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Proof by Contradiction

MAT

Transition to Higher Mathematics

Fall 2014

Outline

(^1) Proving Statements with Contradiction

2 Proving Conditional Statements by Contradiction

Motivating Example

Proposition

For all integers n, if n^3 + 5 is odd then n is even.

Motivating Example

Proposition

For all integers n, if n^3 + 5 is odd then n is even.

Proof.

Let n be any integer and suppose, for the sake of contradiction, that n^3 + 5 and n are both odd. In this case integers j and k exist such that n^3 + 5 = 2k + 1 and n = 2j + 1. Substituting for n we have

2 k + 1 = n^3 + 5 2 k + 1 = (2j + 1)^3 + 5 2 k + 1 = 8j^3 + 3(2j)^2 (1) + 3(2j)(1)^2 + 1^3 + 5 2 k = 8j^3 + 12j^2 + 6j + 5.

(Continued next slide)

Proof by Contradiction

This is an example of proof by contradiction. To prove a statement P is true, we begin by assuming P false and show that this leads to a contradiction; something that always false.

Many of the statements we prove have the form P ⇒ Q which, when negated, has the form P ⇒ ∼Q. Often proof by contradiction has the form

Proposition

P ⇒ Q.

Proof.

Assume, for the sake of contradiction P is true but Q is false.

· · ·

Since we have a contradiction, it must be that Q is true.

Proof:

2 is irrational

Proof.

Suppose

2 is rational. Then integers a and b exist so that

2 = a/b. Without loss of generality we can assume that a and b have no factors in common (i.e., the fraction is in simplest form). Multiplying both sides by b and squaring, we have 2 b^2 = a^2 so we see that a^2 is even. This means that a is even (how would you prove this?) so a = 2m for some m ∈ Z. Then

2 b^2 = a^2 = (2m)^2 = 4m^2

which, after dividing by 2, gives b^2 = 2m^2 so b^2 is even. This means b = 2n for some n ∈ Z.

(Continued next slide)

Quantifications and Contradiction

Sometimes we need prove statements of the form

∀x, P(x).

These are often particularly well suited to proof by contradiction as the negation of the statement is

∃x, ∼P(x)

so all that is necessary to complete the proof is to assume there is an x that makes ∼P(x) true and see that it leads to a contradiction.

Quantifications and Contradiction

Proposition

There exist no integers a and b for which 18 a + 6b = 1.

This could be written as “∀a, b ∈ Z, 18 a + 6b 6 = 1.” Negating this yields “∃a, b ∈ Z, 18 a + 6b = 1.”

Proof.

Assume, for the sake of contradiction, that integers a and b can be found for which 18a + 6b = 1. Dividing by 6 we obtain

3 a + b =

This is a contradiction, since by the closure properties 3a + b is an integer but 1/6 is not. Therefore, it must be that no integers a and b exist for which 18a + 6b = 1.

Practice

Use a direct proof, a contrapositive proof, or a proof by contradiction to prove each of the following propositions.

Proposition

Suppose a, b ∈ Z. If a + b ≥ 19 , then a ≥ 10 or b ≥ 10.

Proposition

Suppose a, b, c, d ∈ Z and n ∈ N. If a ≡ b (mod n) and c ≡ d (mod n), then a + c ≡ b + d (mod n).

Proposition

Suppose n is a composite integer. Then n has a prime divisor less than or equal to

n.