n0 ) ∈ S .
d To show that S ⊆ T we will need to recall the following proposition on Division by
the GCD (Proposition 18). Proposition 25 (Division by the GCD)
Let a and b be integers. If gcd(a, b) = d = 0, then gcd ab
d, d =1 Let’s begin our analysis of S ⊆ T . How do we show that S ⊆ T ? We choose a
representative element in S and show that it is in T , that is, that it satisﬁes the
deﬁning property of T . Speciﬁcally, we must show that an arbitrary solution (x, y )
b
has the form (x0 + d n, y0 − a n).
d
Let (x, y ) be an arbitrary solution. Then (x, y ) ∈ S and we must show (x, y ) ∈ T .
Let (x0 , y0 ) be a particular solution to the linear Diophantine equation ax + by = c.
The existence of (x0 , y0 ) is assured by the hypothesis. Let’s do the obvious thing and
substitute both solutions into the equation.
ax + by = c
ax0 + by0 = c
Eliminating c and factoring gives
a(x − x0 ) = −b(y − y0 )
a
b
We know that d = gcd(a, b) is a common factor of a and b so
and
are both
d
d
integers. Dividing the previous equation by d gives
a
b
(x − x0 ) = − (y − y0 )
d
d
b
Using Division by the GCD, gcd a , d = 1. Since
d
Coprimeness and Divisibility that b
d divides a (x − x0 ) we know from
d b
divides (x − x0 )
d
By the deﬁnition of divisibility, there exists an n ∈ Z so that
x − x0 = n (9.1) b
b
⇒ x = x0 + n
d
d b
Also, substituting n d for x − x0 in Equation (9.1) yields a
y = y0 − n
d
b
So every solution is of the form (x, y ) = (x0 + d n0 , y0 − a n0 ) and so
d 66 Chapter 9 Linear Diophantine Equations (x, y ) ∈ T
A very condensed proof of Linear Diophantine Equation Theorem, Part 2 might look
like the following. Notice the lack of mention of sets. Theorem 26 (Linear Diophantine Equation Theorem, Part 2, (LDET 2))
Let gcd(a, b) = d = 0. If x = x0 and y = y0 is one particular integer solution to the
linear Diophantine equation ax + by = c, then the complete integer solution is
b
a
x = x0 + n, y = y0 − n, ∀n ∈ Z.
d
d b
Proof: Substitution shows that integers of the form x = x0 + n d , y = y0 − a n, n ∈ Z
d
are solutions. Now, let (x, y ) be an arbitrary solution and let (x0 , y0 ) be a particular solution to the
linear Diophantine equation ax + by = c. Then
ax + by = c
ax0 + by0 = c
Eliminating c and factoring gives a(x − x0 ) = −b(y − y0 ) (1). Dividing by d and
b
using Division by the GCD and Coprimeness and Divisibility we have d  (x − x0 ).
b
Hence, there exists an n ∈ Z so that x = x0 + d n (2). Substituting (2) in (1) gives
a
y = y0 − d n0 as needed. Exercise 11 Find all solutions to
1. 35x + 21y = 28
2. 35x − 21y = 28 Chapter 10 Nested Quantiﬁers
10.1 Objectives The technique objectives are:
1. Recognize nested quantiﬁers.
2. Learn how to parse nested quantiﬁers.
3. Learn which techniques to apply to a sentence containing nested quantiﬁers. 10.2 Onto 10.2.1 Deﬁnition We begin with the deﬁnition of an onto function. Deﬁnition
Onto,
Surjective Let S and T be two sets. A function f : S → T is onto (or surjective) if and only
if for every y ∈ T there exists an x ∈ S so that f (x) = y . Often S and T are equal to R or are subsets of R.
Though you may not understand the deﬁnition, the important observation is that
the deﬁnition contains two quantiﬁers. Let’s carefully parse the deﬁnition beginning
with the universal quantiﬁer “For every”. Recall that we must identify the quantiﬁer,
variable, domain and open sentence.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
y
T
there exists an x ∈ S so that f (x) = y The open sentence itself contains a quantiﬁer! So we must again identify the four
parts of this quantiﬁer.
67 68 Chapter 10
Quantiﬁer:
Variable:
Domain:
Open sentence: Nested Quantiﬁers ∃
x
S
f (x) = y REMARK
Because the existential quantiﬁer is “nested” within the universal quantiﬁer, this
deﬁnition is an example of nested quantiﬁers. There are really two basic principles
to keep in mind.
1. Process quantiﬁers from left to right. (This captures the “nested” structure.)
2. Use Object, Construct and Select methods as you proceed from left to right.
Moving from left to right is important. The order of quantiﬁers matters. For example, consider the following statement about the integers.
∀x ∃y, y > x
Translated into prose, this statement can be read as “Given any integer x, there exists
a larger integer y .” This is a true statement. Now let’s make a small modiﬁcation.
We will just change the order of the quantiﬁers. Our new statement is
∃y ∀x, y > x
A translation for this statement would be “There exists an integer y which is larger
than all integers.” A very diﬀerent, and false, statement.
We should be able to determine the structure of any proof that a function is onto.
The order of quantiﬁers is
For all there exists so we would expect the proof to be structured
Select Method Construct Method The Construct Method identiﬁes a mathematical object, shows that the object is
within the domain, and that the object satisﬁes the open sentence. So an onto proof
will look like
Structure of an “Onto” Proof
• Let y ∈ T . (This comes from the Select Method.)
• Consider the object x. (This comes from the Construct Method.)
• First, we show that x ∈ S . (Show that the constructed object is within the
domain.)
• Now we show that f (x) = y . (We show that the open sentence is satisﬁed.) Section 10.2 Onto 69 10.2.2 Reading Let’s work through an example. Notice how closely the proof follows the structure of
an onto proof. Proposition 1 The function f : R → R deﬁned by f (x) = x3 is onto. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let y ∈ R.
2. Consider x = √
3 y. 3. Since y ∈ R, x ∈ R.
√
√
4. But then f (x) = f ( 3 y ) = ( 3 y )3 = y as needed. Let’s perform an analysis of this proof.
Analysis of Proof The deﬁnition of onto uses a nested quantiﬁer.
Hypothesis: f (x) = x3 .
Conclusion: f (x) is onto.
Core Proof Technique: Nested quantiﬁers.
Preliminary Material: Let us remind ourselves of the deﬁnition of the deﬁning property of onto as it applies in this situation.
For every y ∈ R there exists x ∈ R so that f (x) = y .
Sentence 1 Let y ∈ R.
The ﬁrst quantiﬁer in the deﬁnition is a universal quantiﬁer so the author uses
the Select Method. That is, the author chooses an element (y ) in the domain
(R). The author must now show that the open sentence is satisﬁed (there exists
an x ∈ R so that f (x) = y ). The constructed object in this example is not
surprising  we can simply solve for x in y = x3 . In general, though, it can be
diﬃcult to construct a suitable object. Note also that the choice of x depends
on y so that it is not surprising that x is a function of y .
√
Sentence 2 Consider x = 3 y .
The second quantiﬁer in the deﬁnition is a nested existential quantiﬁer so the
author uses the Construction Method. That is, the author constructs an element
(x).
Sentence 3 Since y ∈ R, x ∈ R.
Because this step is usually straightforward, it is often omitted.
√
√
Sentence 4 But then f (x) = f ( 3 y ) = ( 3 y )3 = y as needed.
Here the author conﬁrms that the open sentence is satisﬁed. 70 Chapter 10 10.2.3 Nested Quantiﬁers Discovering Having read a proof, let’s discover one. Proposition 1 Let f : T → U and g : S → T be onto functions. Then f ◦ g is an onto function. Analysis of Proof The deﬁnition of onto uses a nested quantiﬁer.
Hypothesis: f : T → U and g : S → T are both onto functions.
Conclusion: f ◦ g is onto.
Core Proof Technique: Nested quantiﬁers.
Preliminary Material: Let us recast the deﬁnition of onto for f ◦ g . To do
this we need to be cognizant of the fact that f : T → U and g : S → T
and f ◦ g : S → U .
For every y ∈ U there exists x ∈ S so that f (g (x)) = y .
There are three instances of onto in the proposition. Two occur in the hypothesis and
are associated with the functions f and g . The third occurs in the conclusion and is
associated with the function f ◦ g . That is the one that interests us right now. The
deﬁnition of onto begins with a universal qualiﬁer. So we will use the Select Method
applied to f ◦ g .
Proof in Progress
1. Let y ∈ U .
2. To be completed.
3. Hence, there exists x ∈ S so that f (g (x)) = y .
The second quantiﬁer in the deﬁnition is a nested existential quantiﬁer. Since f ◦
g appears in the conclusion, we use the Construction Method. That is, we must
construct an element (x) in the domain S and show that f (g (x)) = y .
Now, this construction seems diﬃcult. We do not know what the sets S , T and U
are and we have no idea what the functions f and g look like. But we have not made
use of our hypotheses at all so let’s see if they can give us any ideas.
Since f : T → U is onto, we know that for any u ∈ U , there exists a t ∈ T so that
f (t) = u.
Since g : S → T is onto, we know that for any t ∈ T , there exists an s ∈ S so that
g (s) = t.
How does y ﬁt in? Observe that y ∈ U . But f : T → U and is onto, so there exists a
t ∈ T so that f (t ) = y . Since t ∈ T and g : S → T is onto, there exists an s ∈ S so
that g (s ) = t .
But what have we constructed? If we let x = s then we have an element that maps
from S to T and then from T to U for which f (g (s )) = y . Let’s record these.
Proof in Progress Section 10.3 Limits 71
1. Let y ∈ U .
2. Since f : T → U is onto, there exists a t ∈ T so that f (t ) = y .
3. Since t ∈ T and g : S → T is onto, there exists an s ∈ S so that g (s ) = t .
4. Hence, there exists s ∈ S so that f (g (s )) = f (t ) = y .
5. Hence, there exists x ∈ S so that f (g (x)) = y .
Notice that our last two lines are essentially duplicates. When doing rough work, this
is common. However, when writing up a proof, such duplications should be removed,
consistent notation should be enforced and omitted steps should be included. In this
case, the proof is almost done for us.
Proof: Let y in U . Since f : T → U is onto, there exists a t ∈ T so that f (t ) = y .
Since t ∈ T and g : S → T is onto, there exists an s ∈ S so that g (s ) = t . Hence,
there exists s ∈ S so that f (g (s )) = f (t ) = y . 10.3 Limits 10.3.1 Deﬁnition Almost everyone who takes a calculus course encounters the notion of a limit. When
we write
lim f (x) = L
x→a we informally mean that we can make the values of f (x) arbitrarily close to L by
taking x suﬃciently close to, but not equal to a. But formally we need to be more
explicit about what “arbitrarily” and “suﬃciently” mean. That leads to the infamous
ε − δ deﬁnition of a limit. Deﬁnition
Limit The limit of f (x), as x approaches a, equals L means that for every real number
ε > 0 there exists a real number δ > 0 such that
0 < x − a < δ ⇒ f (x) − L < ε Though you may not understand the deﬁnition, the important observation is that
the deﬁnition contains two quantiﬁers. Let’s carefully parse the deﬁnition beginning
with the universal quantiﬁer “For every”. Recall that we must identify the quantiﬁer,
variable, domain and open sentence.
Quantiﬁer:
Variable:
Domain:
Open sentence: ∀
ε
real numbers > 0
there exists a real number δ > 0 such that 0 < x − a < δ ⇒ f (x) − L < The open sentence itself contains a quantiﬁer! So we must again identify the four
parts of the quantiﬁer. 72 Chapter 10
Quantiﬁer:
Variable:
Domain:
Open sentence: Nested Quantiﬁers ∃
δ
real numbers > 0
0 < x − a < δ ⇒ f (x) − L < ε Because the existential quantiﬁer is “nested” within the universal quantiﬁer, this
deﬁnition is another example of nested quantiﬁers. 10.3.2 Reading A Limit Proof Before we begin our example, we should be able to determine the structure of any
limit proof. The order of quantiﬁers is
For all there exists so we would expect the proof to be structured
Select Method Construct Method Moreover, the choice of δ will depend on the choice of ε and so δ will be a function
of ε. The Construct Method identiﬁes a mathematical object, shows that the object
is within the domain, and that the object satisﬁes the open sentence. Moreover, the
open sentence is an implication with hypothesis 0 < x − a < δ (ε) and conclusion
f (x) − L < ε. We assume that the hypothesis is true and show that the conclusion
is true. So a limit proof will look like
Structure of a “Limit” Proof
• Let ε > 0 be a real number. (This comes from the Select Method.)
• Consider the real number δ (ε). (This comes from the Construct Method.)
• First, we show that δ (ε) > 0. (This shows δ is within the domain.)
• Now let 0 < x − a < δ (ε). (This is the hypothesis of the open sentence in the
deﬁnition of limit.)
• We show that f (x) − L < ε. (This is the conclusion of the open sentence.)
The diﬃculty lies in ﬁnding a suitable choice of δ (ε). Let’s analyze a proof where
someone else has made the choice of δ (ε) for us. Proposition 1 Let m = 0 be a real number.
lim mx + b = ma + b x→a Proof: (For reference purposes, each sentence of the proof is written on a separate
line.) Section 10.3 Limits 73
1. Let ε > 0 be a real number.
2. Consider the real number δ (ε) =
3. Since ε > 0 and m > 0, δ (ε) = ε
.
m ε
> 0.
m 4. Now
0 < x − a < δ (ε) ⇒ 0 < x − a < ε
m ⇒ mx − a < ε
⇒ m(x − a) < ε
⇒ m(x − a) + (b − b) < ε
⇒ (mx + b) − (ma + b) < ε
⇒ f (x) − L < ε
as required. Analysis of Proof As usual, we begin with the hypothesis and the conclusion.
Hypothesis: m = 0 is a real number.
Conclusion: limx→a mx + b = ma + b.
Core Proof Technique: Nested quantiﬁers.
Preliminary Material: Deﬁnition of a limit. Notice how closely this proof
follows the model proof.
Sentence 1 Let ε > 0 be a real number.
The deﬁnition of limit begins with a universal quantiﬁer so the ﬁrst proof technique is the Select Method, just as in the model proof.
ε
Sentence 2 Consider the real number δ (ε) =
.
m
The next quantiﬁer is an existential quantiﬁer in the conclusion and so we use
the Construct Method. This again follows the pattern of the model proof.
ε
The constructed object is the real number δ (ε) =
. The author gives no
m
indication why that particular value was chosen or how it was derived.
ε
Sentence 3 Since ε > 0 and m > 0, δ (ε) =
> 0.
m
After an object is constructed, the Construct Method requires that the object
be in the domain and that it satisfy the open sentence. Sentence 3 of the proof
shows that δ is in the domain, the set of real numbers greater than zero.
Sentence 4 Now . . .
Sentence 4 demonstrates that δ satisﬁes the open sentence. The hypothesis of
the open sentence is 0 < x − a < δ (ε) and the conclusion is f (x) − L <
ε. The chain of reasoning begins with the hypothesis, and after arithmetic
manipulation, arrives at the conclusion. 74 Chapter 10 10.3.3 Nested Quantiﬁers Discovering a Limit Proof We will prove Proposition 1 2 If f (x) = e−1/x then
lim f (x) = 0. x→0 You might object that the function is not even deﬁned at 0, which is true. But the
deﬁnition of limx→a f (x) does not require f to be deﬁned at a. As usual, we begin
by explicitly identifying our hypothesis and conclusion.
2 Hypothesis: f (x) = e−1/x Conclusion: limx→0 f (x) = 0
This is a standard limit proof so our model proof has all of the structure we need.
Proof in Progress
1. Let ε > 0 be a real number.
2. Consider the real number δ (ε).
3. First, we show that δ (ε) > 0.
4. Now let 0 < x − a < δ (ε).
5. We show that f (x) − L < ε.
The problem is: How do we construct a suitable δ ? Because ε is not numerically
speciﬁed, our construction for δ will be a function of ε. Now is the time to go to scrap
paper. Since we need
2
e−1/x  < ε
2 we begin there and look for a way to get to 0 < x < δ (ε). e−1/x > 0 for all x so we
do not need the absolute value signs.
1
e1/x2 <ε
2 Now divide by ε (we are using the hypothesis that ε = 0) and multiply by e1/x (we
2
are using the fact that e1/x > 0) to get the following.
1
2
< e1/x
ε
Taking the natural log gives
ln 1
ε < 1
x2 This is hopeful. We can invert the fractions to get
x2 < 1
ln(1/ε) Section 10.3 Limits 75
and since x2 > 0 we now have
0 < x2 < 1
ln(1/ε) Taking square roots gives
0 < x < 1
ln(1/ε) And this is precisely the form we want. Our constructed delta is
δ= 1
ln(1/ε) This looks great. Unfortunately, we have made a dangerous assumption, that is
ln(1/ε) > 0. This is only true when ε < 1. However, it is mathematical practice
to consider ε as small, much smaller than one. We will adopt standard practice and
ignore the case ε ≥ 1 though details can be given for it as well.
We have already worked out the math so now we are in a position to write out the
proof. Take a minute to read the proof.
Proof: Let ε > 0. Since ε is small, we assume ε < 1. Consider δ = 1
ln(1/ε) . Since ε < 1, 1/ε > 1 which implies ln(1/ε) > 0 and so δ > 0. Now
0 < x <
as required. 1
1
⇒ 0 < x2 <
⇒ ln
ln(1/ε)
ln(1/ε) 1
ε < 1
1
2
2
⇒ < e1/x ⇒ e−1/x  < ε
x2
ε Chapter 11 Practice, Practice, Practice:
Quantiﬁers and Sets
11.1 Objectives This class provides an opportunity to practice working with quantiﬁers and sets. 11.2
Exercise 1 Exercises For each of the following statements, identify each quantiﬁer, its parts and your
approach to a proof of the statement.
1. For every integer a, 2  a(a + 1).
2. If n is an integer, then 8  (52n + 7).
3. If there exist integer solutions to the Diophantine equation ax2 + by 2 = c, then
gcd(a, b)  c. Exercise 2 For each of the following deﬁnitions, identify each quantiﬁer, its parts and the proof
techniques that you would use to prove that a speciﬁc object satisﬁes the deﬁnition.
1. Saying that the function f of one real variable is bounded above means that
there is a real number y such that for every real number x, f (x) ≤ y .
2. Saying that a set of real numbers S is bounded means that there is a real
number M > 0 such that for every element s ∈ S , s < M .
3. Saying that the function f of one real variable is continuous at the point x
means that for every real number ε > 0 there is a real number δ > 0 such that,
for all real numbers y with x − y  < δ , f (x) − f (y ) < . 76 Section 11.2 Exercises Exercise 3 77 Prove each of the following propositions.
1. Suppose a and b are ﬁxed integers. Then
{ax + by  x, y ∈ Z} = {n · gcd(a, b)  n ∈ Z}.
2. An integer p > 1 is called a prime if its only positive divisors are 1 and p;
otherwise it is called composite.
Let a and b be ﬁxed integers. If p is a prime and p  ab, then p  a or p  b. Exercise 4 Solve the following problems.
1. Find the complete solution to 7x + 11y = 3.
2. Find the complete solution to 35x − 42y = 14.
3. Find the complete solution to 28x + 60y = 10.
4. For what value of c does 8x + 5y = c have exactly one solution where both x
and y are strictly positive? Exercise 5 The proof of the following statement is incomplete. Identify two sets used in the
statement and proof (they are used implicitly) and state why the proof is incomplete.
Let a, b, c ∈ R, a = 0 and b2 − 4ac ≥ 0. Then the quadratic equation
ax2 + bx + c = 0
has the solution
x= −b ± √ b2 − 4ac
2a Proof: To show that a particular value is a solution, it is enough to substitute that
value into the equation and show that the equation is satisﬁed. Consider
√
−b + b2 − 4ac
x=
2a
Substitution gives
√
√
2
−b + b2 − 4ac
−b + b2 − 4ac
a
+b
+c
2a
2a
√
√
a(b2 − 2b b2 − 4ac + b2 − 4ac) −b2 + b b2 − 4ac
=
+
+c
4a2
2
√
√a
b2 − 2b b2 − 4ac + b2 − 4ac −2b2 + 2b b2 − 4ac 4ac
=
+
+
4a
4a
4a
=0 78 Chapter 11 Practice, Practice, Practice: Quantiﬁers and Sets Since a similar result holds for
x=
the proposition holds. −b − √ b2 − 4ac
2a Chapter 12 Congruence
12.1 Objectives The content objectives are:
1. Deﬁne a is congruent to b modulo m.
2. Read a proof of Congruence is an Equivalence Relation.
3. Discover the proof of Properties of Congruence.
4. Read the proof of Congruences and Division.
5. Read the proof of Congruent Iﬀ Same Remainder.
6. Do examples. 12.2 Congruences 12.2.1 Deﬁnition of Congruences One of the diﬃculties in working out properties of divisibility is that we don’t have
an “arithmetic” of divisibility. Wouldn’t it be nice if we could resolve problems about
divisibility in much the same way that we usually do arithmetic: add, subtract,
multiply and divide?
Carl Friedrich Gauss (1777  1855) was the greatest mathematician of the last two
centuries. In a landmark work, Disquisitiones Arithmeticae, published when Gauss
was 23, he introduced congruences and provided a mechanism to treat divisibility
with arithmetic. Deﬁnition
Congruent Let m be a ﬁxed positive integer. If a, b ∈ Z we say that a is congruent to b modulo
m, and write
a ≡ b (mod m)
if m  (a − b). If m (a − b), we write a ≡ b (mod m). 79 80 Chapter 12 Example 1 Congruence Verify each of the following
1. 20 ≡ 2 (mod 6)
2. 2 ≡ 20 (mod 6)
3. 20 ≡ 8 (mod 6)
4. −20 ≡ 4 (mod 6)
5. 24 ≡ 0 (mod 6)
6. 5 ≡ 3 (mod 7) REMARK
One already useful trait of this deﬁnition is the number of equivalent ways we have
to work with it.
a ≡ b (mod m)
⇐⇒ m  (a − b)
⇐⇒ ∃k ∈ Z
⇐⇒ ∃k ∈ Z 12.3 a − b = km
a = km + b Elementary Properties Another extraordinarily useful trait of this deﬁnition is that it behaves a lot like
equality. Equality is an equivalence relation. That is, it has the following three
properties:
1. reﬂexivity, a = a.
2. symmetry, If a = b then b = a.
3. transitivity, If a = b and b = c, then a = c.
Most relationships that you can think of do not have these three properties. The
relation greater than fails reﬂexivity. The relation divides fails symmetry. The nonmathematical relation is a parent of fails transitivity. Proposition 1 (Congruence Is An Equivalence Relation (CER))
Let a, b, c ∈ Z. Then
1. a ≡ a (mod m). Section 12.3 Elementary Properties 81 2. If a ≡ b (mod m), then b ≡ a (mod m).
3. If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m) These may seem obvious but as the earlier examples showed, many relations do not
have these properties. So, a proof is needed. We will give a condensed proof for all
of them, and then an analysis for part 3.
Proof: We show each part in turn.
1. Because a − a = 0 and m  0, the deﬁnition of congruence gives a ≡ a (mod m).
2. Since a ≡ b (mod m), m  (a − b) which in turn implies that there exists k ∈ Z
so that km = a − b. But if km = a − b, then (−k )m = b − a and so m  (b − a).
By the deﬁnition of congruence, b ≡ a (mod m).
3. Since a ≡ b (mod m), m  (a − b). Since b ≡ c (mod m), m  (b − c). Now,
by the Divisibility of Integer Combinations, m  ((1)(a − b) + (1)(b − c)) so
m  (a − c). By the deﬁnition of congruence, a ≡ c (mod m). Analysis of Proof We will prove part 3 of the proposition Congruence Is An Equivalence Relation.
Hypothesis: a, b, c ∈ Z, a ≡ b (mod m) and b ≡ c (mod m).
Conclusion: a ≡ c (mod m).
Sentence 1 Since a ≡ b (mod m), m  (a − b).
The author is working forward from the hypothesis using the deﬁnition of congruence.
Sentence 2 Since b ≡ c (mod m), m  (b − c).
The author is working forward from the hypothesis using the deﬁnition of congruence.
Sentence 3 Now, by the Divisibility of Integer Combinations, m  ((1)(a − b)+(1)(b −
c)) so m  (a − c).
Here it is useful to keep in mind where the author is going. The question “How
do I show that one number is congruent to another number?” has the answer,
in this case, of showing that m  (a − c) so the author needs to ﬁnd a way of
generating a − c. And a − c follows nicely from an application of the Divisibility
of Integer Combinations.
Sentence 4 By the deﬁnition of congruence, a ≡ c (mod m).
The author is working forward from m  (a − c) using the deﬁnition of congruence. 82 Proposition 2 Chapter 12 Congruence (Properties of Congruence (PG))
If a ≡ a (mod m) and b ≡ b (mod m), then
1. a + b ≡ a + b (mod m)
2. a − b ≡ a − b (mod m)
3. ab ≡ a b (mod m) We will discover a proof of the third part and leave the ﬁrst two parts as exercises.
As usual we begin by identifying the hypothesis and the conclusion.
Hypothesis: a ≡ a (mod m) and b ≡ b (mod m)
Conclusion: ab ≡ a b (mod m)
Let’s consider the question “How do we show that two numbers are congruent to one
another?” The obvious abstract answer is “Use the deﬁnition of congruent.” We may
want to keep in mind, however, that there are several equivalent forms.
a ≡ b (mod m)
⇐⇒ m  (a − b)
⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b It is not at all clear which is best or whether, in fact, several could work. Since
the conclusion of part three involves the arithmetic operation of multiplication, and
we don’t have multiplication properties for equivalence or divisibility, it makes sense
to consider either the third or fourth of the equivalent forms. There isn’t much to
separate them. I’ll choose the last form and see how it works. So, the answer to “How
do we show that two numbers are congruent to one another?” in the notation of this
proof is “We must ﬁnd an integer k so that ab = km + a b . Let’s record that.
Proof in Progress
1. To be completed.
2. Since there exists k so that ab = km + a b , ab ≡ a b (mod m).
The problem is how to ﬁnd k . There is no obvious way backwards here so let’s start
working forward. The two hypotheses a ≡ a (mod m) and b ≡ b (mod m) can be
restated in any of their equivalent forms. Since we have already decided that we
would work backwards with the fourth form, it makes sense to use the same form
working forwards. That gives two statements.
Proof in Progress
1. Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1). Section 12.3 Elementary Properties 83 2. Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2).
3. To be completed.
4. Since there exists k so that ab = km + a b , ab ≡ a b (mod m).
But now there seems to be a rather direct way to produce an ab and an a b which
we want for the conclusion. Just multiply equations (1) and (2) together. Doing that
produces
ab = m2 jh + mjb + a mh + a b = (mjh + jb + a h)m + a b
If we let k = mjh + jb + a h then k is an integer and satisﬁes the property we needed
in the last line of the proof, that is ab = km + a b . Let’s record this.
Proof in Progress
1. Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1).
2. Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2).
3. Multiplying (1) by (2) gives ab = m2 jh + mjb + a mh + a b = (mjh + jb +
a h)m + a b .
4. Since there exists k so that ab = km + a b , ab ≡ a b (mod m).
Lastly, we write a condensed proof. Note that the reader of the proof is expected to
be familiar with the equivalent forms.
Proof: Since a ≡ a (mod m), there exists an integer j such that a = mj + a (1).
Since b ≡ b (mod m), there exists an integer h such that b = mh + b (2). Multiplying
(1) by (2) gives
ab = m2 jh + mjb + a mh + a b = (mjh + jb + a h)m + a b .
Since mjh + jb + a h is an integer, ab ≡ a b (mod m). Exercise 1 Prove the remainder of the Properties of Congruence proposition.
There are four arithmetic operations with integers, but analogues to only three have
been given. It turns out that division is problematic. A statement of the form
ab ≡ ab (mod m) ⇒ b ≡ b (mod m) seems natural enough, simply divide by a. This works with the integer equation
ab = ab . But consider the case where m = 12, a = 6, b = 3 and b = 5. It is indeed
true that
18 ≡ 30 (mod 12)
and so
6×3≡6×5 (mod 12) but “dividing” by 6 gives the clearly false statement
3≡5 (mod 12). Division works only under the speciﬁc conditions of the next proposition. 84 Chapter 12 Proposition 3 Congruence (Congruences and Division (CD))
If ac ≡ bc (mod m) and gcd(c, m) = 1, then a ≡ b (mod m). Before we read the proof, let’s look at an example. Example 2 Examples of division in congruence relations.
1. 8 × 7 ≡ 17 × 7 (mod 3) ⇒ 8 ≡ 17 (mod 3)
2. For 6 × 3 ≡ 6 × 5 (mod 12), CD cannot be invoked. Why? Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Since ac ≡ bc (mod m), m  (ac − bc). That is, m  c(a − b).
2. By the proposition Coprimeness and Divisibility, m  (a − b).
3. Hence, by the deﬁnition of congruence a ≡ b (mod m). Exercise 2 Analyze the proof of the proposition on Congruences and Division. We now give one more statement that is equivalent to a ≡ b (mod m). Proposition 4 (Congruent Iﬀ Same Remainder (CISR))
a ≡ b (mod m) if and only if a and b have the same remainder when divided by m. Because this proposition is an “if and only if” proposition, there are two parts to the
proof: a statement and its converse. We can restate the proposition to make the two
parts more explicit. Proposition 5 (Congruent Iﬀ Same Remainder (CISR))
1. If a ≡ b (mod m), then a and b have the same remainder when divided by m.
2. If a and b have the same remainder when divided by m, then a ≡ b (mod m). Section 12.3 Elementary Properties 85 In practice, the two statements are not usually written out separately. The authors
assume that you do that whenever you read “if and only if”. Many “if and only if”
proofs begin with some prefatory material that will help both parts of the proof. For
example, they often introduce notation that will be used in both parts.
Let’s look at a proof of the Congruent Iﬀ Same Remainder proposition. Before we
do an analysis, make sure that you can identify
1. prefatory material (if any exists)
2. the proof of a statement
3. the proof of the converse of the statement
Proof: The Division Algorithm applied to a and m gives
a = q1 m + r1 , where 0 ≤ r1 < m
The Division Algorithm applied to b and m gives
b = q2 m + r2 , where 0 ≤ r2 < m
Subtracting the second equation from the ﬁrst gives
a − b = (q1 − q2 )m + (r1 − r2 ), where − m < r1 − r2 < m
If a ≡ b (mod m), then m  (a − b) and there exists an integer h so that hm = a − b.
Hence
a − b = (q1 − q2 )m +(r1 − r2 ) ⇒ hm = (q1 − q2 )m +(r1 − r2 ) ⇒ r1 − r2 = m(h − q1 + q2 )
which implies m  (r1 − r2 ). But, −m < r1 − r2 < m so r1 − r2 = 0.
Conversely, if a and b have the same remainder when divided by m, then r1 = r2 and
a − b = (q1 − q2 )m so a ≡ b (mod m). The prefatory material is quoted below.
The Division Algorithm applied to a and m gives
a = q1 m + r1 , where 0 ≤ r1 < m
The Division Algorithm applied to b and m gives
b = q2 m + r2 , where 0 ≤ r2 < m
Subtracting the second equation from the ﬁrst gives
a − b = (q1 − q2 )m + (r1 − r2 ), where − m < r1 − r2 < m
The proof of Statement 1 is 86 Chapter 12 Congruence If a ≡ b (mod m), then m  (a − b) and there exists an integer h so that
hm = a − b. Hence
a−b = (q1 −q2 )m+(r1 −r2 ) ⇒ hm = (q1 −q2 )m+(r1 −r2 ) ⇒ r1 −r2 = m(h−q1 +q2 )
which implies m  (r1 − r2 ). But, −m < r1 − r2 < m so r1 − r2 = 0.
The proof of the converse of Statement 1, Statement 2, is
Conversely, if a and b have the same remainder when divided by m, then
r1 = r2 and a − b = (q1 − q2 )m so a ≡ b (mod m).
We will do an analysis of the proof of Statement 1. An analysis of the proof Statement
2 is left as an exercise.
Analysis of Proof In many “if and only if” statements one direction is much easier
than the other. In this particular case, we are starting with the harder of the
two directions.
Hypothesis: a ≡ b (mod m).
Conclusion: a and b have the same remainder when divided by m.
Sentence 1 If a ≡ b (mod m), then m  (a − b) and there exists an integer h so that
hm = a − b.
Here the author is working forwards using two deﬁnitions. The deﬁnition of
congruence allows the author to assert that “If a ≡ b (mod m), then m  (a−b)”.
The deﬁnition of divisibility allows the author to assert that “m  (a − b) [implies
that] there exists an integer h so that hm = a − b.”
Sentence 2 Hence
a−b = (q1 −q2 )m+(r1 −r2 ) ⇒ hm = (q1 −q2 )m+(r1 −r2 ) ⇒ r1 −r2 = m(h−q1 +q2 )
which implies m  (r1 − r2 ).
This is mostly arithmetic. The author begins with a − b = (q1 − q2 )m + (r1 − r2 )
from the prefatory paragraph, substitutes hm for a − b, isolates r1 − r2 and
factors out an m from the remaining terms. Since h − q1 + q2 is an integer, the
author deduces that m  (r1 − r2 ).
Sentence 3 But, −m < r1 − r2 < m so r1 − r2 = 0.
This part is not so obvious. The author is working with two pieces of information. The prefatory material provides −m < r1 − r2 < m. Sentence 2 provides
m  (r1 − r2 ). Now, what are the possible values of r1 − r2 ? Certainly r1 − r2 can
be zero but are there any other possible choices? If there were another choice it
would be of the form mx with x = 0. But that would make r1 − r2 = xm > m
or r1 − r2 = xm < −m both of which are impossible because −m < r1 − r2 < m.
Hence, r1 − r2 = 0.
The conclusion does not say r1 − r2 = 0. It says that a and b have the same
remainder when divided by m. Since r1 and r2 are those remainders, and
r1 −r2 = 0 ⇒ r1 = r2 , the author leaves it to the reader to deduce the conclusion. Section 12.3 Elementary Properties Exercise 3 87 Perform an analysis of the proof of Statement 2. REMARK
The proposition Congruent Iﬀ Same Remainder gives us another part to our chain of
equivalent statements.
a ≡ b (mod m)
⇐⇒ m  (a − b)
⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b ⇐⇒ a and b have the same remainder when divided by m The propositions covered in this lecture are surprisingly powerful. Consider the following example. Example 3 What is the remainder when 347 is divided by 7?
Solution: You could attempt to compute 347 with your calculator but it might
explode. Here is a simpler way. First, recognize that the remainder when 347 is divided
by 7 is just 347 (mod 7). Now observe that 32 ≡ 2 (mod 7) and 33 ≡ 3 × 2 ≡ 6 ≡ −1
(mod 7). But then
347 ≡ 345 32 ≡ (33 )15 32 ≡ (−1)15 32 ≡ (−1)(2) ≡ −2 ≡ 5
Hence, the remainder when 347 is divided by 7 is 5. (mod 7) Chapter 13 Modular Arithmetic
13.1 Objectives The content objectives are:
1. Deﬁne the congruence class modulo m.
2. Construct Zm and perform modular arithmetic. Highlight the role of additive
and multiplicative identities, and additive and multiplicative inverses.
3. State Fermat’s Little Theorem.
4. Read a proof to a corollary of Fermat’s Little Theorem.
5. Discover a proof to the Existence of Inverses in Zp . 13.2 Modular Arithmetic In this section we will see the creation of a number system which will likely be new
to you. Deﬁnition The congruence class modulo m of the integer a is the set of integers Congruence
Class [a] = {x ∈ Z  x ≡ a Example 4 (mod m)} For example, when m = 4
[0]
[1]
[2]
[3] =
=
=
= {x ∈ Z  x ≡ 0 (mod m)}
{x ∈ Z  x ≡ 1 (mod m)}
{x ∈ Z  x ≡ 2 (mod m)}
{x ∈ Z  x ≡ 3 (mod m)} =
=
=
= {. . . , −8, −4, 0, 4, 8, . . .}
{. . . , −7, −3, 1, 5, 9, . . .}
{. . . , −6, −2, 2, 6, 10, . . .}
{. . . , −5, −1, 3, 8, 11, . . .} 88 =
=
=
= {4k  k ∈ Z}
{4k + 1  k ∈ Z}
{4k + 2  k ∈ Z}
{4k + 3  k ∈ Z} Section 13.2 Modular Arithmetic 89 REMARK
Note that congruence classes have more than one representation. In the example
above [0] = [4] = [8] and, in fact [0] has inﬁnitely many representations. If this seems
strange to you, remember that fractions are another example of where one number
has inﬁnitely many representations. For example 1/2 = 2/4 = 3/6 = · · · . Deﬁnition We deﬁne Zm to be the set of m congruence classes Zm Zm = {[0], [1], [2], . . . , [m − 1]}
and we deﬁne two operations on Zm , addition and multiplication, as follows:
[a] + [b] = [a + b]
[a] · [b] = [a · b] Though the deﬁnition of these operations may seem obvious there is a fair amount
going on here.
1. Sets are being treated as individual “numbers”. Modular addition and multiplication are being performed on congruence classes which are sets.
2. The addition and multiplication symbols on the left of the equals signs are in
Zm and those on the right are operations in the integers.
3. We are assuming that the operations are welldeﬁned. That is, we are assuming
that these operations make sense even when there are multiple representatives
of a congruence class.
Just as there were addition and multiplication tables in grade school for the integers,
we have addition and multiplication tables in Zm . Example 5 Addition and multiplication tables in Z4 +
[0]
[1]
[2]
[3] Exercise 4 [0]
[0]
[1]
[2]
[3] [1]
[1]
[2]
[3]
[0] [2]
[2]
[3]
[0]
[1] [3]
[3]
[0]
[1]
[2] Write out the addition and multiplication tables in Z5 ·
[0]
[1]
[2]
[3] [0]
[0]
[0]
[0]
[0] [1]
[0]
[1]
[2]
[3] [2]
[0]
[2]
[0]
[2] [3]
[0]
[3]
[2]
[1] 90 Chapter 13 13.2.1 Modular Arithmetic [0] ∈ Zm By looking at the tables for Z4 and Z5 it seems that [0] ∈ Zm behaves just like 0 ∈ Z.
In Z
∀a ∈ Z, a + 0 = a
∀a ∈ Z, a · 0 = 0
and in Zm
∀[a] ∈ Zm , [a] + [0] = [a]
∀[a] ∈ Zm , [a] · [0] = [0]
This actually follows from our deﬁnition of addition and multiplication in Zm .
∀[a] ∈ Zm , [a] + [0] = [a + 0] = [a]
∀[a] ∈ Zm , [a] · [0] = [a · 0] = [0] 13.2.2 [1] ∈ Zm In a similar fashion, by looking at the multiplication tables for Z4 and Z5 it seems
that [1] ∈ Zm behaves just like 1 ∈ Z. In Z
∀a ∈ Z, a · 1 = a
and in Zm
∀[a] ∈ Zm , [a] · [1] = [a]
This follows from our deﬁnition of multiplication in Zm .
∀[a] ∈ Zm , [a] · [1] = [a · 1] = [a] 13.2.3 Subtraction in Zm Many of us think of subtraction as independent from the other three arithmetic operations of addition, multiplication and division. In fact, subtraction is just addition
of the inverse. Now, what’s an inverse? To answer that question we must ﬁrst deﬁne
an identity. Deﬁnition
Identity Given a set and an operation, an identity is, informally, “something that does nothing”. More formally, given a set S and an operation designated by ◦, an identity is
an element e ∈ S so that
∀a ∈ S, a ◦ e = a The element e has no eﬀect. Having something that does nothing is extremely useful
 though parents might not say that of teenagers. Section 13.2 Modular Arithmetic Example 1 91 Here are examples of sets, operations and identities.
• The set of integers with the operation of addition has the identity 0.
• The set of rational numbers excluding 0 with the operation of multiplication
has the identity 1.
• The set of real valued functions with the operation of function composition has
the identity f (x) = x.
• The set of integers modulo m with the operation of modular addition has the
identity [0]. Deﬁnition The element b ∈ S is an inverse of a ∈ S if a ◦ b = b ◦ a = e. Inverse Example 2 Here are examples of inverses.
• Under the operation of addition, the integer 3 has inverse −3 since 3 + (−3) =
(−3) + 3 = 0.
• Under the operation of multiplication, the rational number
34
43
4 · 3 = 3 · 4 = 1. 3
4 has inverse 4
3 since • Under the operation of function composition ln x has the inverse ex since
ln(ex ) = eln x = x
• Under the operation of modular addition, [3] has the inverse [−3] in Z7 since
[3] + [−3] = [−3] + [3] = [0]. When the operation is addition, we usually denote the inverse by −a. Otherwise, we
typically denote the inverse of a by a−1 . This does cause confusion. Many students
interpret a−1 as the reciprocal. This works for real or rational multiplication but fails
in other contexts like function composition. We will use −a to mean the inverse of a
under addition and a−1 to mean the inverse under all other operations.
Let’s return to Zm . The identity under addition in Zm is [0] since
∀[a] ∈ Zm , [a] + [0] = [a] Deﬁnition
Subtraction We will deﬁne subtraction as addition of the inverse. Thus
[a] − [b] = [a] + [−b] = [a − b] 92 Chapter 13 13.2.4 Modular Arithmetic Division in Zm Division is related to multiplication in the same way that subtraction is related to
addition. So ﬁrst, we must identify the multiplicative identity in Zm . Since
∀[a] ∈ Zm , [a][1] = [a]
we know that [1] is the identity under multiplication in Zm .
Inverses are more problematic with multiplication. Looking at the multiplication table
for Z5 we see that [2]−1 = [3] since [2][3] = [6] = [1]. But what is the inverse of [2]
in Z4 ? It doesn’t exist! Looking at the row containing [2] in the multiplication table
for Z4 we cannot ﬁnd [1]. Unlike addition in Zm where every element has an additive
inverse, it is not always the case that a nonzero element in Zm has a multiplicative
inverse.
We deﬁne division analogously to subtraction. Deﬁnition
Division Division by a ∈ Zm is deﬁned as multiplication by the multiplicative inverse of
a ∈ Zm , assuming that the multiplicative inverse exists. 13.3 Extending Equivalencies REMARK
Since
[a] = {x ∈ Z  x ≡ a (mod m)} we can extend our list of equivalent statements to
[a] = [b] in Zm
⇐⇒ a ≡ b (mod m)
⇐⇒ m  (a − b)
⇐⇒ ∃k ∈ Z a − b = km ⇐⇒ ∃k ∈ Z a = km + b ⇐⇒ a and b have the same remainder when divided by m 13.4 Fermat’s Little Theorem Pierre de Fermat conjectured what we now call Fermat’s Last Theorem. He also
proved a much smaller but extremely useful result called Fermat’s Little Theorem. Theorem 1 (Fermat’s Little Theorem (F T))
If p is a prime number that does not divide the integer a, then
ap−1 ≡ 1 (mod p) Section 13.4 Fermat’s Little Theorem 93 A proof of Fermat’s Little Theorem is available in the Appendix. Add proof of F T
for the appendix. We will examine two corollaries. Corollary 2 For any integer a and any prime p
ap ≡ a (mod p) Proof: Let a ∈ Z and let p be a prime. If p a, then ap−1 ≡ 1 (mod p). Multiplying
both sides of the equivalence by a gives ap ≡ a (mod p). If p  a, then a ≡ 0 (mod p)
and ap ≡ 0 (mod p). Thus ap ≡ a (mod p).
Let’s make sure we understand the proof.
Analysis of Proof There are two important items to note: the use of nested quantiﬁers in the hypothesis and the use of cases in the proof.
Hypothesis: a ∈ Z, p is a prime
Conclusion: ap ≡ a (mod p)
Core Proof Technique: Select Method
Preliminary Material: Fermat’s Little Theorem
Sentence 1 Let a ∈ Z and let p be a prime.
The hypotheses contain two universal quantiﬁers, so we use the Select Method
twice, once for integers and once for primes.
Sentence 2 If p a, then ap−1 ≡ 1 (mod p).
The author breaks up the proof into two parts depending on whether or not p
divides a. The author will need two distinct cases because the approach diﬀers
based on the case. In the case where p does not divide a, the author uses F T.
Sentence 3 Multiplying both sides of the equivalence by a gives ap ≡ a (mod p).
This is just modular arithmetic.
Sentence 4 If p  a, then a ≡ 0 (mod p) and ap ≡ 0 (mod p). Thus ap ≡ a (mod p).
This is the second case where p does divide a. Both ap and a are congruent to
zero mod p so they are congruent to each other. Corollary 3 (Existence of Inverses in Zp (INV Zp ))
Let p be a prime number. If [a] is any nonzero element in Zp , then there exists an
element [b] ∈ Zp so that [a] · [b] = [1] This corollary is equivalent to stating that every nonzero element of Zp has an inverse.
Let’s discover a proof. As usual, we begin by identifying the hypothesis and the
conclusion. 94 Chapter 13 Modular Arithmetic Hypothesis: p is a prime number. [a] is any nonzero element in Zp .
Conclusion: There exists an element [b] ∈ Zp so that [a] · [b] = [1].
Three points are salient. First, the corollary only states that an inverse exists. It
doesn’t tell us what the inverse is or how to compute the inverse. Second, there are
three quantiﬁers.
1. Let p be a prime number is equivalent to For all primes p. Since this is an
instance of a universal quantiﬁer we would expect to use the Choose Method.
2. [a] is any nonzero element in Zp is another instance of a universal quantiﬁer so
we would expect to use the Choose Method again.
3. There is an existential quantiﬁer in the conclusion so we would expect to use
the Construct Method.
Together these give us the following.
Proof in Progress
1. Let p be a prime number.
2. Let [a] be a nonzero element in Zp .
3. Construct [b] as follows.
4. To be completed.
The third salient point is that this statement is a corollary of Fermat’s Little Theorem.
Now Fermat’s Little Theorem uses congruences, not congruence classes. But we could
restate F T with congruence classes as Theorem 4 (Fermat’s Little Theorem (F T))
If p is a prime number that does not divide the integer a, then
[ap−1 ] = [1] in Zp Now an analogy to real numbers provides the ﬁnal step. In the reals ap−1 = a · ap−2
so why not let [b] = [ap−2 ]? This would give
Proof in Progress
1. Let p be a prime number.
2. Let [a] be a nonzero element in Zp .
3. Consider [b] = [ap−2 ].
4. To be completed. Section 13.4 Fermat’s Little Theorem 95 Now we can invoke Fermat’s Little Theorem but ﬁrst we need to make sure the
hypotheses are satisﬁed.
Proof in Progress
1. Let p be a prime number.
2. Let [a] be a nonzero element in Zp .
3. Consider [b] = [ap−2 ].
4. Since [a] = [0] in Zp , p a and so by F T
[a][b] = [a][ap−2 ] = [ap−1 ] = [1]
A proof might look as follows.
Proof: Let p be a prime number. Let [a] be a nonzero element in Zp . Consider
[b] = [ap−2 ]. Since [a] = [0] in Zp , p a and so by F T
[a][b] = [a][ap−2 ] = [ap−1 ] = [1] REMARK
In summary, if p is a prime number and [a] is any nonzero element in Zp , then
[a]−1 = [ap−2 ] Exercise 1 What is [3]−1 in Z7 ? Chapter 14 Linear Congruences
14.1 Objectives The content objectives are:
1. Deﬁne a linear congruence in the variable x.
2. State and prove the Linear Congruence Theorem.
3. Do examples. 14.2 The Problem One of the advantages of congruence over divisibility is that we have an “arithmetic”
of congruence. This allows us to solve new kinds of “equations”. Deﬁnition
Linear
Congruence A relation of the form
ax ≡ c (mod m)
is called a linear congruence in the variable x. A solution to such a linear
congruence is an integer x0 so that
ax0 ≡ c (mod m) The problem for this lecture is to determine when linear congruences have solutions
and how to ﬁnd them.
Recalling our table of statements equivalent to a ≡ b (mod m) we see that
ax0 ≡ c (mod m) if and only if there exists an integer y0 such that ax0 + my0 = c 96 Section 14.2 The Problem 97 REMARK
Thus
ax ≡ c (mod m) has a solution
⇐⇒ there exists an integer x0 such that ax0 ≡ c (mod m)
⇐⇒ there exists an integer y0 such that ax0 + my0 = c
⇐⇒ gcd(a, m)  c (by the Linear Diophantine Equation Theorem, Part 1) Moreover, the Linear Diophantine Equation Theorem, Part 2 tells us what the solutions to ax + by = c look like. Theorem 5 (Linear Diophantine Equation Theorem, Part 2, (LDET 2))
Let gcd(a, m) = d = 0.
If x = x0 and y = y0 is one particular integer solution to the linear Diophantine
equation ax + my = c, then the complete integer solution is
x = x0 + m
a
n, y = y0 − n, ∀n ∈ Z.
d
d But then, if x0 ∈ Z is one solution to ax ≡ c (mod m) the complete solution will be
m
x ≡ x0 (mod ) where d = gcd(a, m)
d
Equivalently,
m
m
m
x ≡ x0 , x0 + , x0 + 2 , · · · , x0 + (d − 1)
(mod m)
d
d
d
Take note that there are d = gcd(a, m) distinct solutions modulo m.
We record this discussion as the following theorem. Theorem 6 (Linear Congruence Theorem, Version 1, (LCT 1))
Let gcd(a, m) = d = 0.
The linear congruence
ax ≡ c (mod m)
has a solution if and only if d  c.
Moreover, if x = x0 is one particular solution, then the complete solution is
x ≡ x0 (mod m
)
d or, equivalently,
x ≡ x0 , x0 + m
m
m
, x0 + 2 , · · · , x0 + (d − 1)
d
d
d (mod m) 98 Chapter 14 Linear Congruences Another way of considering the same problem is to reframe it in Zm . Since
[a] = {x ∈ Z  x ≡ a (mod m)} solving
ax ≡ c (mod m)
is equivalent to ﬁnding a congruence class [x0 ] ∈ Zm that solves
[a][x] = [c] in Zm
Thus Theorem 7 (Linear Congruence Theorem, Version 2, (LCT 2))
Let gcd(a, m) = d = 0.
The equation
[a][x] = [c] in Zm
has a solution if and only if d  c.
Moreover, if x = x0 is one particular solution, then the complete solution is
[x0 ] , x0 + 14.3 m
m
m
, x0 + 2
, · · · , x0 + (d − 1)
d
d
d in Zm Extending Equivalencies Putting all of this together we have several views of the same problem. REMARK [a][x] = [c] has a solution in Zm
⇐⇒ ax ≡ c (mod m) has a solution
⇐⇒ there exists an integer x0 such that ax0 ≡ c (mod m)
⇐⇒ there exists an integer y0 such that ax0 + my0 = c
⇐⇒ gcd(a, m)  c
Moreover, if x0 , y0 is a particular integer solution to ax + my = c then
a
m
n, y = y0 − n, ∀n ∈ Z
d
d
m
⇐⇒ the complete solution to ax ≡ c (mod m) is x ≡ x0 (mod )
d
m
m
m
⇐⇒ the complete solution to ax ≡ c (mod m) is x ≡ x0 , x0 + , x0 + 2 , · · · , x0 + (d − 1)
(mod m)
d
d
d
m
m
m
⇐⇒ the complete solution to [a][x] = [c] in Zm is [x0 ] , x0 +
, x0 + 2
, · · · , x0 + (d − 1)
in Zm
d
d
d
the complete solution to ax + my = c is x = x0 + Section 14.4 Examples 14.4
Example 1 99 Examples If possible, solve the linear congruence
3x ≡ 5 (mod 6) Solution: Since gcd(3, 6) = 3 and 3 5, there is no solution to 3x ≡ 5 (mod 6) by
the Linear Congruence Theorem, Version 1. Example 2 If possible, solve the linear congruence
4x ≡ 6 (mod 10) Solution: Since gcd(4, 10) = 2 and 2  6, we would expect to ﬁnd two solutions to
4x ≡ 6 (mod 10). Since ten is a small modulus, we can simply test all possibilities
modulo 10.
x (mod 10)
4x (mod 10) 0
0 1
4 2
8 3
2 4
6 5
0 6
4 7
8 8
2 9
6 Hence, x ≡ 4 or 9 (mod 10). Example 3 If possible, solve the linear congruence
3x ≡ 5 (mod 76) Solution: Since gcd(3, 76) = 1 and 1  5, we would expect to ﬁnd one solution
to 3x ≡ 5 (mod 76). We could try all 76 possibilities but there is a more eﬃcient
way. Thinking of our list of equivalencies, solving 3x ≡ 5 (mod 76) is equivalent to
solving 3x + 76y = 5 and that we know how to do that using the Extended Euclidean
Algorithm.
x
y
rq
1
0
76 0
0
1
30
1 −25 1 25
−3 −76 0 3
From the second last row, 76(1) + 3(−25) = 1, or to match up with the order of
the original equation, 3(−25) + 76(1) = 1. Multiplying the equation by 5 gives
3(−125) + 76(5) = 5. Hence
x ≡ −125 ≡ 27 (mod 76) We can check our work by substitution. 3 · 27 ≡ 81 ≡ 5 (mod 76). 100 Chapter 14 Example 4 Linear Congruences Find the inverse of [13] in Z29 .
Solution: By deﬁnition, the inverse of [13] in Z29 is the congruence class [x] so
that [13][x] = [1] in Z29 . Since gcd(13, 29) = 1, we know by the Linear Congruence
Theorem, Version 2 that there is exactly one solution. We could try all 29 possibilities
or recall that solving
[13][x] = [1] in Z29
is equivalent to solving
13x + 29y = 1
and that we know how to do using the Extended Euclidean Algorithm.
x
y
rq
1
0
29 0
0
1
13 0
1
−2 3 2
−4
9
14
13 −29 0 3
From the second last row, 29(−4) + 13(9) = 1, or to match up with the order of the
original equation, 13(9) + 29(−4) = 1. Hence
[13]−1 = [9] in Z29
We can check our work by substitution. [13][9] = [117] = [1] in Z29 . Chapter 15 Chinese Remainder Theorem
15.1 Objectives The content objectives are:
1. Do examples.
2. Discover a proof of the Chinese Remainder Theorem. 15.2 An Old Problem The following problem was posed, likely in the third century CE, by Sun Zi in his
Mathematical Manual and republished in 1247 by Qin Jiushao in the Mathematical
Treatise in Nine Sections.
There are certain things whose number is unknown. Repeatedly divided
by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder
is 2. What will be the number?
The word problem asks us to ﬁnd an integer n that simultaneously satisﬁes the following three linear congruences.
n≡2 (mod 3) n≡3 (mod 5) n≡2 (mod 7) Before we solve this problem, we will begin with two simultaneous congruences whose
moduli are coprime. 101 102 Chapter 15 15.3
Example 5 Chinese Remainder Theorem Chinese Remainder Theorem Solve
n≡2 (mod 5) n≡9 (mod 11) Solution: The ﬁrst congruence is equivalent to
n = 5x + 2 where x ∈ Z (15.1) Substituting this into the second congruence we get
5x + 2 ≡ 9 (mod 11) ⇒ 5x ≡ 7 (mod 11) Have we seen anything like this before? Of course, this is just a linear congruence!
Its solution is
x ≡ 8 (mod 11)
Now x ≡ 8 (mod 11) is equivalent to
x = 11y + 8 where y ∈ Z (15.2) Substituting Equation 15.2 into Equation 15.1 gives the solution
n = 5(11y + 8) + 2 = 55y + 42 for all y ∈ Z
which is equivalent to
n ≡ 42 (mod 55) We can check by substitution. If n = 55y + 42, then n ≡ 2 (mod 5) and n 9
(mod 11). Theorem 8 (Chinese Remainder Theorem (CRT))
If gcd(m1 , m2 ) = 1, then for any choice of integers a1 and a2 , there exists a solution
to the simultaneous congruences
n ≡ a1 (mod m1 ) n ≡ a2 (mod m2 ) Moreover, if n = n0 is one integer solution, then the complete solution is
n ≡ n0 (mod m1 m2 ) Before we begin our discovery of a solution, let’s be clear that there are two things to
prove. First, that a solution exists and second, what a complete solution looks like.
With regards to the ﬁrst part let’s identify, as usual, the hypothesis and the conclusion. Section 15.3 Chinese Remainder Theorem 103 Hypothesis: gcd(m1 , m2 ) = 1.
Conclusion: For any choice of integers a1 and a2 , there exists a solution to the
simultaneous congruences
n ≡ a1 (mod m1 ) n ≡ a2 (mod m2 ) Since there is an existential quantiﬁer in the conclusion, we have to construct a
solution. There is nothing obvious from the statement of the theorem that will help
us, but we have already solved such a problem once in Example 5. Perhaps we could
mimic what we did there.
From the ﬁrst linear congruence
The integer n satisﬁes n ≡ a1 (mod m1 ) if and only if
n = a1 + m1 x for some x ∈ Z
The next thing we did was substitute this expression into the second congruence.
The number n satisﬁes the second congruence if and only if
a1 + m 1 x ≡ a2 (mod m2 ) m 1 x ≡ a2 − a1 (mod m2 ) Have we seen anything like this before? Of course, this is just a linear congruence!
Since gcd(m1 , m2 ) = 1, the Linear Congruence Theorem tells us that this
congruence has a solution, say x = b and that the complete solution is
x = b + m2 y for all y ∈ Z
If we set y = 0 we get x = b and hence n = a1 + m1 b is one particular
solution.
Now let’s consider the second part, a complete solution. Following on what we have
done above, an integer n satisﬁes the simultaneous congruences if and only if
n = a1 + m1 x
= a1 + m1 (b + m2 y )
= (a1 + m1 b) + m1 m2 y for all y ∈ Z
But these are the elements of exactly one congruence class modulo m1 m2 . Hence, if
n = n0 is one solution, then the complete solution is
n ≡ n0 (mod m1 m2 ) 104 Chapter 15 Exercise 2 Chinese Remainder Theorem Using the analysis above, write a proof for the Chinese Remainder Theorem. Question Panel Exercise 3 Solve
n≡2
n≡3 (mod 5) n≡2 (mod 3) n≡3 (mod 5) n≡4 Exercise 4 (mod 3) (mod 11) Solve The exercise above makes it clear that we can solve more than two simultaneous linear
congruences simply by solving pairs of linear congruences successively. We record this
as Theorem 9 (Generalized Chinese Remainder Theorem (GCRT))
If m1 , m2 , . . . , mk ∈ Z and gcd(mi , mj ) = 1 whenever i = j , then for any choice of
integers a1 , a2 , . . . , ak , there exists a solution to the simultaneous congruences
n ≡ a1 (mod m1 ) n ≡ a2
.
.
. (mod m2 ) n ≡ ak (mod mk ) Moreover, if n = n0 is one integer solution, then the complete solution is
n ≡ n0 (mod m1 m2 . . . mk ) Chapter 16 Practice, Practice, Practice:
Congruences
16.1 Objectives The content objectives are:
1. Computational practice.
2. Preparing for RSA. 16.2 Linear and Polynomial Congruences Let’s recall how to solve linear congruences. Example 6 Solve 13x ≡ 1 (mod 60).
Solution: Since gcd(13, 60) = 1 and 1  1 we would expect to ﬁnd one congruence
class as a solution to 13x ≡ 1 (mod 60). Now 13x ≡ 1 (mod 60) is equivalent to the
linear Diophantine equation 13x + 60y = 1 so we can use the EEA.
x
y
rq
1
0
60 0
0
1
13 0
1
−4 8 4
−1
5
51
2
−9 3 1
−3
14
21
5
−23 1 1
−13 60
02
Thus 13(−23) + 60(5) = 1 and so x ≡ −23 ≡ 37 (mod 60) is a solution to 13x ≡ 1
(mod 60). 105 106 Chapter 16 Practice, Practice, Practice: Congruences Though we have eﬃcient means to solve linear congruences, we have no equivalent
means to solve polynomial congruences. Example 7 Solve x2 ≡ 1 (mod 8) by substitution.
Your ﬁrst reaction might be that there are zero, one or two solutions as there would
be in the reals.
Solution:
x (mod 8)
x2 (mod 8) 0
0 1
1 2
4 3
1 4
0 5
1 6
4 7
1 Hence, the solution is x ≡ 1, 3, 5 or 7 (mod 8). Example 8 Solve 36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2 (mod 5). Reduce terms and use Fermat’s
Little Theorem or its corollaries before substitution.
Solution: Since 36 ≡ 1 (mod 5) the term 36x47 reduces to x47 (mod 5). Since 5 ≡ 0
(mod 5) the term 5x9 reduces to 0 (mod 5). Thus,
36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2 (mod 5) reduces to
x47 + x3 + x2 + x + 1 ≡ 2 (mod 5) By Fermat’s Little Theorem, x4 ≡ 1 (mod 5) and so
x47 ≡ (x4 )11 x3 ≡ 111 x3 ≡ x3 (mod 5) and the polynomial congruence further reduces to
x3 + x3 + x2 + x + 1 ≡ 2 (mod 5) or, more simply,
2x3 + x2 + x + 1 ≡ 2
x (mod 5)
2x3 + x2 + x + 1 (mod 5) (mod 5)
0
1 1
0 2
3 3
2 4
4 Hence, the only solution to
36x47 + 5x9 + x3 + x2 + x + 1 ≡ 2
is
x≡3 Example 9 Solve n3 ≡ 127 (mod 165). (mod 5) (mod 5) Section 16.2 Linear and Polynomial Congruences 107 Solution: We could try all 165 possibilities but perhaps there is another way. Observing that 165 = 3 × 5 × 11 and all three factors are relatively prime as pairs, maybe
we could split the problem into three linear congruences and then apply the Chinese
Remainder Theorem. Unfortunately, the polynomial is not linear. Let’s see what
happens anyway.
Since n3 ≡ 127 (mod 165),
n3 ≡ 127 ≡ 1 (mod 3) n3 ≡ 127 ≡ 2 (mod 5) 3 n ≡ 127 ≡ 6 (mod 11) Let’s consider each of the three congruences separately. In the case n3 ≡ 1 (mod 3)
we can use a corollary to Fermat’s Little Theorem. Since n3 ≡ n (mod 3) by FlT,
n3 ≡ 1 (mod 3) reduces to n ≡ 1 (mod 3) which is just the solution to the ﬁrst
congruence.
For the case n3 ≡ 2 (mod 5) we will use a table.
n (mod 5)
n3 (mod 5) 0
0 1
1 2
3 3
2 4
4 The only solution to n3 ≡ 2 (mod 5) is n ≡ 3 (mod 5)
For the case n3 ≡ 6 (mod 11) we will use a table.
n (mod 11)
n3 (mod 11) 0
0 1
1 2
8 3
5 4
9 5
4 6
7 7
2 8
6 9
3 10
10 The only solution to n3 ≡ 6 (mod 11) is n ≡ 8 (mod 11)
Hence, a solution to n3 ≡ 127 (mod 165) can be found by solving the simultaneous
linear congruences
n≡1 (mod 3) n≡3 (mod 5) n≡8 (mod 11) Though these could be solved by eye (note that n ≡ 8 (mod 55) is a solution to the
last two) we will solve these, for practice, by writing out and substituting equations.
From n ≡ 1 (mod 3) we have
n = 3x + 1 where x ∈ Z (16.1) Substituting into the second equation we get
3x + 1 ≡ 3 (mod 5) ⇒ 3x ≡ 2 (mod 5) ⇒ x ≡ 4 (mod 5) Now x ≡ 4 (mod 5) is equivalent to
x = 5y + 4 where y ∈ Z (16.2) 108 Chapter 16 Practice, Practice, Practice: Congruences Substituting Equation 16.2 into Equation 16.1 gives the solution to the ﬁrst two linear
congruences.
n = 3(5y + 4) + 1 = 15y + 13 for all y ∈ Z
which is equivalent to
n ≡ 13 (mod 15) n ≡ 13 (mod 15) Now we need to solve n≡8 (mod 11) From n ≡ 13 (mod 15) we have
n = 15x + 13 where x ∈ Z (16.3) Substituting into the second equation we get
15x+13 ≡ 8 (mod 11) ⇒ 4x+2 ≡ 8 (mod 11) ⇒ 4x ≡ 6 (mod 11) ⇒ x ≡ 7 (mod 11) Now x ≡ 7 (mod 11) is equivalent to
x = 11y + 7 where y ∈ Z (16.4) Substituting Equation 16.4 into Equation 16.3 gives the solution.
n = 15(11y + 7) + 13 = 165y + 118 for all y ∈ Z
which is equivalent to
n ≡ 118 (mod 165) and which is the solution to the original problem n3 ≡ 127 (mod 165).
Checking we have n2 ≡ 1182 ≡ 64 (mod 165) and n3 ≡ 118 × 64 ≡ 127 (mod 165). Example 10 Determine, with justiﬁcation all solutions of the congruence equation
x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 143) Solution: We could simply try all 143 distinct values modulo 143. However, computing numbers like 7061 might be problematic. Have we seen anything like this before?
Note that 143 = 11 × 13. The previous question had a polynomial on the left and
a composite modulus on the right so perhaps we could do now what we did in the
previous exercise, break up the larger problem into several smaller problems.
If x0 is a solution to
x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 143) then x0 is also a solution to
x61 + 26x41 + 11x25 + 5 ≡ 0
x 61 + 26x 41 + 11x 25 +5≡0 (mod 11) (16.5) (mod 13) (16.6) Section 16.3 Linear and Polynomial Congruences 109 Let’s start with the polynomial congruence 16.5.
x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 11) The most obvious thing to do is reduce each term modulo 11. This gives
x61 + 4x41 + 5 ≡ 0 (mod 11) Since 11 is prime, as long as 11 x0 , we can use F T which implies that
x10 ≡ 1 (mod 11) So then
x61 ≡ x60 x1 ≡ (x10 )6 x1 ≡ 16 x1 ≡ x (mod 11) Similarly, x41 ≡ x (mod 11) and so the congruence reduces to
x + 4x + 5 ≡ 0 (mod 11) or
5x + 5 ≡ 0 (mod 11) Since gcd(5, 11) = 1 and 1  (−1), the Linear Congruence Theorem assures us that
exactly one solution exists in this case
x ≡ −1 ≡ 10 (mod 11) We still have to deal with the possibility that 11  x0 . If 11 did divide x0 , then 11
would be a solution to
x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 11) Replacing x by 0 in the above equation, since 11 ≡ 0 (mod 11), gives 5 ≡ 0 (mod 11)
which is false. So 11 x0 . Similarly,
x61 + 26x41 + 11x25 + 5 ≡ 0 (mod 13) reduces to Insert Question Panel.
12x + 5 ≡ 0 (mod 1)3 which has the solution
x≡5 (mod 1)3 Note again that 13 x0 .
But now we have two simultaneous linear congruences
x ≡ 10
x≡5 (mod 11)
(mod 13) Have you seen anything like this before? Insert Question Panel. 110 Chapter 16 16.3 Practice, Practice, Practice: Congruences Preparing for RSA This exercise will help us understand the implementation of the RSA scheme which
we will look at next. In commercial practice the numbers chosen are large but here,
choose numbers small enough to work with by hand.
I will give an example. You follow along but use your own numbers.
1. Choose two distinct primes p and q and let n = pq . I will choose p = 7 and
q = 11 so n = 77.
2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1).
I will choose e = 13 which satisﬁes gcd(13, 60) = 1 and 1 < 13 < 60.
3. Solve
ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1). In my case d = 37. Chapter 17 The RSA Scheme
17.1 Objectives The content objectives are:
1. Illustrate the use of RSA.
2. Prove that the message sent will be the message received. 17.2 Why Public Key Cryptography? In a private key cryptographic scheme, like the substitution cipher or Vigen`re cipher
e
that you have already learned about, participants share a common key. This raises
the problem of how to distribute a large number of keys between users, especially if
these keys need to be changed frequently. For example, there are almost 200 countries
in the world. If Canada maintains an embassy in each country and allows embassies to
communicate with one another, the embassies must exchange a common key between
00
each pair of embassies. That means there are 22 = 19, 900 keys to exchange. Worse
yet, for security reasons, keys should be changed frequently and so 19, 900 keys might
need to be exchanged daily.
In a public key cryptographic scheme, keys are divided into two parts: a public
encryption key which is shared in an open repository of some sort, and a private
decryption key held secretly by each participant. For user A to send a private message
to user B , A would look up B ’s public key, encrypt the message and send it to B .
Since B is the only person who possesses the secret key required for decryption, only
B can read the message.
Such an arrangement solves the key distribution problem. The public keys do not
need to be kept secret and only one per participant needs to be available. Thus, in
our embassy example previously, only 200 keys need to be published.
The possibility of public key cryptography was ﬁrst published in 1976 in a paper
by Diﬃe, Hellman and Merkle. The RSA scheme, named after its discoverers Rivest,
Shamir and Adleman is an example of a commercially implemented public key scheme.
RSA is now widely deployed.
• Add list here.
111 112 Chapter 17 17.3 The RSA Scheme Implementing RSA In RSA, messages are integers. How does one get an integer from plaintext? In much
the same way we did with a Vigen`re cipher, assign a number to each letter of the
e
alphabet and then concatenate the digits together. 17.3.1 Setting up RSA 1. Choose two large, distinct primes p and q and let n = pq .
2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1).
3. Solve
ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1).
4. Publish the public encryption key (e, n).
5. Keep secure the private decryption key (d, n). 17.3.2 Sending a Message To send a message:
1. Look up the recipient’s public key (e, n).
2. Generate the integer message M so that 0 ≤ M < n.
3. Compute the ciphertext C as follows:
Me ≡ C (mod n) where 0 ≤ C < n 4. Send C . 17.3.4 Example All of the computation in this part was done in Maple.
Setting up RSA
1. Choose two large, distinct primes p and q and let n = pq .
Let p be
9026694843 0929817462 4847943076 6619417461
5791443937,
and let q be
7138718791 1693596343 0802517103 2405888327
6844736583
so n is
6443903609 8539423089 8003779070 0502485677 Section 17.3 Implementing RSA 17.3.3 113 Receiving a Message To decrypt a message:
1. Use your private key key (d, n).
2. Compute the messagetext R from the ciphertext C as follows:
Cd ≡ R (mod n) where 0 ≤ R < n 3. R is the original message. 1034536315 4526254586 6290164606 1990955188
1922989980 3977447271.
2. Select an integer e so that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1).
Now (p − 1)(q − 1) is
6443903609 8539423089 8003779070 0502485677
1034536313 8360840952 3666750800 6340495008
2897684191 1341266752.
Choose e as
9573596212 0300597326 2950869579 7174556955
8757345310 2344121731.
It is indeed the case that gcd(e, (p − 1)(q − 1)) = 1 and 1 < e < (p − 1)(q − 1).
3. Solve
ed ≡ 1 (mod (p − 1)(q − 1)) for an integer d where 1 < d < (p − 1)(q − 1).
Solving this LDE gives d as
5587652122 6351022927 9795248536 5522717791
7285682675 6100082011 1849030646 3274981250
2583120946 4072548779.
4. Publish the public encryption key (e, n).
5. Keep secure the private decryption key (d, n).
Sending a Message
To send a message:
1. Look up the recipient’s public key (e, n).
2. Generate the integer message M so that 0 ≤ M < n.
We will let M = 3141592653.
3. Compute the ciphertext C as follows:
Me ≡ C (mod n) where 0 ≤ C < n 114 Chapter 17 The RSA Scheme Computing gives C
4006696554 3080815610 2814019838 8509626485
8151054441 5245547382 5506759308 1333888622
4491394825 3742205367.
4. Send C .
Receiving a Message
To decrypt a message:
1. Use your private key key (d, n).
2. Compute the messagetext R from the ciphertext C as follows:
Cd ≡ R
3. R is the original message.
R = 3141592653. (mod n) where 0 ≤ R < n Section 17.4 Does M = R? 17.4 115 Does M = R? Are we conﬁdent that the message sent is the message received? Theorem 1 (RSA)
If
1. p and q are distinct primes,
2. n = pq
3. e and d are positive integers such that ed ≡ 1 (mod (p − 1)(q − 1)),
4. 0 ≤ M < n
5. M e ≡ C (mod n)
6. C d ≡ R (mod n) where 0 ≤ R < n
then R = M . The proof is long and can appear intimidating but, in fact, it is structurally straightforward if we break it into pieces. The proof is done in four parts.
1. Write R as a function of M , speciﬁcally
R ≡ M M k(p−1)(q−1) (mod n) 2. Show that R ≡ M (mod p). We will do this in two cases: (i) p
p  M.
3. Show that R ≡ M (mod q ).
4. Use the Chinese Remainder Theorem to deduce that R = M .
Proof: First, we will show that
R ≡ M M k(p−1)(q−1) (mod n) Since ed ≡ 1 (mod (p − 1)(q − 1)), there exists an integer k so that
ed = 1 + k (p − 1)(q − 1)
Now
R ≡ Cd (mod n) ≡ (M e )d
≡M ed (mod n)
(mod n) ≡ M 1+k(p−1)(q−1) (mod n) ≡ M M k(p−1)(q−1) (mod n) M and (ii) 116 Chapter 17 The RSA Scheme Second, we will show that R ≡ M (mod p). Suppose that p M . By Fermat’s Little
Theorem,
M p−1 ≡ 1 (mod p)
Hence
M k(p−1)(q−1) ≡ (M p−1 )k(q−1)
≡ 1k(q−1)
≡1 (mod p) (mod p) (mod p) Multiplying both sides by M gives
M M k(p−1)(q−1) ≡ M (mod p) Since
R ≡ M M k(p−1)(q−1) (mod n) ⇒ R ≡ M M k(p−1)(q−1) (mod p) we have
R≡M (mod p) Now suppose that p  M . But then M ≡ 0 (mod p) and so M M k(p−1)(q−1) ≡ 0
(mod p). That is,
M M k(p−1)(q−1) ≡ M (mod p)
Again, since
R ≡ M M k(p−1)(q−1) (mod n) ⇒ R ≡ M M k(p−1)(q−1) (mod p) we have
R≡M (mod p) In either case, we have R ≡ M (mod p).
Third, we will show that R ≡ M (mod q ). But this similar to R ≡ M (mod p).
Fourth and last, we will show that R = M . So far we have generated two linear
congruences that have to be satisﬁed simultaneously.
R≡M (mod p) R≡M (mod q ) Since gcd(p, q ) = 1 we can invoke the Chinese Remainder Theorem and conclude that
R≡M (mod pq ) R≡M (mod n) Since pq = n we have
Now, R and M are both integers congruent to each other modulo n, and both lie
between 0 and n − 1, so R = M . 17.5 How Secure Is RSA? The basic idea behind RSA is that multiplying is easy and factoring is diﬃcult. Hence
it is easy to generate n, which is part of the key, and diﬃcult to factor a large n, say
200 digits, into p and q which would make it easy to decrypt any message. Complete
this. Chapter 18 Just Say No
18.1 Objectives The technique objectives are:
1. To learn how to negate systems.
2. To learn when to use counterexamples.
3. To practice ﬁnding counterexamples. 18.2 Negating Statements You will frequently encounter the negation of statement A. Deﬁnition
Negation The negation of the statement A is the statement NOT A. Because statements
cannot be both true and false, exactly only one of A and NOT A can be true. In some instances, ﬁnding the negation of a statement is easy. For example
A: f (x) has a real root.
NOT A: f (x) does not have a real root.
When A is already negated, a truth table tells us what to do.
A ¬A ¬(¬A)
TF
T
FT
F
Thus, ¬(¬A) = A. Two negatives are a positive, or equivalently, one NOT cancels
another NOT. For example,
A: 7 is not a divisor of 28.
NOT A: 7 is a divisor of 28.
117 118 Chapter 18 Just Say No You have already seen DeMorgan’s Laws when we worked with truth tables. Proposition 2 (De Morgan’s Law’s (DML))
If A and B are statements, then
1. ¬(A ∨ B ) ≡ (¬A) ∧ (¬B )
2. ¬(A ∧ B ) ≡ (¬A) ∨ (¬B ) REMARK
Thus, there is a speciﬁc rule applied when negating a statement containing the word
AND.
A: B AND C
NOT A: (NOT B ) OR (NOT C ) Note that the connecting word has changed from AND to OR and that each term in
the expression has been negated. The brackets are not needed because NOT precedes
OR in logical evaluation, but the brackets are useful to emphasize the change. Here
is a speciﬁc example.
For example,
A: T is isosceles and it has perimeter 42.
NOT A: T is not isosceles or it does not have perimeter 42. REMARK
Similar to the conjunctive AND, a speciﬁc rule is applied when negating a statement
containing the word OR.
A: B OR C
NOT A: (NOT B ) AND (NOT C ) Note that the connecting word has changed from OR to AND and, again, each term
in the expression has been negated. As before, the brackets are not needed because
NOT precedes AND in logical evaluation, but the brackets are useful to emphasize
the change. Here is a speciﬁc example.
For example,
A: T is isosceles or it has perimeter 42.
NOT A: T is not isosceles and it does not have perimeter 42.
A: T is isosceles or it has perimeter 42.
NOT A: T is not isosceles and it does not have perimeter 42. Section 18.3 Negating Statements with Quantiﬁers 18.3 119 Negating Statements with Quantiﬁers Negating statements that contains quantiﬁers is more complicated. We ﬁrst observe
that:
• The negation of a universal statement results in an existential statement.
• The negation of an existential statement results in a universal statement. REMARK
A statement with an existential quantiﬁer looks like
There exists an x in the set S such that P (x) is true.
Its negation is
For every x in the set S , P (x) is false. REMARK
A statement with a universal quantiﬁer looks like
For every x in the set S , P (x) is true.
Its negation is
There exists an x in the set S such that P (x) is false. REMARK
To negate a statement using nested quantiﬁers, do the following.
Step 1 Put the word NOT in front of the entire statement.
Step 2 Move the NOT from left to right replacing quantiﬁers by their opposites and
in each case place the NOT just before the open sentence. Repeat until there
are no quantiﬁers to the right of NOT.
Step 3 When all of the quantiﬁers are to the left of NOT, incorporate the NOT into
the open sentence. Let’s do some examples. 120 Chapter 18 Just Say No Example 1
1. For every x ∈ S , f (x) = 0.
(a) NOT [For every x ∈ S , f (x) = 0.]
(b) There exists x ∈ S such that NOT [f (x) = 0].
(c) There exists x ∈ S such that f (x) = 0.
2. There exists x ∈ S such that f (x) = 0.
(a) NOT [There exists x ∈ S such that f (x) = 0.]
(b) For every x ∈ S , NOT [f (x) = 0].
(c) For every x ∈ S , f (x) = 0.
3. For every x ∈ S and for every f ∈ F , f (x) = 0.
(a) NOT [For every x ∈ S and for every f ∈ F , f (x) = 0.]
(b) There exists x ∈ S such that NOT [for every f ∈ F , f (x) = 0].
There exists x ∈ S and there exists f ∈ F such that NOT [f (x) = 0].
(c) There exists x ∈ S and there exists f ∈ F such that f (x) = 0.
4. There exists x ∈ S such that, for every f ∈ F , f (x) = 0.
(a) NOT [There exists x ∈ S such that for every f ∈ F , f (x) = 0.]
(b) For every x ∈ S , NOT [for every f ∈ F , f (x) = 0].
For every x ∈ S there exists a f ∈ F , NOT [f (x) = 0].
(c) For every x ∈ S there exists a f ∈ F , f (x) = 0. 18.3.1 Counterexamples So far in the course, we have worked on proving that statements are true. How do
we prove that a statement is false? In principle, this is relatively easy. To show that
the statement A is false, we only need to prove that the statement NOT A is true.
Suppose A is the statement:
A: For every x ∈ [−π, π ], sin(x) = 0.
This statement is very similar to our ﬁrst example. NOT A is the statement
NOT A: There exists x ∈ [−π, π ] such that sin(x) = 0.
In this case, NOT A is easy to prove using our construction method. If I consider x =
0, I know that 0 ∈ [−π, π ] and sin(x) = 1 = 0. The number 0 is a counterexample. Deﬁnition In general, if we wish to prove that a universal statement A is false, we show that
Counterexample its negation, which is an existential statement, is true. The particular object which
we use to show that the existential statement is true is called a counterexample of
statement A. Section 18.3 Negating Statements with Quantiﬁers 121 The same idea arises when we want to show that a statement of the form “A implies
B ” is false. It is enough to show a particular instance where A is true and B is false,
or equivalently NOT B is true. For example, consider the following statement. Statement 1 S : If a, b and c are integers, and a  (bc), then a  b and a  c. The hypothesis is
A: a, b and c are integers, and a  (bc)
and the conclusion is
B : a  b and a  c.
To show that S is false, we must ﬁnd a speciﬁc instance where A is true and B is
false. To show that B is false we must show that NOT B is true.
NOT B : a b or a c.
Choosing a = 3, b = 6 and c = 7 we have an instance where the hypothesis A is true
(since 3  42) and the conclusion B is false, equivalently, NOT B is true. The values
a = 3, b = 6 and c = 7 are a counterexample for S . Chapter 19 Contradiction
19.1 Objectives The technique objectives are:
1. Learn how to read and discover proofs by contradiction.
The content objectives are:
1. Read a proof of Prime Factorization.
2. Discover a proof of Inﬁnitely Many Primes. 19.2 How To Use Contradiction We have mostly used the Direct Method to discover proofs, often in conjunction with
one of the methods associated with quantiﬁers. There are times when this is diﬃcult.
A proof by contradiction provides a new method. REMARK
Suppose that we wish to prove that the statement “A implies B ” is true. We assume
that A is true. We must show that B is true. What would happen if B were true, but
we assumed it was false and continued our reasoning based on the assumption that
B was false? Since a mathematical statement cannot be both true and false, it seems
likely we would eventually encounter a mathematically nonsensical statement. Then
we would ask ourselves “How did we arrive at this nonsense?” and the answer would
have to be that our assumption that B was false was wrong and B is, in fact, true. REMARK
A proof by contradiction of the statement “A implies B ” structures proofs in exactly
this way. Proceed as follows. 122 Section 19.2 How To Use Contradiction 123 1. Assume that A is true.
2. Assume that B is false, or equivalently, assume that NOT B is true.
3. Reason forward from A and NOT B to reach a contradiction. Unfortunately, it is not always clear what contradiction to ﬁnd, or how to ﬁnd it.
What is more clear is when to use contradiction. 19.2.1 When To Use Contradiction The general rule of thumb is to use contradiction when the statement NOT B gives
you useful information. There are typically two instances when this is useful. The
ﬁrst instance is when the statement B is one of only two alternatives. For example,
if the conclusion B is the statement f (x) = 0 then the only two possibilities are
f (x) = 0 and f (x) = 0. NOT B is the statement f (x) = 0 which could be useful to
you. The second instance is when B contains a negation. As we saw earlier, NOT B
eliminates the negation. 19.2.2 Reading a Proof by Contradiction Suppose we want to prove the following proposition. Proposition 1 (Prime Factorization (PF))
If n is an integer greater than 1, then n can be expressed as a product of primes. Example 1 The integers 2, 3, 5 and 7 are primes and each is a product unto itself, that is, it is a
product consisting of one factor. The integers 4 = 2 × 2, 6 = 2 × 3 and 8 = 2 × 2 × 2
have been factored as products of primes. Here is a proof.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let N be the smallest integer, greater than 1, that cannot be written as a
product of primes.
2. N is not itself a prime, so we can write N = rs where 1 < r ≤ s < N .
3. Since r and s are less than N , they can be written as a product of primes.
4. But then it follows that N = rs can be written as a product of primes, a
contradiction. 124 Chapter 19 Contradiction Analysis of Proof An interpretation of sentences 1 through 4 follows.
Sentence 1 Let N be the smallest integer, greater than 1, that cannot be written as
a product of primes.
The ﬁrst sentence of a proof by contradiction usually gives the speciﬁc form of
NOT B that the author is going to work with. In this case, the author identiﬁes
that this is a proof by contradiction by assuming the existence of an object which
contradicts the conclusion, an integer N which cannot be written as a product of
primes. Moreover, of all such candidates for N the author chooses the smallest
one. Though it may not be obvious when ﬁrst encountering the proof why the
author would stipulate such a condition, it always has to do with something
needed later in the argument.
Once you know that this is a proof by contradiction, look ahead to ﬁnd the
contradiction. In this case, the contradiction appears in Sentence 4.
Sentence 2 N is not itself a prime, so we can write N = rs where 1 < r ≤ s < N .
If N were prime, then N by itself is a product of primes (with just one factor).
But the author has assumed that N is not a product of primes, hence N is
composite and can be written as the product of two nontrivial factors r and s.
Sentence 3 Since r and s are less than N , they can be written as a product of primes.
This sentence makes it clear why N needs to be the smallest integer that cannot
be written as a product of primes. In order to generate the contradiction, r and
s must be written as products of primes. If it were the case that N was not the
smallest such integer, it might be the case that neither r nor s could be written
as a product of primes.
Sentence 4 But then it follows that N = rs can be written as a product of primes,
a contradiction.
Since both r and s can be written as a product of primes, the product rs = N can
certainly be written as a product of primes. But this contradicts the assumption
in Sentence 1 that N cannot be written as a product of primes.
Since our reasoning is correct, it must be the case that our assumption that there is
an integer which cannot be written as a product of primes is incorrect. That is, every
integer can be written as a product of primes. 19.2.3 Discovering and Writing a Proof by Contradiction Discovering a proof by contradiction can be diﬃcult and often requires several attempts at ﬁnding the path to a contradiction. Let’s see how we might discover a
proof to a famous theorem recorded by Euclid. Proposition 1 (Inﬁnitely Many Primes (INF P))
The number of primes is inﬁnite. We should always be clear about our hypothesis and conclusion. There is no explicit
hypothesis in this case and the conclusion is the statement Section 19.2 How To Use Contradiction 125 Conclusion: The number of primes is inﬁnite.
This statement contains a negation, inﬁnite is an abbreviation of not ﬁnite, and so is a
candidate for a proof by contradiction. Our ﬁrst statement in a proof by contradiction
is a negation of the conclusion so we have
Proof in Progress
1. Assume that the number of primes is ﬁnite. (This is NOT B.)
2. To be completed.
Now comes the tough part. What do we do from here? How do we generate a
contradiction? Well, if the number of primes is ﬁnite, could we somehow use that
assumption to ﬁnd a “new” prime not in our ﬁnite list of primes? Our candidate
should not have any of the ﬁnite primes as a factor. At this point, it sounds like we
need to list our primes.
Proof in Progress
1. Assume that the number of primes is ﬁnite. (This is NOT B.)
2. Label the ﬁnite number of primes p1 , p2 , p3 , . . . , pn .
3. To be completed.
Now we have a way to express a candidate for a “new” prime.
Proof in Progress
1. Assume that the number of primes is ﬁnite. (This is NOT B.)
2. Label the ﬁnite number of primes p1 , p2 , p3 , . . . , pn .
3. Consider the integer N = p1 p2 p3 · · · pn + 1.
4. To be completed.
Clearly N is larger than any of the pi so, by the ﬁrst sentence, N cannot be in the
list of primes. Thus
Proof in Progress
1. Assume that the number of primes is ﬁnite. (This is NOT B.)
2. Label the ﬁnite number of primes p1 , p2 , p3 , . . . , pn .
3. Consider the integer N = p1 p2 p3 · · · pn + 1.
4. Since N > pi ∀i, N is not a prime.
5. To be completed. 126 Chapter 19 Contradiction And this is where we can ﬁnd our contradiction. N has no nontrivial factors since
dividing N by any of the pi leaves a remainder of 1. But that means N cannot
be written as a product of primes, which contradicts the previous proposition. The
contradiction in this proof arises from a result which is inconsistent with something
else we have proved.
Proof in Progress
1. Assume that the number of primes is ﬁnite. (This is NOT B.)
2. Label the ﬁnite number of primes p1 , p2 , p3 , . . . , pn .
3. Consider the integer N = p1 p2 p3 · · · pn + 1.
4. Since N > pi ∀i, N is not a prime.
5. Since N = pi q + 1 for each of the primes pi , no pi is a factor of N . Hence
N cannot be written as a product of primes, which contradicts our previous
proposition.
Putting all of the statements together gives the following proof.
Proof: Assume that there are only a ﬁnite number of primes, say p1 , p2 , p3 , . . . , pn .
Consider the integer N = p1 p2 p3 · · · pn + 1. Since N > pi ∀i, N is not a prime. But
N = pi q + 1 for each of the primes pi , so no pi is a factor of N . Hence N cannot be
written as a product of primes, which contradicts our previous proposition. Chapter 20 Contrapositive
20.1 Objectives The technique objectives are:
1. Deﬁne the contrapositive.
2. Read a proof using the contrapositive.
3. Discover and write a proof using the contrapositive. 20.2 The Contrapositive We begin with an exercise. Exercise 1 Use truth tables to show that A ⇒ B ≡ ¬B ⇒ ¬A. Deﬁnition The statement ¬B ⇒ ¬A is called the contrapositive of A ⇒ B . Contrapositive The logical equivalence between a statement and its contrapositive gives us another
proof technique. Instead of proving “A implies B ” we prove “ NOT B implies NOT
A” using any of the existing techniques. 20.2.1 When To Use The Contrapositive 127 128 Chapter 20 Contrapositive REMARK
This is very similar to contradiction. Use the contrapositive when the statement
NOT A or the statement NOT B gives you useful information. This is most likely
to occur when A or B contains a negation or is one of two possible choices. When
both A and B contain negations, it is highly likely that using the contrapositive will
be productive. 20.3 Reading a Proof That Uses the Contrapositive Consider the following proposition. Proposition 1 Suppose a is an integer. If 32 ((a2 + 3)(a2 + 7)) then a is even. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. We will prove the contrapositive.
2. If a is odd we can write a as 2k + 1 for some integer k .
3. Substitution gives
(a2 + 3)(a2 + 7) = ((2k + 1)2 + 3)((2k + 1)2 + 7)
= (4k 2 + 4k + 1 + 3)(4k 2 + 4k + 1 + 7)
= (4k 2 + 4k + 4)(4k 2 + 4k + 8)
= 4(k 2 + k + 1) × 4(k 2 + k + 2)
= 16(k 2 + k + 1)(k 2 + k + 2)
4. Since one of k 2 + k + 1 or k 2 + k + 2 must be even, and the last line above
shows that a factor of 16 already exists disjoint from (k 2 + k + 1)(k 2 + k + 2),
(a2 + 3)(a2 + 7) must contain a factor of 32. That is 32  ((a2 + 3)(a2 + 7)). Analysis of Proof Since the hypothesis of the proposition contains a negation, and
the conclusion is one of two possible choices, it makes sense to consider the
contrapositive.
Sentence 1 We will prove the contrapositive.
Not all authors will be so obliging as to state the proof technique up front. The
provided proof would also be correct if this sentence was omitted. Correct, but
less easy to understand.
As usual, we begin by identifying the hypothesis and the conclusion.
Hypothesis: A: 32 ((a2 + 3)(a2 + 7)).
Conclusion: B : a is even. Section 20.3 Reading a Proof That Uses the Contrapositive 129 For the contrapositive
Hypothesis: NOT B : a is even.
Conclusion: NOT A: 32  ((a2 + 3)(a2 + 7))
How would we know that the author was using the contrapositive if this sentence
were omitted? The clause “If a is odd” is NOT B so the author is using one of
only two proof techniques that begin this way, contradiction or contrapositive.
Looking ahead to the last line, we see that the author concludes with NOT
A, so this is a proof of the contrapositive. Had the author concluded with a
contradiction, we would know that this is a proof by contradiction.
Sentence 2 If a is odd we can write a as 2k + 1 for some integer k .
This is the statement NOT B . Knowing from Sentence 1 that the author is
using the contrapositive we would expect to see statements moving forward
from the hypothesis of the contrapositive (a is even) or backwards from the
conclusion of the contrapositive (32  ((a2 + 3)(a2 + 7))).
Sentence 3 Substitution gives (a2 + 3)(a2 + 7) = . . . = 16(k 2 + k + 1)(k 2 + k + 2).
This is just arithmetic.
Sentence 4 Since one of k 2 + k + 1 or k 2 + k + 2 must be even, and the last line above
shows that a factor of 16 already exists disjoint from (k 2 + k + 1)(k 2 + k + 2),
(a2 + 3)(a2 + 7) must contain a factor of 32. That is 32  ((a2 + 3)(a2 + 7)).
These sentences establish the conclusion of the contrapositive. Since the contrapositive is true, the original statement is true. 20.3.1 Discovering and Writing a Proof Using The Contrapositive The important observation here is that once you decide to use the contrapositive, all
of your existing skills apply. The diﬃculty is in deciding whether or not to use the
contrapositive.
For our example, we will begin with a deﬁnition. Deﬁnition
Bounded Proposition 1 A set S of real numbers is bounded if there is a real number M > 0 such that, for
all elements x ∈ S , x < M . Suppose that S and T are sets of real numbers with S ⊆ T . If S is not bounded, then
T is not bounded. We should always be clear about our hypothesis and conclusion.
Hypothesis: A: S is not bounded.
Conclusion: B: T is not bounded. 130 Chapter 20 Contrapositive Since both the hypothesis and conclusion are negated, it makes sense to try to prove
the contrapositive “If T is bounded, then S is bounded.” This gives us two statements
in our proof.
Proof in Progress
1. Suppose that T is bounded. (This is just NOT B .)
2. To be completed.
3. Hence, S is bounded. (This is just NOT A.)
Working backwards from the conclusion we can ask “How do we show that S is
bounded?” Using the deﬁnition of bounded, we can write
Proof in Progress
1. Suppose that T is bounded. (This is just NOT B .)
2. To be completed.
3. For every x ∈ S , we have x < M .
4. Hence, S is bounded. (This is just NOT A.)
Now the question becomes “Where can we ﬁnd such an M ?” If we use the deﬁnition
of bounded and work forward from the hypothesis we can write
Proof in Progress
1. Suppose that T is bounded. (This is just NOT B .)
2. Since T is bounded, there exists a real number M > 0 such that, for all x ∈ T ,
x < M .
3. To be completed.
4. For every x ∈ S , we have x < M .
5. Hence, S is bounded. (This is just NOT A.)
Next, we need to connect the two sets and show that the M of the set T is the same
as the M of the set S . But we know
Since x ∈ S and S ⊆ T , x ∈ T .
Combining this with second sentence we have
Since x ∈ S , x ∈ T and so x < M .
Putting all of the statements together gives the following proof.
Proof: We will prove the contrapositive. Suppose that T is bounded. Hence, there
exists a real number M > 0 such that, for all x ∈ T , x < M . Let x ∈ S . Since
S ⊆ T , x ∈ T and so x < M . But then S is bounded as required. Chapter 21 Uniqueness
21.1 Objectives The technique objective is:
1. Learn how to prove a statement about uniqueness in the conclusion. 21.2 Introduction You have already encountered statements that contain the adjective unique. Instead
of the word “unique” you may see “one and only one” or “exactly one”.
Prior to this course you have probably seen statements like the following. Example 1
1. Two lines in the plane which are not parallel will intersect in one and only one
point.
2. There is a unique function f (x) such that f (x) = f (x). And earlier in this course you saw the Division Algorithm. Proposition 2 (Division Algorithm (DA))
If a and b are integers, and b > 0, then there exist unique integers q and r such that
a = qb + r where 0 ≤ r < b. To prove a statement of the form
If . . ., then there is a unique object x in the set S such that P (x) is true.
131 132 Chapter 21 Uniqueness there are basically two approaches.
1. Assume that there are two objects X and Y in the set S such that P (X ) and
P (Y ) are true. Show that X = Y .
2. Assume that there are two distinct objects X and Y in the set S such that
P (X ) and P (Y ) are true. Derive a contradiction.
You can use whichever is easier in the circumstance. 21.3 Showing X = Y The method is as follows.
1. Assume that there are two objects X and Y in the set S such that P (X ) and
P (Y ) are true.
2. Show that X = Y .
For example, let us prove the following statement. Proposition 3 If a and b are integers with a = 0 and a  b, then there is a unique integer k so that
b = ka.
As usual, we begin by explicitly identifying the hypothesis and conclusion.
Hypothesis: a and b are integers with a = 0 and a  b.
Conclusion: There is a unique integer k so that b = ka.
The appearance of “unique” in the conclusion tells us to use one of the two approaches
described in the previous section. In this case, we will assume the existence of two
integers k1 and k2 and show that k1 = k2 .
Proof in Progress
1. Let k1 and k2 be integers such that b = k1 a and b = k2 a. (Note how closely this
follows the standard pattern. k1 corresponds to X . k2 corresponds to Y . Both
come from the set of integers and if P (x) is the statement “b = xa”, then P (X )
and P (Y ) are assumed to be true.
2. To be completed.
3. Hence, k1 = k2 .
The obvious thing to do is equate the two equations to get
k1 a = k2 a
Since a is not zero we can divide both sides by a to get
k1 = k2
A proof might look like the following.
Proof: Let k1 and k2 be integers such that b = k1 a and b = k2 a. But then k1 a = k2 a
and dividing by a gives k1 = k2 . Section 21.4 Finding a Contradiction 21.4 133 Finding a Contradiction The method is as follows.
1. Assume that there are two distinct objects X and Y in the set S such that
P (X ) and P (Y ) are true.
2. Derive a contradiction.
For example, let us prove the following statement. Proposition 4 If m1 = m2 , then there is a unique solution to the simultaneous linear equations
y = m1 x + b1 and y = m2 x + b2 . As usual, we begin by explicitly identifying the hypothesis and conclusion.
Hypothesis: m1 = m2 .
Conclusion: There is a unique solution to the simultaneous linear equations y =
m1 x + b1 and y = m2 x + b2 .
The appearance of “unique” in the conclusion tells us to use one of the two approaches
described in the previous section. In this case, we will assume the existence of two
distinct points of intersection and derive a conclusion.
Proof in Progress
1. Suppose that y = m1 x + b1 and y = m2 x + b2 intersect in the distinct points
(x1 , y1 ) and (x2 , y2 ). (Note again how closely this follows the standard pattern.
(x1 , y1 ) corresponds to X . (x2 , y2 ) corresponds to Y . Both come from the set of
ordered pairs and both satisfy the statement “are a solution to the simultaneous
linear equations y = m1 x + b1 and y = m2 x + b2 .”.
2. To be completed, hence a contradiction.
But now if we substitute (x1 , y1 ) and (x2 , y2 ) into y = m1 x + b1
y1 = m1 x1 + b1 (21.1) y2 = m1 x2 + b1 (21.2) which implies that
y1 − y2 = m1 (x1 − x2 )
Similarly, substituting (x1 , y1 ) and (x2 , y2 ) into y = m2 x + b2 gives
y1 − y2 = m2 (x1 − x2 )
Equating the two expressions for y1 − y2 gives
(m1 − m2 )(x1 − x2 ) = 0 134 Chapter 21 Uniqueness Since m1 = m2 , m1 − m2 = 0 so x1 − x2 = 0. That is, x1 = x2 . But also,
y1 − y2 = m1 (x1 − x2 ) and x1 − x2 = 0
imply
y1 − y2 = 0
That is, y1 = y2 . But then the points (x1 , y1 ) and (x2 , y2 ) are not distinct, a contradiction. Exercise 1 Write a proof for the preceding proposition. 21.5 The Division Algorithm Suppose that in a proof of the Division Algorithm it has already been established that
integers q and r exist and only uniqueness remains. A proposed proof of uniqueness
follows. Proposition 5 (Division Algorithm)
If a and b are integers and b > 0, then there exist unique integers q and r such that
a = qb + r where 0 ≤ r < b Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Suppose that a = q1 b + r1 with 0 ≤ r1 < b. Also, suppose that a = q2 b + r2
with 0 ≤ r2 < b and r1 = r2 .
2. Without loss of generality, we can assume r1 < r2 .
3. Then 0 < r2 − r1 < b and
4. (q1 − q2 )b = r2 − r1 .
5. Hence b  (r2 − r1 ).
6. By Bounds By Divisibility, b ≤ r2 − r1 which contradicts the fact that r2 − r1 < b.
7. Therefore, the assumption that r1 = r2 is false and in fact r1 = r2 .
8. But then (q1 − q2 )b = r2 − r1 implies q1 = q2 . Let’s make sure that we understand every line of the proof. Section 21.5 The Division Algorithm 135 Sentence 1 Suppose that a = q1 b + r1 with 0 ≤ r1 < b. Also, suppose that a = q2 b + r2
with 0 ≤ r2 < b and r1 = r2 .
Since a statement about uniqueness appears in the conclusion, we would expect
one of the two uniqueness methods to be used. In fact, both are used. The
assertion of uniqueness applies to both q and r. Since the author writes r1 = r2 ,
that is, there are distinct values of r1 and r2 , we should look for a contradiction
regarding r. But the author does not assume distinct values of q and so we
would expect that the author will show q1 = q2 .
Sentence 2 Without loss of generality, we can assume r1 < r2 .
“Without loss of generality” is an expression that means the upcoming argument
would hold identically if we made any other choice, so we will simply assume
one of the possibilities.
Sentence 3 Then 0 < r2 − r1 < b and
This is a particularly important line. It comes, in part, from r1 < r2 by subtracting r1 from both sides (this gives 0 < r2 − r1 ) and by remembering that
the largest possible value of r2 is b − 1 and the smallest possible value of r1 is
0, so the largest possible diﬀerence is b − 1, thus r2 − r1 < b
Sentence 4 (q1 − q2 )b = r2 − r1 .
This follows from equating a = q1 b + r1 and a = q2 b + r2 .
Sentence 5 Hence b  (r2 − r1 ).
This follows from the deﬁnition of divisibility.
Sentence 6 By Bounds By Divisibility, b ≤ r2 − r1 which contradicts the fact that
r2 − r1 < b.
Note the importance of the strict inequality in the relation
b ≤ r2 − r1 < b
Sentence 7 Therefore, the assumption that r1 = r2 is false and in fact r1 = r2 .
The contradiction we were looking for. The Division Algorithm states that both
q and r are unique. So far, only the uniqueness of r has been established.
Sentence 7 But then (q1 − q2 )b = r2 − r1 implies q1 = q2 .
And this is where the uniqueness of q is established. Originally, the author
assumed the existence of q1 and q2 and now has shown that they are, in fact,
the same. Chapter 22 Induction
22.1 Objectives The technique objective is:
1. Learn how to use the Principle of Mathematical Induction, sometimes called
Simple Induction.
2. Learn how to use the Principle of Strong Induction, usually called Strong Induction. 22.2 Introduction Induction is a common and powerful technique and should be your ﬁrst choice whenever you encounter a statement of the form
For every integer n ≥ 1, P (n) is true.
where P (n) is a statement that depends on n.
Here are two examples of propositions in this form. Proposition 6 For every integer n ≥ 1 n i2 =
i=1 n(n + 1)(2n + 1)
.
6 Often the clause “For every integer n ≥ 1” is implied and does not actually appear
in the proposition, as in the following version of the same theorem. Proposition 7 The sum of the ﬁrst n perfect squares is n(n+1)(2n+1)
.
6 The second example uses sets, not equations. 136 Section 22.3 Principle of Mathematical Induction Proposition 8 Every set of size n has exactly 2n subsets. 22.3
Deﬁnition
Axiom 137 Principle of Mathematical Induction An axiom of a mathematical system is a statement that is assumed to be true. No
proof is given. From axioms we derive propositions and theorems.
Sometimes axioms are described as selfevident, though many are not. Axioms are
deﬁning properties of mathematical systems. The Principle of Mathematical Induction is one such axiom. Axiom 1 Principle of Mathematical Induction (POMI)
Let P (n) be a statement that depends on n ∈ N.
If
1. P (1) is true, and
2. P (k ) is true implies P (k + 1) is true for all k ∈ N
then P (n) is true for all n ∈ N.
We use the Principle of Mathematical Induction to prove statements of the form
For every integer n ≥ 1, P (n) is true.
The structure of a proof by induction models the deﬁnition of induction. The three
parts of the structure are as follows.
Base Case Verify that P (1) is true. This is usually easy. You will often see the
statement “It is easy to see that the statement is true for n = 1.” It is best to
write this step out completely.
Inductive Hypothesis Assume that P (k ) is true for some integer k ≥ 1. It is best
to write out the statement P (k ).
Inductive Conclusion Using the assumption that P (k ) is true, show that P (k + 1)
is true. Again, it is best to write out the statement P (k + 1) before trying to
prove it. 22.3.1 Why Does Induction Work? The basic idea is simple. We show that P (1) is true. We then use P (1) to show
that P (2) is true. And then we use P (2) to show that P (3) is true and continue
indeﬁnitely. That is
P (1) ⇒ P (2) ⇒ P (3) ⇒ . . . ⇒ P (i) ⇒ P (i + 1) ⇒ . . . 138 Chapter 22 22.3.2 Induction Two Examples of Simple Induction Our ﬁrst example is very typical and uses an equation containing the integer n. Proposition 1 n i2 =
i=1 n(n + 1)(2n + 1)
.
6 Proof: We begin by formally writing out our inductive statement
n i2 = P (n) :
i=1 n(n + 1)(2n + 1)
.
6 Base Case We verify that P (1) is true where P (1) is the statement
1 i2 = P (1) :
i=1 1(1 + 1)(2 × 1 + 1)
.
6 As in most base cases involving equations, we can make our way from the left
side of the equation to the right side of the equation with just a little algebra.
1 i2 = 12 = 1 =
i=1 1(1 + 1)(2 × 1 + 1)
.
6 Inductive Hypothesis We assume that the statement P (k ) is true for some integer
k ≥ 1.
k
k (k + 1)(2k + 1)
P (k ) :
i2 =
.
6
i=1 Inductive Conclusion Now show that the statement P (k + 1) is true.
k+1 i2 = P (k + 1) :
i=1 (k + 1)((k + 1) + 1)(2(k + 1) + 1)
.
6 This is the diﬃcult part. When working with equations, you can often start
with the more complicated expression and decompose it into an instance of P (k )
with some leftovers. That’s what we will do here.
k+1 k i2 =
i=1 i2 + (k + 1)2 partition into P (k ) and other i=1 k (k + 1)(2k + 1)
+ (k + 1)2
use the inductive hypothesis
6
k (k + 1)(2k + 1) + 6(k + 1)2
=
algebraic manipulation
6
(k + 1) 2k 2 + 7k + 6
=
factor out k + 1, expand the rest
6
(k + 1)(k + 2)(2k + 3)
=
factor
6
(k + 1)((k + 1) + 1)(2(k + 1) + 1)
=
6
The result is true for n = k + 1, and so holds for all n by POMI.
= Section 22.3 Principle of Mathematical Induction 139 Our next example does not have any equations. Proposition 2 Let Sn = {1, 2, 3, . . . , n}. Then Sn has 2n subsets. Let’s be very clear about what our statement P (n) is.
P (n): Sn has 2n subsets.
Now we can begin the proof.
Proof: Base Case We verify that P (1) is true where P (1) is the statement
P (1): S (1) has 2 subsets.
We can enumerate all of the sets of S1 easily. They are { } and {1}, exactly two
as required.
Inductive Hypothesis We assume that the statement P (k ) is true for some integer
k ≥ 1.
P (k ): Sk has 2k subsets.
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): Sk+1 has 2k+1 subsets.
The subsets of Sk+1 can be partitioned into two sets. The set A in which no
subset contains the element k + 1, and the complement of A, A, in which every
subset contains the element k + 1. Now A is just the subsets of Sk and so, by
the inductive hypothesis, has 2k subsets. A is composed of the subsets of Sk to
which the element k + 1 is added. So, again by our inductive hypothesis, there
are 2k subsets of A. Since A and A are disjoint and together contain all of the
subsets of Sk+1 , there must be 2k + 2k = 2k+1 subsets of Sk+1 .
The result is true for n = k + 1, and so holds for all n by POMI. 22.3.3 A Diﬀerent Starting Point Some true statements cannot start with “for all integers n, n ≥ 1”. For example,
“2n > n2 ” is false for n = 2, 3, and 4 but true for n ≥ 5. But the basic idea holds. If
we can show that a statement is true for some base case n = b, and then show that
P (b) ⇒ P (b + 1) ⇒ P (b + 2) ⇒ . . . ⇒ P (i) ⇒ P (i + 1) ⇒ . . .
this is also induction. Perhaps this is not surprising because we can always recast
a statement “For every integer n ≥ b, P (n)” as an equivalent statement “For every
integer k ≥ 1, P (k )”. For example, 140 Chapter 22 Induction For every integer n ≥ 5, 2n > n2 .
is equivalent to
For every integer k ≥ 1, 2k+4 > (k + 4)2 .
In this case, we have just replaced n by k + 4 in the statement.
The basic structure of induction is the same. To prove the statement
For every integer n ≥ b, P (n) is true.
the only changes we need to make are that our base case is P (b) rather than P (1),
and that in our inductive hypothesis we assume P (k ) is true for k ≥ b rather than
k ≥ 1.
Here is an example. Proposition 1 For every integer n ≥ 5, 2n > n2 . As usual, let’s be very clear about what our statement P (n) is.
P (n): 2n > n2 .
Now we can begin the proof.
Proof: Base Case We verify that P (5) is true where P (5) is the statement
P (5): 25 > 52
This is just arithmetic.
25 = 32 > 25 = 52
Inductive Hypothesis We assume that the statement P (k ) is true for some integer
k ≥ 5.
P (k ): 2k > k 2
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): 2k+1 > (k + 1)2
2k+1 = 2 × 2k > 2 × k 2 = k 2 + k 2 > k 2 + 2k + 1 = (k + 1)2
The result is true for n = k + 1, and so holds for all n by POMI. 22.4 Strong Induction Sometimes Simple Induction doesn’t work where it looks like it should. We then need
to change our approach a bit. The following example is similar to examples that we’ve
done earlier. Lets try to make Simple Induction work and see where things go wrong. Section 22.4 Strong Induction Proposition 2 141 Let the sequence {xn } be deﬁned by x1 = 0, x2 = 30 and xm = xm−1 + 6xm−2 for
m ≥ 3. Then
xn = 2 · 3n + 3 · (−2)n for n ≥ 1. This seems like a classic case for induction since the conclusion clearly depends on
the integer n. Let’s begin with our statement P (n).
P (n): xn = 2 · 3n + 3 · (−2)n .
Now we can begin the proof.
Proof: Base Case We verify that P (1) is true where P (1) is the statement
P (1): x1 = 2 · 31 + 3 · (−2)1 .
From the deﬁnition of the sequence x1 = 0. The right side of the statement
P (1) evaluates to 0 so P (1) is true.
Inductive Hypothesis We assume that the statement P (k ) is true for k ≥ 1.
P (k ): xk = 2 · 3k + 3 · (−2)k .
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): xk+1 = 2 · 3k+1 + 3 · (−2)k+1 .
xk+1 = xk + 6xk−1
k by the deﬁnition of the sequence
k = 2 · 3 + 3 · (−2) + 6xk−1 by the Inductive Hypothesis Now two problems are exposed. The more obvious problem is what do we do with
xk−1 ? The more subtle problem is whether we can even validly write the ﬁrst line.
When k + 1 = 2 we get
x2 = x1 + 6 x0
and x0 is not even deﬁned.
The basic principle that earlier instances imply later instances is sound. We need to
strengthen our notion of induction in two ways. First, we need to allow for more than
one base case so that we avoid the problem of undeﬁned terms. Second, we need to
allow access to any of the statements P (1), P (2), P (3), ... , P (k ) when showing that
P (k + 1) is true. This may seem like too strong an assumption but is, in fact, quite
acceptable. This practice is based on the Principle of Strong Induction. 142 Chapter 22 Axiom 2 Induction Principle of Strong Induction (POSI)
Let P (n) be a statement that depends on n ∈ P.
If
1. P (1), P (2), . . . , P (b), are true, and
2. P (1), P (2), . . . , P (k ) are all true implies P (k + 1) is true for all k ∈ N,
then P (n) is true for all n ∈ P. Just as before, there are three parts in a proof by strong induction.
Base Cases Verify that P (1), P (2), . . . , P (b) are all true. This is usually easy.
Inductive Hypothesis Assume that P (i) is true for i = 1, 2, 3, . . . , k , k ≥ b. This
is sometimes written as Assume that P (1), P (2), . . . , P (k ) are true.
Inductive Conclusion Using the assumption that P (1), P (2), . . . , P (k ) are true,
show that P (k + 1) is true.
As a rule of thumb, use Strong Induction when the general case depends on more
than one previous case. Though we could use Strong Induction all the time, Simple
Induction is often easier.
Let’s return to our previous proposition. Proposition 3 Let the sequence {xn } be deﬁned by x1 = 0, x2 = 30 and xm = xm−1 + 6xm−2 for
m ≥ 3. Then
xn = 2 · 3n + 3 · (−2)n for n ≥ 1. We will use Strong Induction. Recall our statement P (n).
P (n): xn = 2 · 3n + 3 · (−2)n .
Now we can begin the proof.
Proof: Base Case We verify that P (1) and P (2) are true.
P (1): x1 = 2 · 31 + 3 · (−2)1 .
From the deﬁnition of the sequence x1 = 0. The right side of the statement
P (1) evaluates to 0 so P (1) is true.
P (2): x2 = 2 · 32 + 3 · (−2)2 .
From the deﬁnition of the sequence x2 = 30. The right side of the statement
P (2) evaluates to 30 so P (2) is true. Section 22.4 Strong Induction 143 Inductive Hypothesis We assume that the statement P (i) is true for 1 ≤ i ≤ k ,
k ≥ 2.
P (i): xi = 2 · 3i + 3 · (−2)i .
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): xk+1 = 2 · 3k+1 + 3 · (−2)k+1 .
xk+1 = xk + 6xk−1
k by the deﬁnition of the sequence
k = 2 · 3 + 3 · (−2) + 6(2 · 3 k −1 k−1 + 3 · (−2) ) by the Inductive Hypothesis = 3k−1 [2 · 3 + 6 · 2] + (−2)k−1 [3 · (−2) + 6 · 3] expand and factor = 18 · 3 k −1 k −1 + 12 · (−2) = 2 · 3k+1 + 3 · (−2)k+1 The result is true for n = k + 1, and so holds for all n by POSI. 22.4.1 Interesting Example A triomino is a tile of the form Proposition 1 A 2n × 2n grid of squares with one square removed can be covered by triominoes. As usual, we begin by explicitly stating P (n).
P (n): A 2n × 2n grid of squares with one square removed can be covered
by triominoes.
We will use Simple Induction.
Proof: Base Case We verify that P (1) is true.
P (1): A 2 × 2 grid of squares with one square removed can be covered
by triominoes.
A 2 × 2 grid with one square removed looks like or or or . Each of these can be covered by one triomino.
Inductive Hypothesis We assume that the statement P (i) is true for 1 ≤ i ≤ k .
P (k ): A 2i × 2i grid of squares with one square removed can be covered
by triominoes.
Note that our hypothesis covers every possible position for the empty square
within the grid. 144 Chapter 22 Induction Inductive Conclusion We now show that the statement P (k + 1) is true.
P (k + 1): A 2k+1 × 2k+1 grid of squares with one square removed can
be covered by triominoes.
Consider a 2k+1 × 2k+1 grid with any square removed. Split the 2k+1 × 2k+1 grid in half vertically and horizontally. The missing square occurs in one of the four 2k × 2k subgrids formed. We’ll
start by placing one tile around the centre of the grid, not covering any of the
2k × 2k subgrids where the square is missing: We can now view the grid as being made up of four 2k × 2k subgrids, each with
one square missing. The Inductive Hypothesis tells us that each of these can be
covered by triominoes. Therefore, the whole 2k+1 × 2k+1 grid can be covered.
The result is true for n = k + 1, and so holds for all n by POMI. Chapter 23 Introduction to Primes
23.1 Objectives The technique objectives are:
1. Practice with induction.
2. Practice with arguments of uniqueness.
The content objectives are:
1. Recall the deﬁnition of prime and composite.
2. Discover a proof by induction of the Prime Factorization Theorem.
3. 23.2 Introduction to Primes The second problem that the course focuses on is Fermat’s Last Theorem. Theorem 2 (Fermat’s Last Theorem (FLT))
If n ≥ 3, then there are no solutions to
xn + y n = z n
where x, y and z are positive integers. To make progress on this problem, we need to work with prime numbers. Recall our
deﬁnition of prime number. Deﬁnition
Prime,
Composite An integer p > 1 is called a prime if its only divisors are 1 and p, and composite
otherwise. 145 146 Chapter 23 Example 1 Introduction to Primes The integers 2, 3, 5 and 7 are primes. The integers 4 = 2 × 2, 6 = 2 × 3 and 8 = 2 × 2 × 2
are composite. Note, that by deﬁnition, 1 is not a prime. We have already proved three propositions about primes, one of which is a consequence
of Coprimeness and Divisibility, and the other two were proved in the chapter on
contradiction. Proposition 3 (Primes and Divisibility (PAD))
If p is a prime and p  ab, then p  a or p  b. Proposition 4 (Prime Factorization (PF))
If n is an integer greater than 1, then n can be written as a product of prime factors. Proposition 5 (Inﬁnitely Many Primes (INF P))
The number of primes is inﬁnite. We will prove Prime Factorization again, this time with induction. 23.3 Induction Recall how induction, Strong Induction in this case, works. Axiom 3 Principle of Strong Induction (POSI)
Let P (n) be a statement that depends on n ∈ P.
If
1. P (1), P (2), . . . , P (b), are true, and
2. P (1), P (2), . . . , P (k ) are all true implies P (k + 1) is true
then P (n) is true for all n ∈ P. Recall the three parts in a proof by strong induction.
Base Cases Verify that P (1), P (2), . . . , P (b) are all true.
Inductive Hypothesis Assume that P (i) is true for i = 1, 2, 3, . . . , k where k ≥ b.
Inductive Conclusion Using the assumption that P (1), P (2), . . . , P (k ) are true,
show that P (k + 1) is true. Section 23.4 Fundamental Theorem of Arithmetic 147 We will use Strong Induction to prove Proposition 6 (Prime Factorization (PF))
If n is an integer greater than 1, then n can be expressed as a product of prime factors. First, we formulate our statement P (n) that relies on the integer n.
P (n): n can be expressed as a product of prime factors.
Now we can begin the proof.
Proof: Base Case We verify P (2). Recall that the base case does not need to start
at 1.
P (2): 2 can be expressed as a product of prime factors.
This is trivially true.
Inductive Hypothesis We assume that P (i) is true for i = 2, 3, . . . , k where k ≥ 2.
P (i): i can be expressed as a product of prime factors.
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): k + 1 can be expressed as a product of prime factors.
If k +1 is prime, then k +1 by itself is a product of prime factors. It is a product
with just one factor. In this case, P (k + 1) is true.
If k + 1 is composite, then we can write k + 1 = rs where 1 < r ≤ s < k + 1.
Since r and s are less than k + 1, they can be written as a product of prime
factors by the inductive hypothesis. Hence, k + 1 is a product of prime factors
and P (k + 1) is true in this case also.
The result is true for n = k + 1, and so holds for all n by POSI. 23.4 Fundamental Theorem of Arithmetic In grade school you used prime numbers to write the prime factorization of any
positive integer greater than one. You probably never worried about the possibility
that there might be more than one way to do this. However, in some sets “prime”
factorization is not unique.
√
√
Consider the set S = {a + b 5  a, b ∈ Z}. In S , the number 4 = 4 + 0 5 can be
√
√
factored in two diﬀerent ways, 4 = 2 × 2 and 4 = ( 5 + 1)( 5 − 1). Moreover, 2,
√
√
5 + 1 and 5 − 1 are all prime numbers in S !
Since multiplication in the integers is commutative, the prime factorizations can be
written in any order. For example 12 = 2 × 2 × 3 = 2 × 3 × 2 = 3 × 2 × 2. However, up
to the order of the factors, the factorization of integers is unique. This property is so
basic it is referred to as the Fundamental Theorem of Arithmetic. It is also referred
to as the Unique Factorization Theorem. 148 Chapter 23 Theorem 7 Introduction to Primes (Fundamental Theorem of Arithmetic or
Unique Factorization Theorem (UFT))
If n > 1 is an integer, then n can be written as a product of prime factors and, apart
from the order of factors, this factorization is unique. Observe that the conclusion contains two parts:
1. n can be written as a product of prime factors (which we proved earlier), and
2. apart from the order of factors, this factorization is unique.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. That n can be written as a product of prime factors follows from the proposition
Prime Factorization.
2. Now suppose that n is factored into primes in two ways,
n = p1 p2 . . . pk = q1 q2 . . . q (23.1) where all of the p’s and q ’s are primes.
3. Since p1  n, p1  q1 q2 . . . q .
4. By repeatedly applying the proposition Primes and Divisibility, p1 must divide
one of the q ’s. If necessary, rearrange the q ’s so that p1  q1 .
5. Since q1 is prime, and p1 > 1, it must be the case that p1 = q1 .
6. Dividing Equation 23.1 by p1 = q1 gives
p2 p3 . . . pk = q2 q3 . . . q (23.2) 7. By continuing in this way, we see that each p must be paired oﬀ with one of the
q s until there are no factors on either side.
8. Hence k = and, apart from the order of the factors, the two expressions for n
are the same. Let’s perform an analysis of the proof. As usual, we begin with the hypothesis and
the conclusion.
Hypothesis: n is an integer, n > 1
Conclusion: There are two parts.
1. n can be written as a product of prime factors, and
2. apart from the order of factors, this factorization is unique. Section 23.5 Finding a Prime Factor 149 Core Proof Technique: Uniqueness
Preliminary Material: Primes and Divisibility
Sentence 1 That n can be written as a product of prime factors follows from the
proposition Prime Factorization.
The ﬁrst of the two parts of the conclusion is just the conclusion of a previous
proposition.
Sentence 2 Now suppose that n is factored into primes in two ways,
n = p1 p2 . . . pk = q1 q2 . . . q
where all of the p’s and q ’s are primes.
This is a classic use of the Uniqueness Method. We assume that there are two
representations of the same object, and show that the two representations are,
in fact, identical. One representation of n is the product p1 p2 . . . pk and the
second representation is the product q1 q2 . . . q .
Sentences 3 – 5 Since p1  n, p1  q1 q2 . . . q . By repeatedly applying the proposition
Primes and Divisibility, p1 must divide one of the q ’s. If necessary, rearrange
the q ’s so that p1  q1 . Since q1 is prime, and p1 > 1, it must be the case that
p1 = q1 .
The author shows that the two representations of n are equal by showing that
they have identical factors. Here, the author demonstrates that p1 = q1 .
Sentences 6 – 7 Dividing Equation 23.1 by p1 = q1 gives
p2 p3 . . . pk = q2 q3 . . . q
By continuing in this way, we see that each p must be paired oﬀ with one of the
q s until there are no factors on either side.
This continues the author’s plan of showing that the two representations of n
are equal by showing that they have identical factors.
Sentence 8 Hence k = and, apart from the order of the factors, the two expressions
for n are the same.
This is a typical conclusion to the Uniqueness Method. The two representations
of the same object are identical. 23.5 Finding a Prime Factor The previous proposition does not provide an algorithm for ﬁnding the prime factors
of a positive integer n. The next proposition shows that we do not have to check all
of the prime factors less than n, only those less than or equal to the square root of n. Proposition 8 (Finding a Prime Factor (FPF))
An integer n > 1 is either prime or contains a prime factor ≤ Let’s begin by identifying the hypothesis and the conclusion. √ n. 150 Chapter 23 Introduction to Primes Hypothesis: n is an integer and n > 1.
Conclusion: n is either prime or contains a prime factor ≤ √ n. Before we see a proof, let’s do an example. Example 2 Is 73 a prime number?
Solution: Using Finding a Prime Factor , we can check for divisibility by primes
√
√
≤ 73. Now 73 ≈ 8.544 so any possible prime factor must be less than or equal to
8. The only candidates to check are 2, 3, 5 and 7. Since none of these divide 73, 73
must be prime. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let p be the smallest prime factor of n.
2. Since n is composite we can write n = ab where a and b are positive integers
between 1 and n.
3. Since p is the smallest prime factor, p ≤ a and p ≤ b and so n = ab ≥ p · p = p2 .
√
That is p ≤ n. Analysis of Proof Since or appears in the conclusion, we will use Proof By Elimination. The equivalent statement that is proved is:
If n is an integer greater than 1 and n is not prime, then n contains
√
a prime factor ≤ n.
The word “a” should alert us to the presence of an existential quantiﬁer. We
could reword the statement as
If n is an integer greater than 1 and n is not prime, then there exists
√
a prime factor of n which is ≤ n.
Hypothesis: n is an integer greater than 1 and n is not prime.
√
Conclusion: There exists a prime factor of n which is ≤ n.
Core Proof Technique: Construct Method
Sentence 1 Let p be the smallest prime factor of n.
The conclusion has an existential quantiﬁer and so the author uses the Construct Method. The prime p will be the desired prime factor though it is not
clear yet why “smallest” is important. The proposition on Prime Factorization
guarantees us that a prime factor exists.
Sentence 2 Since n is composite we can write n = ab where a and b are positive
integers between 1 and n.
By the hypotheses of the restated proposition, n > 1 and n is not prime, so n
is composite and can be factored. Section 23.6 Working With Prime Factorizations 151 Sentence 3 Since p is the smallest prime factor, p ≤ a and p ≤ b and so n = ab ≥
√
p · p = p2 . That is p ≤ n.
This is where “smallest” is used. The conclusion follows from arithmetic and
the fact that p is the smallest prime factor. 23.6 Working With Prime Factorizations The next proposition, which we will state but not prove, gives us a means to list all of
the divisors of a positive integer. A proof is available in the Appendix. Add proof. Proposition 9 (Divisors From Prime Factorization (DFPF))
If a > 1 is an integer and
a = p α1 p α2 · · · p αk
12
k
is the prime factorization of a into powers of distinct primes pi , p2 , . . . , pk , then the
positive divisors of a are integers of the form
d = pd1 pd2 · · · pdk where 0 ≤ di ≤ αi for i = 1, 2, . . . , k
12
k Exercise 1 Using Divisors From Prime Factorization, list all of the positive factors of 45. Exercise 2 How many positive divisors are there to the integer a whose prime factorization is
a = p α1 p α2 · · · p αk
12
k Proposition 10 (GCD From Prime Factorization (GCD PF))
If
a = p α1 p α2 · · · p αk
12
k
and
b = pβ1 pβ2 · · · pβ
12
are the prime factorizations of a and b, where some of the exponents may be zero,
then
gcd(a, b) = pd1 pd2 · · · pdk where di = min{αi , βi } for i = 1, 2, . . . , k
12
k Though this method works well enough on small examples, it is much slower than the
Extended Euclidean Algorithm for computing gcds. Exercise 3 Use GCD PF to compute gcd(33 51 74 131 , 52 77 131 232 ). 152 Chapter 23 Exercise 4 Introduction to Primes Use the deﬁnition of gcd to prove GCD From Prime Factorization. Chapter 24 Introduction to Fermat’s Last
Theorem
24.1 Objectives The content objectives are:
1. Provide an historical introduction.
2. Deﬁne gcd(x, y, z ), trivial solutions, Pythagorean triple and primitive Pythagorean
triple.
3. State Extending Coprimeness.
4. Read a proof of Multiples of Pythagorean Triples.
5. Discover a proof of Relative Primeness of Pythagorean Triples.
6. Read a proof of Parity of Pythagorean Triples.
7. State Decomposing nth Powers. 24.2 History of Fermat’s Last Theorem Pierre de Fermat (1601 (?) – 1635) was a brilliant French mathematician. It was his
habit to make notes in the margins of his books and one such note is famous. Fermat
possessed a copy of Bachet’s translation of Diophantus’ Arithmetica. Problem II.8 of
the Arithmetica reads
Partition a given square into two squares.
Diophantus did not require the squares to be integers so we might write Problem II.8
as 153 154 Chapter 24 Introduction to Fermat’s Last Theorem For what positive rational numbers x, y and z is the equation
x2 + y 2 = z 2
satisﬁed?
Adjacent to Problem II.8, and in the margin of his copy of Arithmetica, Fermat wrote
(translated)
It is impossible to separate a cube into two cubes, or a fourth power into
two fourth powers, or in general, any power higher than the second, into
two like powers. I have discovered a truly marvellous proof of this, which
this margin is too narrow to contain.
Fermat was asserting Theorem 11 (Fermat’s Last Theorem)
If n ≥ 3, then
xn + y n = z n
has no solutions when x, y and z are positive integers. No proof was ever published by Fermat, or found among his notes after his death.
It seems very unlikely that he did have a proof and it was not until Andrew Wiles’
publications in 1994 that the Theorem, conjecture really, was proved. Fermat did
prove the case n = 4, as we shall do.
First though, we will clarify our language. Clearly there are solutions to xn + y n = z n .
One solution is x = y = z = 0, another solution is x = 0, y = z . Deﬁnition
Trivial We will say that any solution to xn + y n = z n for which at least one of x, y or z is
zero, is trivial. So we restate Fermat’s Last Theorem as Theorem 12 (Fermat’s Last Theorem)
If n ≥ 3, then
xn + y n = z n
has no nontrivial integer solutions. Our starting point will be a much more familiar problem.
x2 + y 2 = z 2 (24.1) You will recognize this as the equation of the Pythagorean Theorem. Our task is to
identify all positive integer solutions to 24.1. Section 24.3 Pythagorean Triples 24.3 155 Pythagorean Triples We begin with some deﬁnitions. Deﬁnition A Pythagorean triple is a set of nonzero integers x, y and z such that x2 + y 2 = z 2 . Pythagorean
Triple Equivalently, a Pythagorean triple is a nontrivial solution to x2 + y 2 = z 2 .
Now we expand our deﬁnition of gcd. Deﬁnition
Greatest
Common
Divisor Let a, b and c be integers, not all zero. An integer d > 0 is the greatest common
divisor of a, b and c, written gcd(a, b, c), if and only if
1. d  a, d  b and d  c (this captures the common part of the deﬁnition), and
2. if e  a and e  b and e  c then e ≤ d (this captures the greatest part of the
deﬁnition). Deﬁnition A Pythagorean triple is said to be primitive if gcd(x, y, z ) = 1. Primitive
Triple Example 3 Both (6, 8, 10) and (3, 4, 5) are Pythagorean triples. However
• (6, 8, 10) is not a primitive Pythagorean triple since gcd(6, 8, 10) = 2 = 1.
• (3, 4, 5) is a primitive Pythagorean triple since gcd(6, 8, 10) = 1. We leave the proof of the following very useful lemma as an exercise. Lemma 13 (Extending Coprimeness (EC))
If x, y and z are integers, not all zero, and gcd(x, y ) = 1, then gcd(x, y, z ) = 1. Proposition 14 (Multiples of Pythagorean Triples (MPT))
Let d = gcd(x, y, z ). The three integers x, y and z are a Pythagorean triple if and
x
y
z
only if the three integers x1 = , y1 = and z1 = are a Pythagorean triple.
d
d
d This is a simple “if and only if” proof that can be proved using a chain of “if and
only if” statements. 156 Chapter 24 Introduction to Fermat’s Last Theorem Proof:
x, y and z are a Pythagorean triple
⇔ x2 + y 2 = z 2
z2
x2 y 2
+ 2= 2
d2
d
d
2
2
⇔ x 2 + y1 = z1
1
⇔ ⇔ x1 , y1 and z1 are a Pythagorean triple Take ten minutes to prove the following proposition and then compare your proof
with the proof that follows. Proposition 15 (Relative Primeness of Pythagorean Triples (RPPT))
If x, y and z are a primitive Pythagorean triple,
then gcd(x, y ) = gcd(x, z ) = gcd(y, z ) = 1. Proof: We will show that gcd(x, y ) = 1. The other pairs are similar. Suppose to the
contrary that gcd(x, y ) = d > 1. Then there exists a prime p so that p  d. Since
p  x and p  y , p  y 2 , p  (x2 + y 2 ) by the Divisibility of Integer Combinations. Since
x2 + y 2 = z 2 , p  z 2 and so p  z by Primes and Divisibility. But then gcd(x, y, z ) ≥
p > 1 which contradicts the hypothesis that x, y and z are a primitive Pythagorean
triple.
Let us walk through a proof of the following proposition. Proposition 16 (Parity of Pythagorean Triples (PPT))
If x, y and z are a primitive Pythagorean triple, then one of the integers x or y is
even and the other is odd. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. We will proceed by contradiction using two cases: x and y are both even, and
x and y are both odd.
2. Consider the ﬁrst case. Suppose that x and y are both even.
3. But then 2  x and 2  y so 2  (x2 + y 2 ) by our proposition on the Divisibility
of Integer Combinations. Since x2 + y 2 = z 2 , that means 2  z 2 and so 2  z .
But then gcd(x, y, z ) ≥ 2, contradicting the hypothesis that x, y and z are a
primitive Pythagorean triple.
4. Consider the second case. Suppose that x and y are both odd.
5. This implies that x2 ≡ 1 (mod 4) and y 2 ≡ 1 (mod 4) which in turn implies
z 2 = x2 + y 2 ≡ 2 (mod 4) Section 24.3 Pythagorean Triples 157 6. But this is impossible since the square of any integer can only be congruent to
0 or 1 modulo 4.
7. Since the two integers cannot both be even or odd, exactly one must be even
and one must be odd. As usual, we will begin our analysis by identifying the hypothesis, conclusion, core
proof techniques and preliminary material.
Hypothesis: x, y and z are a primitive Pythagorean triple.
Conclusion: One of the integers x or y is even and the other is odd.
Core Proof Technique: Proof by Elimination with two cases being eliminated: x
and y both even and x and y both odd. Each case is dealt with using contradiction. The use of contradiction several times within a proof is common.
Preliminary Material: primitive Pythagorean triple, congruences
Let’s examine each collection of sentences.
Sentence 1. We will proceed by contradiction using two cases: x and y are both
even, and x and y are both odd.
The author indicates the plan of the proof, always a good idea. There are three
possibilities: x and y have opposite parity, x and y are both even or x and y
are both odd. The author will disprove the latter two cases using contradiction,
hence by elimination leaving only opposite parity.
Sentence 2. Consider the ﬁrst case. Suppose that x and y are both even.
This sentence begins the ﬁrst of the two embedded proofs by contradiction.
Sentence 3. But then 2  x and 2  y so 2  (x2 + y 2 ) by our proposition on the
Divisibility of Integer Combinations. Since x2 + y 2 = z 2 , that means 2  z 2 and
so 2  z . But then gcd(x, y, z ) ≥ 2, contradicting the hypothesis that x, y and z
are a primitive Pythagorean triple.
This argument is exactly the same as in the proof of the Relative Primeness of
Pythagorean Triples.
Sentence 4. Consider the second case. Suppose that x and y are both odd.
This sentence begins the second of the two embedded proofs by contradiction.
Sentence 5. This implies that x2 ≡ 1 (mod 4) and y 2 ≡ 1 (mod 4) which in turn
implies
z 2 = x2 + y 2 ≡ 2 (mod 4)
Sentence 6. But this is impossible since the square of any integer can only be congruent to 0 or 1 modulo 4.
This part of proof is quite diﬀerent from the earlier part. Since any odd integer
a can be written in the form 2t + 1, a2 has the form 4t2 + 4t + 1 which is
congruent to 1 (mod 4). Thus z 2 = x2 + y 2 ≡ 2 (mod 4). But how could this
be? If z were odd z 2 ≡ 1 (mod 4) and if If z were even z 2 ≡ 0 (mod 4). 158 Chapter 24 Introduction to Fermat’s Last Theorem Sentence 7. Since the two integers cannot both be even or odd, exactly one must be
even and one must be odd.
Since the cases of x and y both even or x and y both odd have been eliminated,
all that remains is that x and y have opposite parity. REMARK
If x, y , z is a Pythagorean triple, we will assume as a convention that x is even and
y is odd. Notice that this implies that z is odd. (Why?) We conclude with a small proposition that is very useful. The proof appears in the
Appendix. Proposition 17 (Decomposing nth Powers (DNP))
If ab = cn and gcd(a, b) = 1, then there exist integers a1 and b1 so that a = an and
1
b = bn .
1 Example 4 Consider 592, 704 which is just 843 . With n = 3, c = 84, a = 64 and b = 9261, the
hypotheses of the proposition are satisﬁed. Hence, there exist integers a1 and b1 so
that a = 64 = an and b = 9261 = bn . So a1 = 4 and b1 = 21, and 43 213 = 843 .
1
1
Notice that our choice of a and b satisﬁed gcd(a, b) = 1. With a = 32 and b = 18522,
even though ab = cn is still true, the proposition does not apply since gcd(a, b) = 1 Chapter 25 Characterization of Pythagorean
Triples
25.1 Objectives The content objectives are:
1. State and prove the Characterization of Pythagorean Triples theorem.
2. Illustrate the theorem. 25.2 Pythagorean Triples We are now able to characterize all nontrivial, primitive Pythagorean triples. The
proof in this section follows that done by David Burton in Elementary Number Theory,
Seventh Edition. Theorem 18 (Characterization of Pythagorean Triples (CPT))
The complete set of nontrivial, primitive solutions to
x2 + y 2 = z 2
is given by
x = 2st
y = s2 − t2
z = s2 + t2
for integers s > t > 0 such that gcd(s, t) = 1 and s ≡ t (mod 2). Let’s understand what the theorem is saying. Every choice of s and t satisfying
integers s > t > 0 such that gcd(s, t) = 1 and s ≡ t (mod 2) should produce a
Pythagorean triple and these are the only nontrivial, primitive Pythagorean triples.
159 160 Chapter 25 Characterization of Pythagorean Triples The table below lists some primitive Pythagorean triples arising from small values of
s and t.
s t 2
3
4
4
5
5 1
2
1
3
2
4 x
2st
4
12
8
24
20
40 y
s2 − t2
3
5
15
7
21
9 z
s2 + t2
5
13
17
25
29
41 Before we read the proof, let’s do some analysis. The expression “complete solution”
should indicate to us that we are working with sets. So, the ﬁrst step is to identify
which sets are used and what their relationship is.
One set is the collection of nontrivial, primitive Pythagorean triples and can be
deﬁned by
S = {x, y, z ∈ N  x2 + y 2 = z 2 , gcd(x, y, z ) = 1, 2  x}
Note that the use of N is equivalent to nontrivial, that gcd(x, y, z ) = 1 is equivalent
to primitive and 2  x follows our convention that in a primitive Pythagorean triple,
x is even and y and z are odd. The other set is the collection of triples determined
by formula and can be deﬁned by
T = {x, y, z ∈ N 
s, t ∈ N, x = 2st, y = s2 − t2 , z = s2 + t2 , s > t, gcd(s, t) = 1, s ≡ t (mod 2)}
The Characterization of Pythagorean Triples theorem asserts that S = T . We would
expect the proof to show that S = T by showing that S ⊆ T and T ⊆ S , though this
is done implicitly.
Here is the proof. Be sure to identify
1. where S and T appear in the proof,
2. where each of the elements that deﬁne set membership are satisﬁed,
3. where each of the elements that deﬁne set membership are used.
Question Panel walk through of proof here.
Proof: Let x, y and z be a nontrivial, primitive Pythagorean triple. Since x is even
and y and z are odd, z − y and z + y are even. Suppose z − y = 2u and z + y = 2v .
Then
z−y
z+y
u=
and v =
2
2
and
v − u = y and u + v = z
and the equation x2 + y 2 = z 2 may be rewritten as
x2 = z 2 − y 2 = (z − y )(z + y ) Section 25.2 Pythagorean Triples 161 Dividing the preceding equation by 4 gives
x
2 2 = z−y
2 z+y
2 = uv We claim that u and v are relatively prime. Suppose they were not and gcd(u, v ) =
d > 1. Then d  (v − u) and d  (u + v ). But v − u = y and u + v = z so d  y and
d  z which contradicts the fact that y and z are relatively prime. Now we can use
our proposition on Decomposing nth Powers to conclude that u and v are perfect
squares. Hence, for some positive integers s and t
u = t2
v = s2
Using these values of u and v produces
z = v + u = s2 + t2
y = v − u = s2 − t2
x2 = 4vu = 4s2 t2 ⇒ x = 2st
We claim that gcd(s, t) = 1. If d > 1 were a common factor of s and t, d would be a
common factor of y and z contradicting the fact that gcd(y, z ) = 1. Finally, if s and
t are both even or both odd, then y and z are even, a contradiction. Hence, exactly
one of s and t is odd, the other is even. Symbolically, s ≡ t (mod 2).
Conversely, let the natural numbers s and t satisfy s > t, gcd(s, t) = 1, s ≡ t (mod 2).
Using the provided formulas for x, y and z we have
x2 + y 2 = (2st)2 + (s2 − t2 )2 = (s2 + t2 )2 = z 2
so x, y and z are a Pythagorean triple. To see that the triple is nontrivial, we must
show that x, y and z are all positive. Since s, t > 0, x = 2st > 0 and z = s2 + t2 > 0.
Since s > t, y = s2 − t2 > 0. To see that the triple is primitive, assume that
gcd(x, y, z ) = d > 1 and let p be any prime divisor of d. Since one of s and t is odd
and the other is even, z is odd. Since p  z , p = 2. From p  y and p  z , we know that
p  (z + y ) and since z + y = 2s2 , p  2s2 . Hence, p  s. Similarly, p  t. But then p
is a common factor of s and t contradicting gcd(s, t) = 1. Since no such p can exist,
gcd(x, y, z ) = 1 and x, y and z are a primitive triple. Chapter 26 Fermat’s Theorem for n = 4
26.1 Objectives The content objectives are
1. State and prove: The Diophantine equation x4 + y 4 = z 2 has no nontrivial
solution.
2. State and prove: The Diophantine equation x4 + y 4 = z 4 has no nontrivial
solution.
3. Show a reduction of FLT to If p is an odd prime, then the Diophantine equation
xp + y p = z p has no nontrivial solution. 26.2 n=4 Having completely resolved the case of Pythagorean triples, we can now turn our
attention to the one instance of FLT proved by Fermat. Actually, we will prove a
slightly stronger result and the case n = 4 will follow as a corollary. The approach
in this section mostly follows Elementary Number Theory, Seventh Edition by David
Burton. Theorem 19 (FLT, Strong Version of n = 4)
The Diophantine equation x4 + y 4 = z 2 has no nontrivial solution. The proof is demanding but it has a straightforward structure.
1. This is a proof by contradiction. It assumes the existence of a “minimal” solution
x0 , y0 , z0 to x4 + y 4 = z 2 .
2. Using x0 , y0 , z0 the author constructs a nontrivial primitive Pythagorean triple.
3. Using the Characterization of Pythagorean Triples the author ﬁnds various algebraic expressions involving s and t.
162 Section 26.2 n=4 163
4. The author uses these algebraic expressions to construct another nontrivial
primitive Pythagorean triple.
5. Lastly, the author uses this triple to construct a solution x1 , y1 , z1 to x4 +y 4 = z 2
which is “smaller” than x0 , y0 , z0 , hence a contradiction.
Extensive Question Panel here.
Proof: By way of contradiction, suppose there exists a positive integer solution to
x4 + y 4 = z 2 . Of all such solutions, choose any one in which z is smallest. Call this
solution x0 , y0 , z0 . Without loss of generality, we may also assume that gcd(x0 , y0 ) =
1. (Why?) This in turn implies that gcd(x0 , y0 , z0 ) = 1. (Why?)
Since x0 , y0 , z0 is a solution we know
4
2
x4 + y0 = z0
0 which we can rewrite as
x2
0 2 2
+ y0 2 2
= z0 2
But that means that x2 , y0 and z0 are nontrivial primitive solutions of a2 + b2 = c2
0
so we can make use of the Characterization of Pythagorean Triples. In particular, we
2
know that one of x2 and y0 is even. We can assume that x2 is even, hence x0 is even,
0
0
and that there exist integers s and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t
(mod 2) satisfying x2 = 2st
0
2
y0 = s 2 − t 2 z0 = s 2 + t 2
Since s ≡ t (mod 2), exactly one of s and t are even. Suppose s is even and t is odd.
2
2
Now consider the equation y0 = s2 − t2 modulo 4. Because y0 is odd,
2
y0 = s2 − t2 ⇒ 1 ≡ 0 − 1 (mod 4) ⇒ 1 ≡ 3 (mod 4) which is impossible. Therefore s is odd and t is even so we write t = 2r. Then
x2 = 2st ⇒ x2 = 4sr ⇒
0
0 x0
2 2 = sr Now gcd(s, t) = 1 implies that gcd(s, r) = 1 (why?) and so we can use the proposition
on Decomposing nth Powers. Since (x0 /2)2 is a perfect square, s and r must be
2
2
perfect squares and we can write s = z1 and r = w1 for positive integers z1 and w1 .
2
Rewrite y0 = s2 − t2 as 2
t2 + y0 = s2 Because gcd(s, t) = 1 implies gcd(s, t, y0 ) = 1, the triple t, y0 , s is a primitive Pythagorean
triple and we can use the Characterization of Pythagorean Triples again. With t even,
the Characterization of Pythagorean Triples assures us of the existence of integers u
and v so that u > v > 0 and gcd(u, v ) = 1 and u ≡ v (mod 2) satisfying
t = 2uv
y0 = y 2 − v 2
s = u2 + v 2 164 Chapter 26 Fermat’s Theorem for n = 4 Now, observe that
t
2
= r = w1
2
and so by the proposition on Decomposing nth Powers, u and v are perfect squares.
2
Suppose u = x2 and v = y1 where x1 and y1 are positive integers. But then
1
uv = 2
s = u2 + v 2 ⇒ z1 = x2
1 2 2
+ y1 2 and so
2
4
z1 = x4 + y1
1 That is, x1 , y1 , z1 is a solution to x4 + y 4 = z 2 . Since z1 and t are positive
2
0 < z1 ≤ z1 = s ≤ s2 < s2 + t2 = z0 That is,
z1 < z0
But recall that x0 , y0 , z0 is a solution to x4 + y 4 = z 2 with the smallest possible value
of z . But x1 , y1 , z1 is a solution to x4 + y 4 = z 2 with a smaller value of z !
The case n = 4 of Fermat’s Last Theorem follows immediately. Corollary 20 The Diophantine equation x4 + y 4 = z 4 has no positive integer solution. Proof: If x0 , y0 , z0 were a positive integer solution of x4 + y 4 = z 4 , then x0 , y0 ,
2
z0 would be a positive integer solution to x4 + y 4 = z 2 , contradicting the previous
theorem. 26.3 Reducing the Problem It is not necessary to consider every exponent of xn + y n = z n to prove Fermat’s Last
Theorem.
If n > 2, then n is either a power of 2 or divisible by an odd prime p. In the ﬁrst
case, n = 4k for some k ≥ 1 and the equation xn + y n = z n can be rewritten as
xk 4 + yk 4 = zk 4 We have just seen that this equation has no positive integer solution.
In the second case, n = pk for some k ≥ 1 and the equation xn + y n = z n can be
rewritten as
p
p
p
xk + y k = z k
If it could be shown that up + v p = wp has no solution, then there would be no
solution of the form u = xk , v = y k , w = z k and so there would be no solution to
xn + y n = z n . Therefore, Fermat’s Last Theorem reduces to Section 26.4 History Theorem 21 165 (Fermat’s Last Theorem – Reduced)
If p is an odd prime, then the Diophantine equation
xp + y p = z p
has no nontrivial solutions. 26.4 History Awaiting copyright permission. Chapter 27 Problems Related to FLT
27.1 Objectives 1. Read a proof of Squares From the Diﬀerence of Quartics
2. Read a proof of a proposition on the area of Pythagorean triangles. 27.2 x4 − y 4 = z 2 From x4 + y 4 = z 2 , we turn to a closely related Diophantine equation, x4 − y 4 = z 2 .
Our proof is very similar to that of the Strong Version of FLT for n = 4. The approach
in this section mostly follows Elementary Number Theory, Seventh Edition by David
Burton. Proposition 22 (Squares From the Diﬀerence of Quartics (SFDQ))
The Diophantine equation x4 − y 4 = z 2 has no nontrivial solution. Proof: Suppose that there exists a nontrivial solution to x4 − y 4 = z 2 . Of all such
solutions x0 , y0 , z0 , choose any one in which x0 is smallest. Choosing x0 as small as
possible forces x0 to be odd. (Why?)
We now show that we can also assume that gcd(x0 , y0 ) = 1. Suppose gcd(x0 , y0 ) =
d > 1. Then writing dx1 = x0 and dy1 = y0 and substituting into x4 − y 4 = z 2 we
4
2
2
get d4 (x4 − y1 ) = z0 . So d4  z0 , hence d2  z0 . Thus z0 = d2 z1 for some integer z1 .
1
But then
4
2
d4 x4 − d4 y1 = d4 z1
1
so x1 , y1 , z1 is a nontrivial solution to x4 − y 4 = z 2 with 0 < x1 < x0 contradicting
our choice of a minimal x0 .
4
2
If the equation x4 − y0 = z0 is written in the form
0 x2
0 2 2
− y0 then
2
2
z0 + y0 2 2 2
= z0 = x2
0 166 2 Section 27.2 x4 − y 4 = z 2 167 2
and we see that z0 , y0 , x2 constitute a primitive Pythagorean triple.
0 From here there are two cases: y0 odd and y0 even. Consider the case where y0 is
odd. The Characterization of Pythagorean Triples asserts that there exist integers s
and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t (mod 2) satisfying
z0 = 2st (this is forced from y0 odd) 2
y0 = s2 − t2 x2 = s2 + t2
0
Observe that
2
s4 − t4 = (s2 + t2 )(s2 − t2 ) = x2 y0 = (x0 y0 )2
0 so s, t, x0 y0 is a positive solution to x4 − y 4 = z 2 . But
0<s< s2 + t2 = x0 which contradicts the minimality of x0 so y0 cannot be odd.
Now consider the case where y0 is even. The Characterization of Pythagorean Triples
asserts that there exist integers s and t so that s > t > 0 and gcd(s, t) = 1 and s ≡ t
(mod 2) satisfying
2
y0 = 2st (this is forced from y0 even) z0 = s 2 − t 2
x2 = s 2 + t 2
0
Because of the symmetry of expressions for s and t, we may assume that s is even
and t is odd. Consider the relation
2
y0 = 2st Since gcd(s, t) = 1 and s is even, we know that gcd(2s, t) = 1. This allows us to invoke
the proposition on Decomposing nth Powers. That is, 2s and t are each squares of
positive integers, say 2s = w2 and t = v 2 . Because w must be even, set w = 2u to
get s = 2u2 . Therefore
x2 = s2 + t2 = 4u4 + v 4
0
and so 2u2 , v 2 , x0 form a Pythagorean triple. Since gcd(2u2 , v 2 ) = gcd(s, t) = 1,
gcd(2u2 , v 2 , x0 ) = 1 and so the Pythagorean triple is primitive. The Characterization
of Pythagorean Triples asserts that there exist integers a and b so that a > b > 0 and
gcd(a, b) = 1 and a ≡ b (mod 2) satisfying
2u2 = 2ab
v 2 = a2 − b2
x0 = a2 + b2
Now 2u2 = 2ab implies u2 = ab which implies, by the proposition on Decomposing
nth Powers, that a and b are perfect squares. Say a = c2 and b = d2 . And here we
use a pattern we have seen before. Since
v 2 = a2 − b2 = c4 − d4 168 Chapter 27 Problems Related to FLT c, d, v is a positive integer solution to x4 − y 4 = z 2 . But
√
0 < c = a < a2 + b2 = x0
contradicting the minimality of x0 so y0 cannot be even.
Since the integer y0 cannot be either odd r even, it must be the case that our assumption that there is a nontrivial solution is incorrect.
This proposition has an unexpected use in a statement about the areas of Pythagorean
triangles. Deﬁnition A Pythagorean triangle is a right triangle whose sides are of integral length. Pythagorean
Triangle In the margin of his copy of Diophantus’ Arithmetica, Fermat stated and proved a
proposition equivalent to the following. Proposition 23 The area of a Pythagorean triangle can never be equal to a perfect square. Here, perfect square means the square of an integer.
Proof: We will proceed by contradiction. Consider a Pythagorean triangle ABC
where the hypotenuse has length z and the other two sides have lengths x and y , so
that
x2 + y 2 = z 2
(27.1)
The area of
This gives ABC is (1/2)xy and if this were a square we could write (1/2)xy = u2 .
2xy = 4u2 (27.2) Now Equation (27.1) plus Equation (27.2) gives
x2 + y 2 + 2xy = z 2 + 4u2 ⇒ (x + y )2 = z 2 + 4u2
and Equation (27.1) minus Equation (27.2) gives
x2 + y 2 − 2xy = z 2 − 4u2 ⇒ (x − y )2 = z 2 − 4u2
Now multiply these last two equations together to get
(x + y )2 (x − y )2 = (z 2 + 4u2 )(z 2 − 4u2 ) ⇒ x2 − y 2
or
x2 − y 2 2 2 = z 4 − 16u4 = z 4 − (2u)4 But we know by our proposition on the Squares From the Diﬀerence of Quartics that
no nontrivial solution to this equation is possible, hence a contradiction. Chapter 28 Practice, Practice, Practice:
Prime Numbers
28.1 Objectives This class provides an opportunity to practice working with primes. 28.2 Exercises 169 Chapter 29 Complex Numbers
29.1 Objectives The content objectives are :
1. N ⊂ Z ⊂ Q ⊂ R ⊂ C
2. Deﬁne: complex number, C, real part, imaginary part
3. Operations: complex addition, complex multiplication, equality of complex numbers
4. State and prove properties of complex numbers. 29.2 Diﬀerent Equations Require Diﬀerent Number Systems When humans ﬁrst counted, we tallied. We literally made notches on sticks, stones
and bones. Thus the natural numbers, N, were born. But it wouldn’t be long before
the necessity of fractions became obvious. One animal to be shared by four people
(we will assume uniformly) meant that we had to develop the notion of 1/4. Though
it would not have been expressed this way, the equation
4x = 1
does not have a solution in N and so we would have had to extend our notion of
numbers to include fractions, the rationals.
Q= a
 a, b ∈ Z, b = 0
b This is an overstatement historically, because recognition of zero and negative numbers which are permitted in Q were very slow to come. But even these new numbers
would not help solve equations of the form
x2 = 2 170 Section 29.3 Complex Numbers 171 which would arise naturally from isosceles right angled triangles. For this, the notion
of number had to be extended to include irrational numbers, which combined with
the rationals, give us the real numbers.
Eventually, via Hindu and Islamic scholars, western mathematics began to recognize
and accept both zero and negative numbers. Otherwise, equations like
3x = 5x
or
2x + 4 = 0
have no solution. Thus, mathematicians recognized that
N⊂Z⊂Q⊂R
but even R was inadequate because equations of the form
x2 + 1 = 0
had no real solutions.
And so, our number system was extended again. 29.3
Deﬁnition
Complex
Number Complex Numbers A complex number z in standard form is an expression of the form x + yi where
x, y ∈ R. The set of all complex numbers is denoted by
C = {x + yi  x, y ∈ R} Example 5 Some examples are 3 + 4i, 0 + 5i (usually written 5i), 7 − 0i (usually written 7) and
0 + 0i (usually written 0). Deﬁnition For a complex number z = x + yi, the real number x is called the real part and
is written (z ) and the real number y is called the imaginary part and is written
(z ). Real Part,
Imaginary
Part So (3 + 4i) = 3 and (3 + 4i) = 4. If z is a complex number where (z ) = 0, we will
treat z as a real number and we will not write the term containing i. For example,
z = 3 + 0i will be treated as a real number and will be written z = 3. Thus
R⊂C
and so
N⊂Z⊂Q⊂R⊂C
One has to wonder how much further the number system needs to be extended! 172 Chapter 29 Deﬁnition
Equality Deﬁnition Complex Numbers The complex numbers z = x + yi and w = u + vi are equal if and only if x = u and
y = v. Addition is deﬁned as Addition (a + bi) + (c + di) = (a + c) + (b + d)i Example 6
(1 + 7i) + (2 − 3i) = (1 + 2) + (7 − 3)i = 3 + 4i Deﬁnition Multiplication is deﬁned as Multiplication (a + bi) · (c + di) = (ac − bd) + (ad + cb)i Example 7
(1 + 7i) · (2 − 3i) = (1 · 2 − 7 · (−3)) + (1 · (−3) + 7 · 2)i = 23 + 11i
The multiplication symbol is usually omitted and we write zw or (a + bi)(c + di). Exercise 5 Let u = 3 + i and v = 2 − 7i. Compute
1. u + v
2. u − v
3. uv
4. u2 v
5. u3
6. Exercise 6 v
u (write the solution in the form x + yi where x, y ∈ R) Compute
1. i4k for any nonnegative integer k
2. i4k+1 for any nonnegative integer k
3. i4k+2 for any nonnegative integer k
4. i4k+3 for any nonnegative integer k Section 29.3 Complex Numbers 173 The usual properties of associativity, commutativity, identities, inverses and distributivity that we associate with rational and real numbers also apply to complex numbers. Proposition 24 Let u, v, z ∈ C. Then
1. Associativity of addition: (u + v ) + z = u + (v + z )
2. Commutativity of addition: u + v = v + u
3. Additive identity: 0 = 0 + 0i has the property that z + 0 = z
4. Additive inverses: If z = x + yi then there exists an additive inverse of z , written
−z with the property that z + (−z ) = 0. The additive inverse of z = x + yi is
−z = −x − yi.
5. Associativity of multiplication: (u · v ) · z = u · (v · z )
6. Commutativity of multiplication: u · v = v · u
7. Multiplicative identity: 1 = 1 + 0i has the property that z · 1 = z whenever
z = 0.
8. Multiplicative inverses: If z = x + yi = 0 then there exists a multiplicative
inverse of z , written z −1 , with the property that z · z −1 = 1. The multiplicative
inverse of z = x + yi is z −1 = xx−yi2 .
2 +y
9. Distributivity: z · (u + v ) = z · u + z · v We will only prove the eighth property.
Proof: We only need to demonstrate that z −1 = xx−yi2 is welldeﬁned and that
2 +y
−1 = 1. Since z = 0, x2 + y 2 = 0 and so z −1 is welldeﬁned. Now we simply use
z·z
complex arithmetic.
x + yi · x − yi
x2 + xy − xy − y 2 i2
x2 + y 2
=
=2
=1
x2 + y 2
x2 + y 2
x + y2 Chapter 30 Properties Of Complex Numbers
30.1 Objectives The content objectives are:
1. Deﬁne conjugate and modulus
2. State and prove several properties of complex numbers. 30.2
Deﬁnition Conjugate The complex conjugate of z = x + yi is the complex number Conjugate z = x − yi
The conjugate of z = 2 + 3i is z = 2 − 3i. Proposition 25 (Properties of Conjugates (PCJ))
If z and w are complex numbers, then
1. z + w = z + w
2. zw = z w
3. z = z
4. z + z = 2 (z )
5. z − z = 2i (z ) Exercise 7 Prove each part of the Properties of Conjugates proposition.
(Hint: begin with “Let z = x + yi and w = u + vi.”) 174 Section 30.3 Modulus 175 Exercise 8 Prove: Let z ∈ C. The complex number z is a real number if and only if z = z . Exercise 9 Prove: Let z ∈ C and z = 0. The complex number z is purely imaginary ( (z ) = 0)
if and only if z = −z . Example 8 For z = i deﬁne
z+i
z−i
Prove that w is a real number if and only if z is zero or purely imaginary.
w= Solution:
W is real ⇐⇒ w = w
z+i
z−i
⇐⇒
=
z−i
z+i
⇐⇒ z z − 1 + (z + z )i = z z − 1 − (z + z )i
⇐⇒ z + z = 0
⇐⇒ (z ) = 0 ⇐⇒ z is zero or purely imaginary 30.3
Deﬁnition The modulus of the complex number z = x + yi is the nonnegative real number Modulus Example 9 Modulus z  = x + yi = The modulus of z = 2 − 5i is z  = x2 + y 2 (22 ) + (−5)2 = √ 29. Given two real numbers, say x1 and x2 , we can write either x1 ≤ x2 or x2 ≤ x1 .
However, given two complex numbers, z1 and z2 , we cannot meaningfully write z1 ≤
z2 or z2 ≤ z1 . But since the modulus of a complex number is a real number, we
can meaningfully write z1  ≤ z2 . The modulus gives us a means to compare the
magnitude of two complex numbers, but not compare the numbers themselves.
If Proposition 26 (z ) = 0, then the modulus corresponds to the absolute values of real numbers. (Properties of Modulus (PM))
If z and w are complex numbers, then
1. z  = 0 if and only if z = 0 176 Chapter 30 Properties Of Complex Numbers 2. z  = z 
3. zz = z 2
4. zw = z w
5. z + w ≤ z  + w. This is the triangle inequality. Exercise 10 Prove each of the parts of the Properties of Modulus proposition. Chapter 31 Graphical Representations of
Complex Numbers
31.1 Objectives The content objectives are:
1. Deﬁne complex plane, polar coordinates, polar form.
2. Convert between Cartesian and polar form.
3. Multiplication in polar form. 31.2
31.2.1
Deﬁnition
Complex
Plane The Complex Plane
(x, y ) The notation z = x + yi suggests a nonalgebraic representation. Each complex
number z = x + yi can be thought of as a point (x, y ) in a plane with orthogonal
axes. Label one axis the real axis and the other axis the imaginary axis. The
complex number z = x + yi then corresponds to the point (x, y ) in the plane. This
interpretation of the plane is called the complex plane or the Argand plane. 177 178 Chapter 31 Graphical Representations of Complex Numbers Figure 31.2.1: The Complex Plane Exercise 1 Plot the following points in the complex plane.
1. 4 + i
2. −2 + 3i
3. −2 − i 31.2.2 Modulus Recall that the modulus of the complex number z = x + yi is the nonnegative real
number
z  = x + yi = x2 + y 2
There are a couple of geometric points to note about the modulus of z = x + yi.
The Pythagorean Theorem is enough to prove that z  is the distance from the origin
to z in the complex plane, and that the distance between z and w = u + vi is just
z − w = (x − u)2 + (y − v )2 . Exercise 1 Sketch all of the points in the complex plane with modulus 1. 31.3 Polar Representation There is another way to represent points in a plane which is very useful when working
with complex numbers. Instead of beginning with the origin and two orthogonal axes,
we begin with the origin O and a polar axis which is a ray leaving from the origin.
The point P (r, θ) is plotted so that the distance OP is r, and the counter clockwise
angle of rotation from the polar axis, measured in radians, is θ.
Note that this allows for multiple representations since (r, θ) identiﬁes the same point
as (r, θ + 2πk ) for any integer k . Section 31.4 Converting Between Representations 179 Figure 31.3.1: Polar Representation The obvious question is how to go from one to the other. 31.4 Converting Between Representations Simple trigonometry allows us to convert between polar and Cartesian coordinates. Figure 31.4.1: Connecting Polar and Cartesian Representations Given the polar coordinates (r, θ), the corresponding Cartesian coordinates (x, y ) are
x = r cos θ
y = r sin θ
Given the Cartesian coordinates (x, y ), the corresponding polar coordinates are determined by
r=
x
r
y
sin θ =
r cos θ = x2 + y 2 180 Chapter 31 Exercise 2 Graphical Representations of Complex Numbers For each of the following polar coordinates, plot the point and convert to Cartesian
coordinates.
1. (1, 0)
2. (2, π/2)
3. (3, π )
4. (2, 7π/2)
5. (4, π/4)
6. (4, 4π/3) Exercise 3 For each of the following Cartesian coordinates, plot the point and convert to polar
coordinates.
1. (1, 0)
2. (0, 1)
3. (−1, 0)
4. (0, −1)
5. (1, 1)
6. (−1, 1)
√
7. (2, −2 3) From our earlier description of conversions, we can write the complex number
z = x + yi
as
z = r cos θ + ri sin θ = r(cos θ + i sin θ) Deﬁnition The polar form of a complex number z is Polar Form z = r(cos θ + i sin θ)
where r is the modulus of z and the angle θ is called an argument of z . The
expression cos θ + i sin θ is frequently abbreviated to cis θ and so we write z = rcis θ. Example 1 The following are representations of complex numbers in both Cartesian and polar
form. Section 31.4 Converting Between Representations 181 1. 1 = cis 0
√
2. −1 + i = 2cis 3π/4
√
3. −1 − 3i = 2cis 4π/3 One of the advantages of polar representation is that multiplication becomes very
straightforward. Proposition 1 (Polar Multiplication of Complex Numbers (PMCN))
If z1 = r1 (cos θ1 + i sin θ1 ) and z2 = r2 (cos θ2 + i sin θ2 ) are two complex numbers in
polar form, then
z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) Example 2
√
√
( 2cis 3π/4) · (2cis 4π/3) = 2 2cis 3π 4π
+
4
3 √
= 2 2cis 25π
12 √
= 2 2cis Proof:
z1 z2 = r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 )
= r1 r2 ((cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(cos θ1 sin θ2 + sin θ1 cos θ2 ))
= r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )) π
12 Chapter 32 De Moivre’s Theorem
32.1 Objectives The content objectives are:
1. State and prove De Moivre’s Theorem and do examples.
2. Derive Euler’s Formula. 32.2
Theorem 2 De Movre’s Theorem (De Movre’s Theorem (DMT))
If θ ∈ R and n ∈ Z, then
(cos θ + i sin θ)n = (cos nθ + i sin nθ) Example 3 √
√
Consider the complex number z = 1/ 2 + i/ 2 which, in polar form is z = cos π/4 +
i sin π/4. By De Moivre’s Theorem,
z 10 = (cos π/4 + i sin π/4)10 = cos 10π/4 + i sin 10π/4 = cos π/2 + i sin π/2 = i. Proof: We will prove DeMoivre’s Theorem using three cases.
1. n = 0
2. n > 0
3. n < 0 182 Section 32.3 De Movre’s Theorem 183 For the case n = 0, DeMoivre’s Theorem reduces to (cos θ + i sin θ)0 = (cos 0 + i sin 0).
By convention z 0 = 1 so the left hand side of the equation is 1. Since cos 0 = 1 and
sin 0 = 0, the right hand side also evaluates to 1.
For the case n > 0 we will use induction.
P (n): (cos θ + i sin θ)n = (cos nθ + i sin nθ).
Base Case We verify that P (1) is true where P (1) is the statement
P (1): (cos θ + i sin θ)n = (cos 1θ + i sin 1θ).
This is trivially true.
Inductive Hypothesis We assume that the statement P (k ) is true for some k ≥ 1.
P (k ): (cos θ + i sin θ)k = (cos kθ + i sin kθ).
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): (cos θ + i sin θ)k+1 = (cos(k + 1)θ + i sin(k + 1)θ)
(cos θ + i sin θ)k+1 = (cos θ + i sin θ)k (cos θ + i sin θ) by separating out one factor
= (cos kθ + i sin kθ)(cos θ + i sin θ) by the Inductive Hypothesis
= (cos(k + 1)θ + i sin(k + 1)θ) Polar Multiplication of Complex Numbers
Lastly, for the case n < 0 we will use complex arithmetic. Since n < 0, n = −m for
some m ∈ P.
(cos θ + i sin θ)n = (cos θ + i sin θ)−m
1
=
(cos θ + i sin θ)m
1
=
(cos mθ + i sin mθ)
= cos mθ − i sin mθ
= cos(−mθ) + i(sin(−mθ))
= cos nθ + i sin nθ Corollary 3 If z = r(cos θ + i sin θ) and n is an integer,
z n = rn (cos nθ + i sin nθ) 184 Chapter 32 32.3 De Moivre’s Theorem Complex Exponentials If you were asked to ﬁnd a realvalued function y with the property that
dy
= ky and y = 1 when x = 0
dx
for some constant k , you would choose
y = ekx
And if you were asked to ﬁnd the derivative of f (θ) = cos θ + i sin θ where i was
treated as any other constant you would almost certainly write
df (θ)
= − sin θ + i cos θ
dθ
but then and so Deﬁnition df (θ)
= − sin θ + i cos θ = i(cos θ + i sin θ) = if (θ)
dθ
df (θ)
= if (θ) and f (θ) = 1 when θ = 0
dθ By analogy, we deﬁne the complex exponential function by Complex
Exponential eiθ = cos θ + i sin θ As an exercise, prove the following. Proposition 4 (Properties of Complex Exponentials (PCE)) eiθ · eiφ = ei(θ+φ)
eiθ n = einθ The polar form of a complex number z can now be written as
z = reiθ
where r = z  and θ is an argument of z .
Out of this arises one of the most stunning formulas in mathematics. Setting r = 1
and θ = π we get
eiπ = cos π + i sin π = −1 + 0i = −1
That is
eiπ + 1 = 0
Who would have believed that e, i, π, 1 and 0 would have such a wonderful connection! Chapter 33 Roots of Complex Numbers
33.1 Objectives The content objectives are:
1. State and prove the Complex nthe Roots Theorem and do examples. 33.2
Deﬁnition Complex nth Roots If a is a complex number, then the complex numbers that solve Complex
Roots zn = a
are called the complex nth roots. De Moivre’s Theorem gives us a straightforward
way to ﬁnd complex nth roots of a. Theorem 5 (Complex nth Roots Theorem (CNRT))
If r(cos θ + i sin θ) is the polar form of a complex number a, then the solutions to
z n = a are
√
n
r cos θ + 2kπ
n + i sin θ + 2kπ
n for k = 0, 1, 2, . . . , n − 1 √
The modulus n r is the unique nonnegative nth root of r. This theorem shows that
any complex number, including the reals, has exactly n diﬀerent complex nth roots. Example 4 Find all the complex fourth roots of −16. Solution: We will use the Complex nth Roots Theorem. First, we write −16 in
polar form as
−16 = 16(cos π + i sin π )
185 186 Chapter 33 Roots of Complex Numbers Using the Complex nth Roots Theorem the solutions are
√
4 16 cos = 2 cos π + 2kπ
π + 2kπ
+ i sin
4
4
π kπ
π kπ
+ i sin
+
+
4
2
4
2 for k = 0, 1, 2, 3
for k = 0, 1, 2, 3 The four distinct roots are given below
π
π
+ i sin
4
4
3π
+ i sin
When k = 1, z1 = 2 cos
4
5π
+ i sin
When k = 2, z2 = 2 cos
4
7π
+ i sin
When k = 3, z3 = 2 cos
4
When k = 0, z0 = 2 cos =2
3π
4
5π
4
7π
4 i
1
√ +√
2
2
−1
=2 √ +
2
−1
=2 √ +
2
1
=2 √ +
2 =
i
√
2
−i
√
2
−i
√
2 √
√
2+i 2
√
√
=− 2+i 2
√
√
=− 2−i 2
= √ √
2−i 2 Graphing these solutions is illuminating. Figure 33.2.1: The Fourth Roots of 16 Note that the solutions are uniformly distributed around a circle whose radius is √
n r. Proof: As usual, when showing that a complete solution exists we work with two
sets: the set S of solutions and the set T of speciﬁc representation of the solution.
We then show that S = T by mutual inclusion. Our two sets are
S = {z ∈ C  z n = a} Section 33.3 Complex nth Roots 187 and
T= √
n r cos θ + 2kπ
n + i sin θ + 2kπ
n k = 0, 1, 2, . . . , n − 1 where a = r(cos θ + i sin θ).
We begin by showing that T ⊆ S . Let t = √
n r cos θ + 2kπ
n be an element of T . Now
θ + 2kπ n
n
= r(cos(θ + 2kπ ) + i sin(θ + 2kπ ))
√
n tn = r n cos = r(cos θ + i sin θ) De Moivre’s Theorem trigonometry =a
Hence, t is a solution of z n = a, that is, t ∈ S .
Now we show that S ⊆ T . Let w = s(cos φ + i sin φ) be an nth root of a. Since
a = r(cos θ + i sin θ) we have
wn = a
⇐⇒ (s(cos φ + i sin φ))n = r(cos θ + i sin θ)
⇐⇒ sn (cos nφ + i sin nφ) = r(cos θ + i sin θ) De Moivre’s Theorem Now two complex numbers in polar form are equal if and only if their moduli are
equal and their arguments diﬀer by an integer multiple of 2π . So
√
sn = r ⇒ s = n r
and θ + 2πk
n
where k ∈ Z. Hence, the nth roots of a are of the form
nφ − θ = 2πk ⇒ φ = √
n r cos θ + 2kπ
n + i sin θ + 2kπ
n for k ∈ Z But this is k ∈ Z, not k = 0, 1, 2, . . . , n − 1. Since w is an nth root of a, there exists
an integer k0 so that
w= √
n r cos θ + 2k0 π
n + i sin θ + 2k0 π
n √
n r cos θ + 2k1 π
n + i sin θ + 2k1 π
n If we can show that
w= if and only if k0 ≡ k1 (mod n) whenever r = 0, then w ∈ T . Now
k0 ≡ k1
⇐⇒ k0 − k1 = n
⇐⇒ 2πk0 − 2πk1 = 2πn
2πk0 2πk1
−
= 2π
⇐⇒
n
n
θ + 2πk0 θ + 2πk1
⇐⇒
−
= 2π
n
n (mod n)
for some ∈Z for some ∈Z for some ∈Z for some ∈Z 188 Chapter 33 33.3 Roots of Complex Numbers More Examples Exercise 4 An nth root of unity is a complex number that solves z n = 1. Find all of the sixth
roots of unity. Express them in standard form and graph them in the complex plane. Exercise 5 Find the square roots of −2i. Express them in standard form and graph them in the
complex plane. Chapter 34 An Introduction to Polynomials
34.1 Objectives The content objectives are:
1. Deﬁne polynomial, coeﬃcient, F[x], degree, zero polynomial, linear polynomial,
quadratic polynomial, cubic polynomial, equal, sum, diﬀerence, product, quotient,
remainder, divides and factor.
2. Deﬁne operations on polynomials.
3. State the Division Algorithm for Polynmials.
4. Do examples. 34.2 Polynomials Our number systems were developed in response to the need to ﬁnd solutions to real
polynomials. We are now able to solve all equations of the form
a2 x2 + a1 x + a0 = 0
or
xn − a0 = 0
whether the coeﬃcients are real or complex. In fact, a great deal more is known.
Let F be a ﬁeld. Roughly speaking, a ﬁeld is a set of numbers that allows addition,
subtraction, multiplication and division. The rational numbers Q, the real numbers
R, the complex numbers C and the integers modulo a prime p, Zp are all ﬁelds. The
integers are not a ﬁeld because we cannot divide 2 by 4 and get an integer. Since
division is just multiplication by an inverse, Z6 is not a ﬁeld since [3] has no inverse. 189 190 Chapter 34 Deﬁnition An Introduction to Polynomials A polynomial in x over F is an expression of the form Polynomial an xn an−1 xn−1 + · · · + a1 x + a0
where all of the ai belong to F.
The ai are called coeﬃcients. We use F[x] to denote the set of polynomials in x
with coeﬃcients from F. Example 5
1. x2 + 7x − 1 ∈ R[x]
2. x3 − 7ix + (5 − 2i) ∈ C[x]
3. [3]x5 + [2]x3 + [6] ∈ Z7 [x]
It is important to be clear about what ﬁeld the coeﬃcients come from. The polynomial
x2 +1 belongs to both R[x] and C[x] but the equation x2 +1 = 0 has complex solutions
but no real solutions. Deﬁnition If an = 0 in the polynomial Degree et al an xn an−1 xn−1 + · · · + a1 x + a0
then the polynomial is said to have degree n. The zero polynomial has all of its
coeﬃcients zero and its degree is not deﬁned. Polynomials of degree 1 are called linear
polynomials, of degree 2, quadratic polynomials, and of degree 3 cubic polynomials. 34.3 Operations on Polynomials We very frequently use f (x) to denote an element of F[x] and write
n f (x) = an xn an−1 xn−1 + · · · + a1 x + a0 = ai xi
i=0 Let f (x), g (x) ∈ F[x] where
n
n f (x) = an x an−1 x n−1 ai xi + · · · + a1 x + a0 =
i=0
n g (x) = bn xn bn−1 xn−1 + · · · + b1 x + b0 = bi xi
i=0 Deﬁnition
Equal The polynomials f (x) and g (x) are equal if and only if ai = bi for all i. Section 34.3 Operations on Polynomials 191 Polynomials can be added, subtracted and multiplied as algebraic expressions exactly
as you have done in high school. Deﬁnition The sum of the polynomials f (x) and g (x) is deﬁned as Sum max(n,m) (ai + bi )xi f (x) + g (x) =
i=0 where any “missing” terms have coeﬃcient zero. Example 6
1. In R[x], if f (x) = x2 + 7x − 1 and g (x) = 3x4 − x3 + 4x2 − x + 5 then
f (x) + g (x) = 3x4 − x3 + 5x2 + 6x + 4.
2. In C[x], if f (x) = x3 − 7ix + (5 − 2i) and g (x) = (4 + 3i)x + (7 + 7i)x then
f (x) + g (x) = x3 + (4 − 4i)x + (12 + 5i)x.
3. In Z7 [x], if f (x) = [3]x5 + [2]x3 + [6] and g (x) = [2]x4 + [5]x3 + [2]x2 + [4] then
f (x) + g (x) = [3]x5 + [2]x4 + [2]x2 + [3]. Deﬁnition The diﬀerence of the polynomials f (x) and g (x) is deﬁned as Diﬀerence max(n,m) (ai − bi )xi f (x) − g (x) =
i=0 where any “missing” terms have coeﬃcient zero. Exercise 6 Find the diﬀerence of each of the pairs of polynomials given in Example 6. The deﬁnition of the product of two polynomials looks more complicated than it is. Deﬁnition The product of the polynomials f (x) and g (x) is deﬁned as Product m+n ci xi f (x) · g (x) =
i=0 where i ci = a0 bi + a1 bi−1 + · · · + ai−1 b1 + ai b0 = aj bi−j
j =0 192 Chapter 34 Example 7 An Introduction to Polynomials In R[x], if f (x) = x2 + 7x − 1 and g (x) = 3x + 2 then
f (x) · g (x) = Add long multiplication example here.
Now we run into the same issue we had with the integers, division. Though it makes
sense to say that x − 3 divides x2 − 9 since x2 − 9 = (x − 3)(x + 3) what do we do
when there is a remainder? Just as we had a division algorithm for integers, we have
a division algorithm for polynomials. Proposition 6 (Division Algorithm for Polynomials (DAP))
If f (x) and g (x) are polynomials in F[x] and g (x) is not the zero polynomial, then
there exist unique polynomials q (x) and r(x) in F[x] such that
f (x) = q (x)g (x) + r(x) Deﬁnition
Quotient,
Remainder where deg r(x) < deg g (x) or r(x) = 0 The polynomial q (x) is called the quotient polynomial. The polynomial r(x) is
called the remainder polynomial. If r(x) = 0, we say that g (x) divides f (x) or
f (x) is a factor of f (x) and we write g (x)  f (x). Add long division examples over several ﬁelds here. Exercise 7 For each f (x) and g (x), ﬁnd the quotient and remainder polynomials.
1. Let f (x) and g (x) be the real polynomials f (x) = 2x4 + 6x3 − x + 4 and
g (x) = x2 + 3.
2. Let f (z ) and g (z ) be the complex polynomials
f (z ) = iz 3 + (2 + 4i)z 2 + (3 − i)z + (4 − 8i) and g (z ) = iz + (2 − 2i). Chapter 35 Factoring Polynomials
35.1 Objectives The content objectives are:
1. Deﬁne polynomial equation, solution and root.
2. State the Fundamental Theorem of Algebra 35.2
Deﬁnition
Polynomial
Equation Polynomial Equations A polynomial equation is an equation of the form
an xn an−1 xn−1 + · · · + a1 x + a0 = 0
which will often be written as f (x) = 0. An element c ∈ F is called a root or zero of
the polynomial f (x) if f (c) = 0. That is, c is a solution of the polynomial equation
f (x) = 0. The history of mathematics is replete with exciting and sometimes bizarre stories of
mathematicians as they looked, in vain, for an algorithm that would ﬁnd a root of an
arbitrary polynomial. We can now prove that no such algorithm exists. It is known
though, that a root exists for every complex polynomial. This was proved in 1799 by
the brilliant mathematician Karl Friedrich Gauss. Theorem 7 (Fundamental Theorem of Algebra (FTA))
For all complex polynomials f (z ) with deg(f (z )) ≥ 1, there exists a z0 ∈ C so that
f (z0 ) = 0. Ironically, we can prove a root exists, we just can’t construct one in general. The
proof of this theorem is demanding and is left for later courses.
We can use the Division Algorithm for Polynomials to help though. Recall
193 194 Chapter 35 Proposition 8 Factoring Polynomials (Division Algorithm for Polynomials (DAP))
If f (x) and g (x) are polynomials in F[x] and g (x) is not the zero polynomial, then
there exist unique polynomials q (x) and r(x) in F[x] such that
f (x) = q (x)g (x) + r(x) Proposition 9 where deg r(x) < deg g (x) or r(x) = 0 (Remainder Theorem (RT))
The remainder when the polynomial f (x) is divided by (x − c) is f (c). Example 8 Find the remainder when f (z ) = 3z 12 − 8iz 5 + (4 + i)z 2 + z + 2 − 3i is divided by z + i.
Solution: One could do the painful thing and carry out long division. Another
possibility is to use the Remainder Theorem and compute f (−i).
f (−i) = 3(−i)12 − 8i(−i)5 + (4 + i)(−i)2 + (−i) + 2 − 3i
= 3 − 8i(−i) + (4 + i)(−1) − i + 2 − 3i
= 3 − 8 − 4 − i − i + 2 − 3i
= −7 − 5i
The remainder is −7 − 5i. Proof: By the Division Algorithm for Polynomials, there exist unique polynomials
q (x) and r(x) such that
f (x) = q (x)(x − c) + r(x) where deg r(x) < 1 or r(x) = 0 Therefore, the remainder r(x) is a constant (which could be zero) which we will write
as r0 . Hence
f (x) = q (x)(x − c) + r0
Substituting x = c into this equation gives f (c) = r0 . Corollary 10 (Factor Theorem 1 (FT1))
The linear polynomial (x − c) is a factor of the polynomial f (x) if and only if f (c) = 0. Equivalently, Corollary 11 (Factor Theorem 2 (FT2))
The linear polynomial (x − c) is a factor of the polynomial f (x) if and only if c is a
root of the polynomial f (x). Section 35.2 Polynomial Equations 195 How do we go about actually factoring polynomials? In general, this is hard to do.
There are no formulas for roots if the polynomial has degree ﬁve or more. But if the
polynomial has integer coeﬃcients, we have a good starting point. Theorem 12 (Rational Roots Theorem (RRT))
Let f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 be a polynomial with integer
coeﬃcients. If p is a rational root with gcd(p, q ) = 1, then p  a0 and q  an .
q In order to ﬁnd a rational root of f (x), we only need to examine a ﬁnite set of
rational numbers, those whose numerator divides the constant term and those whose
denominators divide the leading coeﬃcient. Note that the theorem only suggests
those rational numbers that might be roots. It does not guarantee that any of these
numbers are roots. Example 9 If possible, ﬁnd a rational root of f (x) = 2x4 + x3 + 6x + 3. Solution: We will use the Rational Roots Theorem. The divisors of 2 are ±1 and
±2. The divisors of 3 are ±1 and ±3. Hence, the candidates for rational roots are
3
1
±1, ± , ±3, ±
2
2
Now test each of these candidates.
x
1 −1
f (x) 12 −2
Thus, the only root is
Proof: If 1
2
25
4 −1
2 0 3
−3
210 120 3
2
51
2 −3
2
3
4 −1
2. p
is a root of f (x) then
q
an p
q n + an−1 p
q n−1 + · · · + a2 p
q 2 + a1 p
q + a0 = 0 Multiplying by q n gives
an pn + an−1 pn−1 q + · · · + a2 p2 q n−2 + a1 pq n−1 + a0 q n = 0
and
an pn = −q an−1 pn−1 + · · · + a2 p2 q n−3 + a1 pq n−2 a0 q n−1
Since all of the symbols in this equation are integers, both the right hand side and
left hand side are integers. Since q divides the the right hand side, q divides the left
hand side, that is
q  an p n
Since gcd(p, q ) = 1 we can repeatedly use the proposition on Coprimeness and Divisibility to show that q  an . In a similar way, we can show that p  a0 . 196 Chapter 35 Factoring Polynomials Exercise 8 Is x + 1 a factor of x10 + 1, or x9 + 1. Can you make a statement about when x + 1
divides or does not divide x2n + 1, or x2n+1 + 1 for n a positive integer? Exercise 9 If x + (2 + i) is a root of f (x) = x4 + 4x3 + 2x2 − 12x − 15, factor f (x) into products
of real polynomials and complex polynomials of lowest degree. Exercise 10 Prove that x = √
n p is irrational for any integer n > 1. Complex roots of real polynomials come in conjugate pairs. Chapter 36 The Shortest Path Problem
36.1 Objectives The technique objectives are:
1. Abstract from a map to a graph.
2. Formulate an algorithm.
3. Extend plausible uses. 36.2 The Problem Suppose you are living in downtown Toronto (the pink dot on the map) on a coop
work term and you want to escape the intense July heat by going to Sibbald Point
Provincial Park (the blue dot on the map) to swim in Lake Simcoe. See Figure 36.2.1.
You could take Highway 404 past the 401, past the 407 up to the end of Highway 404,
and then take a minor road to Highway 48 and go north from there. But perhaps it
would be better to take Lakeshore Drive to Highway 48 and go straight north.
Your task is to ﬁnd an algorithm, a strategy, to ﬁnd the shortest route between
downtown Toronto and Sibbald Point Provincial Park. 197 198 Chapter 36 The Shortest Path Problem Figure 36.2.1: Sibbald Point Provincial Park Section 36.3 Abstraction 36.3 199 Abstraction Let’s focus on what’s important in the problem. Looking at the map there is, for our
purpose, lots of information that is not important: colours, parking locations, where
the Green Belt is, towns not along the way. What is really important are locations
where we might change directions, routes between those locations, and distances.
We’ll highlight locations on the map as grey dots and connections between locations
as solid teal lines. See Figure 36.3.1. Figure 36.3.1: Locations and Connections 200 Chapter 36 The Shortest Path Problem But since we don’t need the rest of the detail, let’s omit it and include only locations,
connections and distances. See Figure 36.3.2. 20
45 15 120
60
60 10
10
10 10 30
5 20
25 Figure 36.3.2: The Essentials 36.4 Algorithm With a partner, draw a random map and attempt to discover an algorithm that will
ﬁnd the shortest route from one location to another. [Note to instructor: solicit some
algorithms and have the class assess whether the algorithm might work.] 36.5 Extensions This problem is set as minimal distances between two points. But perhaps instead of
distance we could use time or cost. And instead of a person travelling we could have
couriers delivering packages, or electrical signals carrying phone calls. In fact, there
are surprising uses as well including managing cutting stock in steel mills and ﬁnding
optimal schedules for construction projects. Chapter 37 Paths, Walks, Cycles and Trees
37.1 Objectives The technique objectives are:
1. Practice with by contradiction.
2. Practice with unqueness.
The content objectives are:
1. Deﬁne graph, walk, path, cycle, and tree.
2. Construct diagrams corresponding to graphs.
3. Observe: Any walk can be decomposed into at most one path and a collection of
cycles.
4. Prove: There is a unique path between every pair of vertices in a tree. 37.2
Deﬁnition
Graph The Basics A graph G is a pair (V, E ) where V is a ﬁnite, nonempty set, and E is a set of
unordered pairs of elements of V . The elements of V are called vertices and the
elements of E called edges. It is often very useful to represent a graph as a drawing where vertices correspond to
points and edges correspond to lines between vertices. Graphs may be represented
by more than one diagram as illustrated in Example 10. 201 202 Chapter 37 Example 10 Paths, Walks, Cycles and Trees Let G = (V, E ) where
V = {1, 2, 3, 4, 5, 6, 7}
and
E = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 1}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}, {1, 2}} . 7
2 1 6 7 3
1 5 2 3 4 5 6 4 Figure 37.2.1: Two representations of the same graph Deﬁnition
Adjacent,
Incident If edge e = {u, v }, then we say that u and v are adjacent vertices, and that edge e
is incident with vertices u and v . We can also say that the edge e joins u and v .
Vertices adjacent to a vertex u are called neighbours of u. A graph is completely speciﬁed by the pairs of vertices that are adjacent, and the
only function of a line in the diagram is to indicate that two vertices are adjacent. Deﬁnition
Walk A walk W is a nonempty sequence of edges
W = {{v0 , v1 }, {v1 , v2 }, {v2 , v3 }, . . . , {vn−1 , vn }} .
Since vi−1 and vi uniquely determine an edge e of a walk, we will usually just list the
vertices. Thus
W = v0 , v1 , v2 , v3 , . . . , vn−1 , vn . Deﬁnition
Path Deﬁnition
Cycle If v0 = s and vn = t in the walk W , we call W an stwalk. If no vertex in the walk
is repeated, that is, if v0 , v1 , v2 , . . . , vn are all distinct, then W is called a path. If v0 = vn and v0 , v1 , v2 , . . . , vn−1 are all distinct, then W is called a cycle. Section 37.2 The Basics 203 2 1 6 7 5 3 4 Figure 37.2.2: The bold lines indicate the walk W = 1, 6, 7, 3, 4, 7, 3, 2. 2 1 6 7 5 3 4 Figure 37.2.3: The bold lines indicate the path P = 1, 6, 7, 3, 2 2 1 6 7 5 3 4 Figure 37.2.4: The bold lines indicate the cycle C = 7, 3, 4, 7 Note that the walk W = 1, 6, 7, 3, 4, 7, 3, 2 can be decomposed into the path P =
1, 6, 7, 3, 2 and the cycle C = 7, 3, 4, 7. In fact, we can always perform this kind of
decomposition for walks but before we state the appropriate theorem, we need to
deﬁne a few more terms. Deﬁnition
Collection,
Decomposed By a collection we mean a family
an stwalk. If s = t, we say that W
if, for every edge e the number of
of times e occurs in cycles of C . If of objects where repetition is allowed. Let W be
can be decomposed into a collection C of cycles
times e occurs in W is the same as the number
s = t, we say that W can be decomposed into 204 Chapter 37 Paths, Walks, Cycles and Trees an stpath P and a collection C of cycles if, for every edge e the number of times e
occurs in W is the same as the number of times e occurs in P and the cycles of C . We will state, but not prove, the following proposition. Proposition 13 (Walk Decomposition (WD))
Let W be an stwalk.
1. If s = t, then W can be decomposed into a nonempty collection of cycles.
2. If s = t and a vertex is not repeated in W , then W is a path.
3. If s = t and a vertex is repeated in W , then W can be decomposed into a path
and a nonempty collection of cycles. You may wonder what the diﬀerence is between the deﬁnition of decomposition and
the proposition Walk Decomposition. The deﬁnition allows for the possibility that
some walks cannot be decomposed. The proposition states that all walks can be
decomposed. Deﬁnition
Connected To say that a graph G is connected means that there is a path between any two
vertices of G. We will assume for this course that all of our graphs are connected,
though in general, that is not a safe assumption. 37.3 Trees A tree is a very special and incredibly useful kind of graph. Deﬁnition A tree is a connected graph with no cycles. Tree We will prove several propositions about trees starting with this one. Proposition 14 (Unique Paths in Trees (UPT))
There is a unique path between every pair of vertices in a tree. We normally begin our proofs by explicitly identifying the hypothesis and the conclusion. Unique Paths in Trees is not in “If A, then B .” form, so let’s ﬁrst restate it.
Recall that the hypothesis is what we get to start with, and the conclusion is what
we must show. We start with a tree. Call it T . We must show that there is a unique
path between every pair of vertices in T . Hence, we could restate Unique Paths in
Trees as Section 37.3 Trees 205 2 1 6 7 5 3 4 Figure 37.3.1: A tree Proposition 15 (Unique Paths in Trees (UPT))
If T is a tree, then there is a unique path between every pair of vertices in T . Working forwards and backwards to prove this proposition will be problematic. So,
it’s time for a diﬀerent technique, proof by contradiction. Normally, when we wish
to prove that the statement “A implies B ” is true, we assume that A is true and
show that B is true. What would happen if B were true, but we assumed it was
false and continued our reasoning based on the assumption that B was false? Since
a mathematical statement cannot be both true and false, it seems likely we would
eventually encounter a mathematically nonsensical statement. Then we would ask
ourselves “How did we arrive at this nonsense?” and the answer would have to be
that our assumption that B was false was wrong and B is, in fact, true.
Proofs by contradiction have the following structure.
1. Assume that A is true.
2. Assume that B is false, or equivalently, assume that NOT B is true.
3. Reason forward from A and NOT B to reach a contradiction.
We will prove Unique Paths in Trees by contradiction.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Suppose that u and v are any two distinct vertices of T .
2. Since T is connected, there is at least one path connecting u to v .
3. Now suppose that there are two distinct uv paths, P1 = x0 , x1 , x2 , . . . , xn and
P2 = y0 , y1 , y2 , . . . , ym . Thus u = x0 = y0 and v = xn = ym .
4. We can construct a walk W beginning with u and ending at u that consists
of “walking” from u to v in P1 , then from v to u “backwards” in P2 . More
speciﬁcally,
W = x0 , x1 , x2 , . . . , xn , ym−1 , ym−2 , ym−3 , . . . , y0 . 206 Chapter 37 Paths, Walks, Cycles and Trees 5. By Part (1) of Proposition 13, W can be decomposed into a nonempty collection
of cycles.
6. But then the tree T contains cycles, contradicting the deﬁnition of a tree. Analysis of Proof We will begin by explicitly identifying the hypothesis and the
conclusion.
Hypothesis: T is a tree.
Conclusion: There is a unique path between every pair of vertices in T .
Core Proof Technique: Contradiction.
Preliminary Material: Deﬁnition of tree.
Sentence 1 Suppose that u and v are any two distinct vertices of T .
The conclusion contains a universal quantiﬁer, every. Let’s ﬁrst identify the
components of the universal quantiﬁer.
Objects:
Universe of discourse:
Certain property:
Something happens: Vertices u and v
Vertices of the tree T
None speciﬁed.
There is a unique path between u and v . Since we are using a universal quantiﬁer in the conclusion of the proposition,
the author uses Choose method.
Sentence 2 Since T is connected, there is at least one path connecting u to v .
Before the author can show that there is a unique path, the author must ﬁrst
show that a path exists.
Sentence 3 Now suppose that there are two distinct uv paths, P1 = x0 , x1 , x2 , . . . , xn
and P2 = y0 , y1 , y2 , . . . , ym . Thus u = x0 = y0 and v = xn = ym .
The author is negating the conclusion and so is going to use on of two techniques,
Contradiction or Contrapositive. Since the author hasn’t indicated which, it is
useful to look ahead in the proof to ﬁnd out. The last sentence of the proof
makes it clear that the author is using Contradiction.
Sentence 4 We can construct a walk W beginning with u and ending at u that consists of “walking” from u to v in P1 , then from v to u “backwards” in P2 . More
speciﬁcally,
W = x0 , x1 , x2 , . . . , xn , ym−1 , ym−2 , ym−3 , . . . , y0 .
Sentence 5 By Part (1) of Proposition 13, W can be decomposed into a nonempty
collection of cycles.
The diﬃcult part in proofs by contradiction is ﬁnding a contradiction. In Sentence 4 the author constructs a walk and in Sentence 5 the author shows that
the walk contains cycles. But cycles don’t exist in trees and so
Sentence 6 But then the tree T contains cycles, contradicting the deﬁnition of a
tree.
This is also an example of working with uniqueness. Chapter 38 Trees
38.1 Objectives The technique objectives are:
1. Induction.
The content objectives are:
1. Deﬁne degree.
2. Prove Two Vertices of Degree One.
3. Prove: Number of Vertices in a Tree. 38.2
Deﬁnition
Degree Proposition 16 Properties of Trees Let G be a graph. The number of edges incident with a vertex v is called the degree
of v and is denoted by deg(v ). In Figure 38.2.1, vertex a has degree 3 and vertex b
has degree 2. (Two Vertices of Degree One (TVDO))
If T is a tree with at least two vertices, then T has at least two vertices of degree one. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Find a longest path P = w0 w1 w2 . . . wn in T , say between nodes u = w0 and
v = wn .
2. Since any edge in the tree constitutes a path, P must contain at least one edge
so u = v .
207 208 Chapter 38 Trees i s h
g f b e d
c
a s Figure 38.2.1: Graph corresponding to Toronto  Sibbald Point map 3. Thus, the vertex wn−1 in the path is adjacent to v but distinct from v .
4. If deg(v ) > 1, there must be another vertex, w, distinct from wn−1 and adjacent
to v .
5. If w is in P , then a cycle would exist but trees do not have cycles. Hence, w is
not in P .
6. If w is not in P , then we could add edge {v, w} to P to get a path longer than
P , contradicting the assumption that P is a longest path in T .
7. Hence, deg(v ) = 1.
8. Similarly, deg(u) = 1 and so two vertices of degree one exist in T . Analysis of Proof We will begin by explicitly identifying the hypothesis and the
conclusion.
Hypothesis: T is a tree with at least two vertices.
Conclusion: T has at least two vertices of degree one.
Core Proof Technique: Construction and Contradiction (three times!). Section 38.2 Properties of Trees 209 Preliminary Material: Deﬁnition of tree and of degree.
Sentence 1 Find a longest path P = w0 w1 w2 . . . wn in T , say between nodes u = w0
and v = wn .
The conclusion contains an existential quantiﬁer, has. Let’s ﬁrst identify the
components of the existential quantiﬁer.
Objects:
Universe of discourse:
Certain property:
Something happens: Two vertices (unnamed)
Vertices of the tree T
None speciﬁed.
Both vertices have degree 1. Since the proposition contains an existential quantiﬁer in the conclusion, the
author is using the Construct method. This sentence serves two purposes. First,
it implicitly identiﬁes the two objects that will be constructed, u and v . And
second, it sets up the contradictions that will be needed later.
Sentence 2 Since any edge in the tree constitutes a path, P must contain at least
one edge so u = v .
Given that the author intends to show that u and v are distinct vertices of degree
one, the author must ﬁrst establish that u = v . Also, the following argument
will require that the path contain an edge.
Sentence 3 Thus, the vertex wn−1 in the path is adjacent to v but distinct from v .
The author is setting up the contradiction, though it is not at all clear from
here how that contradiction will be displayed.
Sentence 4 If deg(v ) > 1, there must be another vertex, w, distinct from wn−1 and
adjacent to to v .
From the analysis of the ﬁrst sentence, the author intends to show that v has
degree one. That means this sentence indicates the author is going to proceed
by contradiction.
Sentence 5 If w is in P , then a cycle would exist but trees do not have cycles. Hence,
w is not in P .
This is a miniature proof by contradiction of the statement “If deg(v ) > 1 and
w is adjacent to v , then w is not in P .” Sentence 5 begins with the negation
of the conclusion and ﬁnds a contradiction quickly. If w is in P , then the walk
constructed by taking the subpath from w to v in P and adding the edge {v, w}
yields a cycle, but trees do not contain cycles by deﬁnition.
Sentence 6 If w is not in P , then we could add edge {v, w} to P to get a path longer
than P , contradicting the assumption that P is a longest path in T .
This is another miniature proof by contradiction, this time of the statement “If
deg(v ) > 1 and w is adjacent to v , then w is in P .”
Sentence 7 Hence, deg(v ) = 1.
Assuming that deg(v ) > 1 leads to an adjacent vertex w being both in P and
not in P , a contradiction. Since the author’s reasoning is correct, it must be the
case that the assumption deg(v ) > 1 is false. Since T is connected, deg(v ) > 0
so deg(v ) = 1. 210 Chapter 38 Trees Sentence 8 Similarly, deg(u) = 1 and so two vertices of degree one exist in T .
Similarly is a useful but dangerous word in proofs. If the conditions really are
similar, then using “similarly” spares tedious eﬀort in checking the details. However, if the conditions are not similar, the use of “similarly” could be masking
a fatal error.
In this case, the argument is identical when w1 replaces wn−1 . Proposition 17 (Number of Vertices in a Tree (NVT))
Let T = (V, E ) be a tree. Then V  = E  + 1. Since V is an integer, we could consider all trees with one vertex, two vertices, three
vertices and so on, this seems like a perfect case for induction. Let’s be very clear
about what our statement P (n) is.
P (n): Let T = (V, E ) be a tree with n vertices. Then n = E  + 1.
Now we can begin the proof.
Proof: Base Case We verify that P (1) is true where P (1) is the statement
P (1): Let T = (V, E ) be a tree with one vertex. Then 1 = E  + 1.
This is equivalent to stating that E  = 0. Since a tree with one vertex has no
edges, the base case is true.
Inductive Hypothesis We assume that the statement P (k ) is true for k ≥ 2.
P (k ): Let T = (V, E ) be a tree with k vertices. Then k = E  + 1.
Inductive Conclusion Now show that the statement P (k + 1) is true.
P (k + 1): Let T = (V, E ) be a tree with k + 1 vertices. Then k + 1 =
E  + 1.
By Proposition 16, we know that there is at least one vertex of degree one in T .
Let’s call such a vertex v . Since deg(v ) = 1, v is adjacent to only one vertex,
say u. Deleting the vertex v and the edge {u, v } creates a new tree T where T
has k vertices and E  − 1 edges. By our Inductive Hypothesis therefore,
k = (E  − 1) + 1 ⇒ k = E .
But T has one more vertex and more edge than T so
k + 1 = E  + 1
as required.
The result is true for n = k + 1, and so holds for all n by the Principle of
Mathematical Induction. Chapter 39 Dijkstra’s Algorithm
39.1 Objectives The content objectives are:
1. Be able to execute Dijkstra’s Algorithm. 39.2 Dijkstra’s Algorithm [Note to instructors: How you start this depends on how you ended the class that
introduced the Shortest Path Problem.]
Let’s look at a formal expression for solving the shortest path problem.
We can think of the “Require” statement as the preconditions to the algorithm, or
the hypothesis to a proposition. In this case, we require a graph with nonnegative
weights on the edges, and a starting vertex s. We can think of the “Ensure” statement
as the postconditions to the algorithm or the conclusion to a proposition. In this
case, the algorithm should terminate with a tree of shortest paths rooted at s, and
the distances of a shortest path from s to each node.
Though our original problem talked about distances, the values we assign to the edges
of the graph could also be time or capacity or costs. The convention is to call these
values weights, which is why the function from the edges to the real numbers is named
w.
Let’s watch the algorithm in operation. Our example appears in Figure 39.2.1.
The initialization steps of the algorithm set the distance to s at 0, and the provisional
distances to all other vertices at inﬁnity. By abuse of notation, we will treat inﬁnity
as a real number. We will record distances as numeric labels in blue near the vertices.
The set V initially contains only s and E is empty. We will show the nodes in V
as bold circles and the edges in E as bold lines. Note that at every stage of the
algorithm, T = (V , E ) is a tree of shortest paths to the vertices in V . See Figure
39.2.2
Now the algorithm examines each edge with one vertex in V and one vertex not in
V . If using the edge creates a shorter path to a vertex not in V , then the provisional
211 212 Chapter 39 Dijkstra’s Algorithm Algorithm 2 Dikstra’s Algorithm
Require: G = (V, E ); w : E → R; w({u, v }) ≥ 0, ∀{u, v } ∈ E ; and a designated
node s.
Ensure: T = (V , E ) is a tree rooted at s of shortest paths from s to every other
node; d : V → R gives the distance of a shortest path to v , ∀v ∈ V .
{Initialize}
d(s) ← 0
d(v ) ← ∞, ∀v ∈ V, v = s
V ← {s}
E ←∅
T ← (V , E )
repeat
for every edge {u, v } ∈ E where u ∈ V and v ∈ V do
if d(v ) < d(u) + w({u, v }) then
d(v ) ← d(u) + w({u, v })
end if
end for
Choose a y ∈ V so that d(y ) = min{d(w)  w ∈ V }
For the y just chosen, choose {x, y } ∈ E where x ∈ V and d(y ) = d(x)+ w({x, y })
V ← V ∪ {y }
E ← E ∪ {{x, y }}
T ← (V , E )
until V = V
3 s 1 b 9 2 c 3 4 a 1 d Figure 39.2.1: Graph G with weights distance to that vertex is updated. Figure 39.2.3 shows the results of the update.
Edges and distances involved in the updates are shown in green. The inﬁnite values
previously assigned to vertices a, b and c have been crossed out.
Continuing with the update, choose the vertex not in V with the smallest provisional
distance. In this iteration, the choice is b. Add b to V and {s, b} to E . This update
is shown in Figure 39.2.4. The nodes in V are shown as bold circles and the edges
in E as bold lines. Note that T = (V , E ) is a tree of shortest paths to the vertices
in V .
We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, b}, V = V and so
we continue. Again, the algorithm examines each edge with one vertex in V and one
vertex not in V . If using the edge creates a shorter path to a vertex in V , then the Section 39.2 Dijkstra’s Algorithm 213
∞ 0
3 s 1 ∞ a 9 b 2 c 3 4 ∞ 1 d ∞
Figure 39.2.2: After initialization 0
3 s 1 ∞1 9 b ∞3 a 2 ∞9 c 3 4 1 d ∞
Figure 39.2.3: First update of d 0 3
3 s 1 1 b a 9 2 c 3 4 9 1 d ∞
Figure 39.2.4: End of ﬁrst iteration provisional distance to that vertex is updated. Figure 39.2.5 shows the results of the
update. Edges and distances involved in the updates are shown in green.
Now choose the vertex not in V with the smallest provisional distance. In this
iteration, the choice is a. Add a to V and {s, a} to E . This update is shown in 214 Chapter 39 0 3
3 s 1 1 b Dijkstra’s Algorithm a 9 2 c 3 4 94 1 d ∞5
Figure 39.2.5: Second update of d Figure 39.2.6. The nodes in V are shown as bold circles and the edges in E as bold
lines. Again, note that T = (V , E ) is a tree of shortest paths to the vertices in V .
0 3
3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.6: End of second iteration We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, a, b}, V = V and
so we continue. Again, the algorithm examines each edge with one vertex in V and
one vertex not in V . If using the edge creates a shorter path to a vertex in V , then
the provisional distance to that vertex is updated. In this iteration, no updates to
provisional distances took place. Figure 39.2.7 shows the results of the update. Edges
and distances involved in the updates are shown in green.
Now choose the vertex not in V with the smallest provisional distance. In this
iteration, the choice is c. Add c to V and {b, c} to E . This update is shown in
Figure 39.2.8. Again, note that T = (V , E ) is a tree of shortest paths to the
vertices in V .
We repeat this until V = V . Since V = {s, a, b, c, d} and V = {s, a, b, c}, V = V and
so we continue. Again, the algorithm examines each edge with one vertex in V and
one vertex not in V . If using the edge creates a shorter path to a vertex in V , then
the provisional distance to that vertex is updated. Figure 39.2.9 shows the results of
the update. Section 39.2 Dijkstra’s Algorithm 215 0 3
3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.7: Third update of d 0 3
3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.8: End of third iteration 0 3
3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 39.2.9: Fourth update of d Now choose the vertex not in V with the smallest provisional distance. In this
iteration, the choice is d. Add d to V . But now both and {b, d} and {c, d} match the
condition to be added to E . Which one should be added or should both be added?
It is only necessary to choose one, say {b, d}. This update is shown in Figure 40.3.1.
Again, note that T = (V , E ) is a tree of shortest paths to the vertices in V . 216 Chapter 39 0 3
3 s 1 1 b Dijkstra’s Algorithm a 9 2 c 3 4 4 1 d 5 Figure 39.2.10: End of fourth iteration and termination of the algorithm Now, ﬁnally V = V and the algorithm terminates. Exercise 11 (Note to instructors: you may wish to do another example before
assigning this exercise.)
Turn to a neighbour and create a random small graph, say of 6 vertices, and run
Dijkstra’s algorithm on your graph. 39.3 Certiﬁcate of Optimality Based on our experiments when we began this section, the example we did together,
and your own examples, it seems that we have lots of empirical evidence that Dijkstra’s
algorithm works. But evidence is not a proof. Moreover, if a colleague were to provide
us with a graph, edge weights and a proposed tree of shortest paths, it would be nice to
have a certiﬁcate of optimality. Simply running the algorithm again might reproduce
an existing error in the computer program that runs the algorithm.
Let’s consider the two objects the algorithm is supposed to produce.
1. A tree of shortest paths rooted at s.
2. A function d : V → R which gives the distance of a shortest path to v , ∀v ∈ V .
We won’t prove that Dijkstra’s algorithm produces these two objects, though we will
certainly think about it. In the next couple of classes we will prove a theorem that
allows us to certify that the output of Dijkstra’s algorithm is, in fact, correct.
Let’s look at the algorithm more closely. Would we expect the algorithm to always
produce a tree? That is, is T = (V , E ) a tree in every iteration? If there is some
iteration where it is not a tree, then the end product will not be a tree because the
algorithm only adds edges. The algorithm never deletes edges.
The algorithm will have V  − 1 iterations because we add a vertex to V at each
iteration and V begins with s. We also add an edge at each iteration so we end up
with V  = E  + 1. Proposition 17 is suggestive but not conclusive. It says that for Section 39.3 Certiﬁcate of Optimality 217 a tree T = (V, E ), V  = E  + 1. It does not say that V  = E  + 1 implies that the
graph (V, E ) is a tree.
Let’s consider the construction of T . A tree is deﬁned as a connected graph with no
cycles so let’s ask ourselves “Can the algorithm create a cycle in T ?” Suppose that
it did and the cycle occurred when the edge {u, v } was added. That means both u
and v already had to exist in V , but the edge that is added always contains a vertex
not in V . Hence, no cycles exist in T . As for connectedness, this makes sense since,
at each iteration an edge is added to an already connected graph constructed in the
previous iteration.
More problematic is guaranteeing that T is a tree of shortest paths.
Let’s look at d more closely. Suppose {u, v } ∈ E and the path in E from s encounters
u before it encounters v . Then d(u) = d(v ) + w({u, v }). That is not a surprise. That
is how the algorithm adds edges to E . Now look at the exercise that you just
completed. Examine any edge at all in E , say {x, y }. My guess is that you will see
d(y ) ≤ d(x) + w({x, y }).
This is what will help us generate a certiﬁcate of optimality. Chapter 40 Certiﬁcate of Optimality  Path
40.1 Objectives The content objectives are:
1. Deﬁne weight, distance potentials, feasible distance potentials, equality edges and
tree of shortest paths.
2. Use a certiﬁcate of optimality to test that a proposed solution is optimal.
3. Prove A Path Shorter Than A Walk.
4. Prove Feasible Potentials.
5. Prove Certiﬁcate of Optimality for a Path.
6. Prove Shortest Paths Give Feasible Potentials.
7. Prove Shortest Path Optimality.
8. Prove Trees of Shortest Paths. 40.2 Certiﬁcate of Optimality Recall that a certiﬁcate consists of a theorem and data. If the data satisfy the
hypothesis of the theorem, the theorem guarantees that the desired property holds.
The data will be a tree T and a function d : V → R, exactly what is produced by
Dijkstra’s algorithm. Our task is to ﬁnd a theorem that will say “If the data satisfy
a certain property, then
1. T is a tree of shortest paths rooted at s.
2. d : V → R gives the distance of a shortest path to v , ∀v ∈ V .” 218 Section 40.3 Weighted Graphs 40.3 219 Weighted Graphs Suppose that G = (V, E ) is a connected graph with weights w : E → R. Let us also
suppose that w({u, v }) ≥ 0, for every edge of E .
Let W = v0 v1 v2 . . . vn be a walk in G. We deﬁne the weight of W to be the sum
of the weights of all arcs in W . If the edge {u, v } occurs more than once in W , its
weight is counted for each occurrence in W . More formally,
n−1 w({vi , vi+1 }) w(W ) =
i=0 We have been using this deﬁnition implicitly. The distance of a trip from downtown
Toronto to Sibbald Point Provincial Park is the sum of distances of each part of the
trip. Dijkstra’s algorithm also uses this deﬁnition implicitly. Proposition 18 (A Path Shorter Than A Walk (PSTW))
Let G = (V, E ) be a connected graph with nonnegative real weights. Let W be an
stwalk with s = t. Then there exists an stpath with w(P ) ≤ w(W ). Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Part 3 of Proposition 13 states that W can be decomposed into an stpath P
and a collection of cycles C1 , C2 , . . . , Cr .
2. Now r w(W ) = w(P ) + w(Ci ).
i=1 3. Since w(Ci ) ≥ 0 for all i = 1, 2, 3, . . . , r, w(P ) ≤ w(W ). Analysis of Proof We will begin by explicitly identifying the hypothesis and the
conclusion.
Hypothesis: G = (V, E ) is a connected graph with nonnegative real weights.
W is an stwalk with s = t.
Conclusion: There exists an stpath with w(P ) ≤ w(W ).
Core Proof Technique: Construct method.
Preliminary Material: Deﬁnitions related to weighted graphs.
Sentence 1 Part 3 of Proposition 13 states that W can be decomposed into an stpath
P and a collection of cycles C1 , C2 , . . . , Cr .
The conclusion contains an existential quantiﬁer so the author uses the Construct method. Let’s ﬁrst identify the components of the existential quantiﬁer. 220 Chapter 40
Quantiﬁer:
Variable:
Domain:
Open sentence: Certiﬁcate of Optimality  Path ∃
A path P
All paths in G = (V, E )
w(P ) ≤ w(W ) The author must construct an stpath P and does so using Part 3 of Proposition
13. The author will now show that w(P ) ≤ w(W ).
Sentence 2 Now r w (W ) = w (P ) + w(Ci ).
i=1 This is the numeric implication of Proposition 13.
Sentence 3 Since w(Ci ) ≥ 0 for all i = 1, 2, 3, . . . , r, w(P ) ≤ w(W ).
This is arithmetic.
The proof is very simple and relies very heavily on the fact that w(Ci ) ≥ 0 for all
i = 1, 2, 3, . . . , r. What if the hypothesis “nonnegative real weights” were simply
“nonnegative real weights”? Exercise 12 Show the necessity of “nonnegative” in the hypothesis of Proposition 18. That is,
ﬁnd a counterexample to the statement:
Let G = (V, E ) be a connected graph with nonnegative real weights. Let W be an
stwalk with s = t. Then there exists an stpath with w(P ) ≤ w(W ). You might argue that this is irrelevant because you never encounter negative distances.
This may be true of distances, but this is not true of costs. Subsidies and rebates do,
in fact, create negative cost edges in models.
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and
d : V → R. The components of d are called distance potentials. We say that
distance potentials are feasible when
d(u) + w({u, v }) ≥ d(v ) for all uv ∈ E.
Edges for which d(u) + w({u, v }) = d(v ) are called equality edges. Proposition 19 (Feasible Potentials (FP))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R, d : V → R
be feasible distance potentials and W an stwalk. Then w(W ) ≥ d(t)−d(s). Moreover,
w(W ) = d(t) − d(s) if and only if every arc of W is an equality edge. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Suppose W = v0 v1 v2 . . . vk where s = v0 and t = vk . Section 40.3 Weighted Graphs 221 2. The feasible distance potentials satisfy
d(v0 ) + w({v0 , v1 }) ≥ d(v1 )
d(v1 ) + w({v1 , v2 }) ≥ d(v2 )
d(v2 ) + w({v2 , v3 }) ≥ d(v3 )
.
.
.
d(vk−1 ) + w({vk−1 , vk }) ≥ d(vK )
3. Adding these inequalities together gives
d(v0 ) + d(v1 ) + d(v2 ) + . . . + d(vk−1 ) + w({v0 , v1 }) + w({v1 , v2 }) + . . . + w({vk−1 , vk })
≥d(v1 ) + d(v2 ) + d(v3 ) + . . . + d(vk ).
4. This simpliﬁes to
d(v0 ) + w(W ) ≥ d(vk )
or
w(W ) ≥ d(t) − d(s).
5. Moreover, w(W ) ≥ d(t) − d(s) if and only if every inequality above holds with
equality, that is, every edge in W is an equality edge. This is a straightforward proof so no analysis is provided. Theorem 20 (Certiﬁcate of Optimality for a Path (OPT P))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and
let s be a designated vertex and let P be an stpath. If there exist feasible distance
potentials d : V → R such that every edge of P is an equality edge, then P is a
shortest stpath. Before we examine the proof, let’s see how the theorem works as part of the certiﬁcate.
Recall the tree and function d that resulted from our example of running Dijkstra’s
algorithm.
The dark edges indicate the tree and the blue labels adjacent to the vertices give d.
Observe the sdpath P = sbd. All of the hypotheses of Theorem 20 are satisﬁed.
G is a connected graph with nonnegative weights. A vertex s has been designated.
P = sbd is an sdpath. By examining each edge of G we can conﬁrm that d are
feasible distance potentials. By examining each edge of P we can conﬁrm that every
edge of P is an equality edge. Hence, by Theorem 20, P is a shortest sdpath.
Now to the proof.
Proof: (For reference, each sentence of the proof is written on a separate line.)
1. By the ﬁrst part of the conclusion of Proposition 19, every stwalk has weight
at least w(t) − w(s). 222 Chapter 40 Certiﬁcate of Optimality  Path 0 3
3 s 1 1 b a 9 2 c 3 4 4 1 d 5 Figure 40.3.1: Tree and d 2. By the second part of the conclusion of Proposition 19, w(P ) = w(t) − w(s).
3. Since the weight of every walk W is bounded below by w(t) − w(s), and P is a
path that achieves that bound, P must be a shortest stpath. Analysis of Proof We will begin by explicitly identifying the hypothesis and the
conclusion.
Hypothesis: G = (V, E ) is a connected graph with nonnegative weights w :
E → R. s is a designated vertex and P is an stpath. There exist feasible
distance potentials d : V → R such that every edge of P is an equality
edge.
Conclusion: P is a shortest stpath.
Core Proof Technique: ForwardBackwards. Existential quantiﬁers occur in
the hypothesis.
Preliminary Material: Accumulated knowledge about weighted graphs.
Sentence 1 By the ﬁrst part of the conclusion of Proposition 19, every stwalk has
weight at least w(t) − w(s).
Since is is a form of the existential quantiﬁer, the hypothesis “P is an stpath”
allows the author to assume the existence of P . What the author must show
is not that P exists, or that P is an stpath, but rather that P is a shortest
stpath. The ﬁrst sentence of the proof places an upper bound on w(P ).
Sentence 2 By the second part of the conclusion of Proposition 19, w(P ) = w(t) −
w(s).
The hypotheses of the current theorem include “There exist feasible distance
potentials d : V → R such that every edge of P is an equality edge.” The
existential quantiﬁer in this hypothesis allows the author to assume the existence
of feasible distance potentials and equality edges. These are needed to invoke
19. Section 40.3 Weighted Graphs 223 Sentence 3 Since the weight of every walk W is bounded below by w(t) − w(s), and
P is a path that achieves that bound, P must be a shortest stpath.
Since no walk, and hence no path, can be shorter than w(t) − w(s), and w(P ) =
w(t) − w(s), P must be a shortest stpath. Proposition 21 (Shortest Paths Give Feasible Potentials (SPGFP))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and a
designated node s. If d : V → R is deﬁned as the length of a shortest path from s to
v for all vertices in V , then d are feasible distance potentials. Proof: (For reference purposes, each sentence of the proof is written on a separate
line.)
1. By contradiction, suppose that d are not feasible distance potentials. Then
there exists {u, v } ∈ E such that d(u) + w({u, v }) < d(v ).
2. Let P be a shortest supath. By the deﬁnition of d, w(P ) = d(u).
3. Consider the walk W constructed by appending the edge {u, v } to the path P .
4. By Proposition 18, there exists an sv path P with w(P ) ≤ w(W ).
5. But w(W ) = w(P ) + w({u, v }) = d(u) + w({u, v }) < d(v ).
6. But then w(P ) < d(v ) so d(v ) cannot be the length of a shortest sv path, a
contradiction. Now we show that the converse of the certiﬁcate of optimality for paths also holds. Theorem 22 (Feasible Distance Potentials and Equality Edges)
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and let
s be a designated vertex. If P is a shortest stpath, then there exist feasible distance
potentials d : V → R such that every edge of P is an equality edge. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let d : V → R be deﬁned as the length of a shortest path from s to v for all
vertices in V . By 21, these are feasible distance potentials.
2. Hence, w(P ) = d(t) = d(t) − 0 = d(t) − d(s).
3. But then 19 implies that every edge of P is an equality edge. 224 Chapter 40 Certiﬁcate of Optimality  Path Together, the theorem on the optimality of paths (Theorem 20) and the existence of
feasible distance potentials (Theorem 22) gives Theorem 23 (Shortest Path Optimality (SPO))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and let
s be a designated vertex. P is a shortest stpath if and only if there exist feasible
distance potentials such that every edge of P is an equality edge. 40.4 Certiﬁcate of Optimality  Tree We have dealt so far with paths, but Dijkstra’s algorithm produces a tree, not a path.
Fortunately, similar theorems hold. Theorem 24 (Trees of Shortest Paths (TSP))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R. Let s be
a designated vertex and let T be a spanning tree rooted at s. If there exist feasible
distance potentials such that every edge of T is an equality edge, then T is a tree of
shortest paths rooted at s. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. Let us assume that there exist feasible distance potentials such that every edge
of T is an equality edge.
2. For every node v in V , there is an stpath in T that satisﬁes the hypotheses of
Theorem 20.
3. Hence, T is a tree of shortest paths rooted at s. Theorem 24 requires a spanning tree, feasible potentials and equality arcs. How do
we know that these exist? Theorem 25 (Existence of Trees of Shortest Paths (ETSP))
Let G = (V, E ) be a connected graph with nonnegative weights w : E → R and let s
be a designated vertex. Then there exists a tree of shortest paths rooted at s. Proof: (For reference, each sentence of the proof is written on a separate line.)
1. For every node v ∈ V , let P (v ) be a shortest stpath in G and let d(v ) =
w(P (v )). Section 40.4 Certiﬁcate of Optimality  Tree 225 2. Since d(v ) is the length of a shortest path to v , Proposition 21, tells us that d
is a set of feasible distance potentials.
3. We know from Proposition 19 that every edge in a shortest sv path is an equality
arc. So, every edge of P (v ) is an equality edge for every v ∈ V .
4. Let
E= P (v ).
v ∈V 5. The edges of E contain a path consisting of equality arcs from s to every v ∈ V .
Delete from E enough edges to produce a tree T .
6. But then 24 applies and T is a tree of shortest paths rooted at s. Chapter 41 Appendix
Proposition 26 (Decomposing nth Power (DNP))
If ab = cn and gcd(a, b) = 1, then there exist integers a1 and b1 so that a = an and
1
b = bn .
1 Proof: Without loss of generality, we may assume that a > 1 and b > 1. If
a = pk1 pk2 · · · pkr
r
12
jj
j
b = q11 q22 · · · qss are the prime factorizations of a and b, then no px can occur among the qy otherwise
the gcd(a, b) > 1. As a result, the prime factorization of ab is
jj
j
ab = pk1 pk2 · · · pkr q11 q22 · · · qss
r
12 Let us suppose that c can be factored into primes as
c = ul1 ul2 · · · ult
t
12
Then ab = cn can be written as
jj
j
pk1 pk2 · · · pkr q11 q22 · · · qss = unl1 unl2 · · · unlt
r
t
12
1
2 This implies that each px and qy equals some uh and that the corresponding exponents
are equal. That is kx = nlh (or jy = nlh ). This implies that all of the exponents of
the px and qy are divisible by n. Thus, we can choose
k /n k /n a = p1 1 p2 2 j /n j /n b = q11 q22
and a = an and b = bn as needed.
1
1 226 · · · pkr /n
r j
· · · qss /n ...
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This note was uploaded on 04/02/2012 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.
 Winter '08
 ANDREWCHILDS
 Math

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