Chapter 3 Elementary Number Theory and Methods of Proof

3.6

Indirect Argument

Reductio Ad Absurdum

• • Argument by contradiction Illustration in proof of innocence – Suppose I did commit the crime. Then at the time of the crime, I would have had to be at the scene of the crime.

– In fact, at the time of the crime, I was meeting with 20 people far from the crime scene, as they will testify.

– This contradicts the assumption that I committed the crime, since it is impossible to be in two places at one time. Hence, that assumption is false.

Proofs

• Direct Proof – start with hypothesis of a statement and make one deduction after another until the conclusion is reached.

• Indirect Proof (argument by contradiction) – show that a given statement is not true leads to the contradiction.

Example

• • Use proof by contradiction to show that there is no greatest integer.

– Starting point: Suppose not. Suppose that there is a greatest integer, N. N≥n for all integers.

– To Show: This supposition leads logically to a contradiction.

– Proof: • Suppose not. Suppose that there is a greatest integer N.

• N ≥ n for every integer n. Let M = N + 1. M is an integer under the addition property of integers. • Thus M > N. N is not the greatest integer; therefore a contradiction.

Theorem 3.6.1

Even and Odd Integer

• Theorem 3.6.2

– – There is no integer that is both even and odd Proof: • Suppose not. That is, suppose there is an integer n that is both even and odd.

• n = 2a (even) and n = 2b + 1 (odd) • • 2a = 2b + 1, 2a – 2b = 1, 2(a – b) = 1 • (a – b) = ½ • Since a and b are integers then a – b should result in an integer. Thus a-b is integer and a-b is not, contradiction.

Sum of Rational and Irrational

• Theorem 3.6.3

– The sum of any rational number and any irrational number is irrational – Proof: • Suppose not. Suppose there is a rational number r and irrational number s such that r + s is rational.

• • r = a/b (definition rational) r + s = c/d, for integers a,b,c,d with b≠0 and d≠0 • a/b + s = c/d, s = c/d – a/b • • s = (bc – ad)/bd (integer) hence, s is quotient of integers and therefore, rational • If s is rational it contradicts supposition that it is irrational

Argument by Contrapositive

• • Argument by contrapositive is based on the equivalence of the statement and contrapositive. If the contrapositive is true then the statement is true.

Proof by Contrapositive

• Method of Proof 1. Express the statement to be provided in the form: ∀ x in D, if P(x) then Q(x) 2. Rewrite the statement in the contrapositive form: ∀ x in D, if Q(x) is false then P(x) is false 3. Prove the contrapositive by a direct proof.

1. Suppose x is a (particular but arbitrarily chosen) element of D such that Q(x) is false.

2. Show that P(x) is false

Example

• If the square of an integer is even, then the integer is even.

– Prove that for all integers n, if n 2 is even then n is even.

– Contrapositive: ∀ integers n, n is not even then n 2 not even.

– is Proof: • Suppose n is any odd integer. n = 2k + 1 • • n 2 n 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 = 2r + 1 (by definition n 2 is odd) + 2k) + 1 • hence, the contrapositive is true therefore the statement must be true—”if n is even then n 2 is even”