= 0, such that s = t/u. Can we use this
information to show that r + s is rational?
The obvious next step is to add r = p/q
and s = t/u, to obtain r + s = p q + t u = pu
+ qt qu . Because q = 0 and u
= 0, it follows that qu
(ii) 1 n is even, (iii) n3 is odd, (iv) n2 + 1
is even. 1.8 Proof Methods and Strategy
Introduction In Section 1.7 we introduced
many methods of proof and illustrated
how each method can be used. In this
section we continue this effort.We will
introduce s
= 0 and a and b have no common factors
(so that the fraction a/b is in lowest
terms.) (Here, we are using the fact that
every rational number can be written in
lowest terms.) Because 2 = a/b, when
both sides of this equation are squared, it
follows that 2
divided both sides by a b. The error is
that a b equals zero; division of both
sides of an equation by the same quantity
is valid as long as this quantity is not zero.
EXAMPLE 16 What is wrong with this
proof? Theorem: If n2 is positive, then
n is positi
n is a perfect square, then n + 2 is not a
perfect square. 9. Use a proof by
contradiction to prove that the sum of an
irrational number and a rational number
is irrational. 10. Use a direct proof to
show that the product of two rational
numbers is ration
|x|y|. Because |xy|=|x|y| holds in each of
the four cases and these cases exhaust all
possibilities, we can conclude that |xy|=|
x|y|, whenever x and y are real numbers.
LEVERAGING PROOF BY CASES The
examples we have presented illustrating
proof by cases
check only a relatively small number of
instances of a statement. Computers do
not complain when they are asked to
check a much larger number of instances
of a statement, but they still have
limitations. Note that not even a computer
can check all instanc
pigeonhole principle, a combinatorial
technique that we will cover in depth in
Section 6.2. EXAMPLE 9 Show that at least
four of any 22 days must fall on the same
day of the week. Solution: Let p be the
proposition At least four of 22 chosen
days fall on
odd. Solution: This theorem has the form
p if and only if q, where p is n is odd
and q is n2 is odd. (As usual, we do not
explicitly deal with the universal
quantification.) To prove this theorem, we
need to show that p q and q p are
true. We have already
this kind, we will be able to quickly
formulate conjectures and prove
theorems without first developing a
theory. We will conclude the section by
discussing the role of open questions. In
particular, we will discuss some
interesting problems either that h
QC: 1/1 T1: 2 CH01-7T Rosen-2311T
MHIA017-Rosen-v5.cls May 13, 2011
15:27 94 1 / The Foundations: Logic and
Proofs EXAMPLE 4 Use a proof by cases to
show that |xy|=|x|y|, where x and y are
real numbers. (Recall that |a|, the absolute
value of a, equals a
is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Here a is the
integer obtained by subtracting the final
decimal digit of n from n and dividing by
10. Next, note that (10a + b)2 = 100a2 +
20ab + b2 = 10(10a2 + 2b) + b2, so that
the final decimal digit of n2 is the sam
= 0. Consequently, we have expressed r + s
as the ratio of two integers, pu + qt and
qu, where qu = 0. This means that r + s is
rational. We have proved that the sum of
two rational numbers is rational; our
attempt to find a direct proof succeeded.
EXAMP
9. EXAMPLE 9 What is wrong with this
proof? Theorem: If x is a real number,
then x2 is a positive real number. Proof:"
Let p1 be x is positive, let p2 be x is
negative, and let q be x2 is positive. To
show that p1 q is true, note that when x
is positive,
89 Mistakes in Proofs There are many
common errors made in constructing
mathematical proofs. We will briefly
describe some of these here.Among the
most common errors are mistakes in
arithmetic and basic algebra. Even
professional mathematicians make such
considered all six cases, we can conclude
that the final decimal digit of n2, where n
is an integer is either 0, 1, 2, 4, 5, 6, or 9.
Sometimes we can eliminate all but a
few examples in a proof by cases, as
Example 6 illustrates. EXAMPLE 6 Show
that the
theorems of the form x(P (x) Q(x):
direct proof and proof by contraposition.
We have also given examples that show
how each is used. However, when you are
presented with a theorem of the form
x(P (x) Q(x), which method should
you use to attempt to prove i
Hardy brought Ramanujan to Cambridge
and collaborated on important joint
papers, establishing new results on the
number of partitions of an integer. Hardy
was interested in mathematics education,
and his book A Course of Pure
Mathematics had a profound ef
are even. Solution: We will use proof by
contraposition, the notion of without loss
of generality, and proof by cases. First,
suppose that x and y are not both even.
That is, assume that x is odd or that y is
odd (or both). Without loss of generality,
we
also possible to give an existence proof
that is nonconstructive; that is, we do not
find an element a such that P (a) is true,
but rather prove that xP (x) is true in
some other way. One common method of
giving a nonconstructive existence proof
is to use
is true, namely, we assume that n is odd.
By the definition of an odd integer, it
follows that n = 2k + 1, where k is some
integer. We want to show that n2 is also
odd. We can square both sides of the
equation n = 2k + 1 to obtain a new
equation that expr
covered. The problem of proving a
theorem is analogous to showing that a
computer program always produces the
output desired. No matter how many
input values are tested, unless all input
values are tested, we cannot conclude that
the program always produc
contraposition can succeed when we
cannot easily find a direct proof.
EXAMPLE 3 Prove that if n is an integer
and 3n + 2 is odd, then n is odd. Solution:
We first attempt a direct proof. To
construct a direct proof, we first assume
that 3n + 2 is an odd i
and 2 does not divide both a and b, p
must be false. That is, the statement p,
2 is irrational, is true. We have proved
that 2 is irrational. Proof by
contradiction can be used to prove
conditional statements. In such proofs, we
first assume the negation
= j with 1 i n and 1 j n. (Note that
there are n2 n such conditional
statements.) When we prove that a group
of statements are equivalent, we can
establish any chain of conditional
statements we choose as long as it is
possible to work through the chain t
= 0 such that r = p/q. A real number that
is not rational is called irrational.
EXAMPLE 7 Prove that the sum of two
rational numbers is rational. (Note that if
we include the implicit quantifiers here,
the theorem we want to prove is For
every real number
SRINIVASA RAMANUJAN (18871920)
The famous mathematical prodigy
Ramanujan was born and raised in
southern India near the city of Madras
(now called Chennai). His father was a
clerk in a cloth shop. His mother
contributed to the family income by
singing at
m = n. 29. Prove or disprove that if m
and n are integers such that mn = 1, then
either m = 1 and n = 1, or else m = 1 and
n = 1. 30. Show that these three
statements are equivalent, where a and b
are real numbers: (i) a is less than b, (ii)
the average o
as 2l for some integer l. This is circular
reasoning because this statement is
equivalent to the statement being proved,
namely, n is even. Of course, the result
itself is correct; only the method of proof
is wrong. Making mistakes in proofs is
part of th
number; it is the smallest number
expressible as the sum of cubes in two
different ways. EXAMPLE 11 A
Nonconstructive Existence Proof Show
that there exist irrational numbers x and y
such that xy is rational. Solution: By
Example 10 in Section 1.7 we know