Chapter 3 Elementary Number Theory and Methods of
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 = 2 a (even) and n = 2 b + 1 (odd) • 2 a = 2 b + 1, 2 a – 2 b = 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 is not even. – Proof: • • Suppose n is any odd integer. n = 2 k + 1 n 2 = (2 k + 1)2 = 4 k 2 + 4 k + 1 = 2(2 k 2 + 2 k) + 1 n 2 = 2 r + 1 (by definition n 2 is odd) hence, the contrapositive is true therefore the statement must be true—”if n is even then n 2 is even”
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