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1 Indirect Argument: Contradiction and Contraposition

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2 Method of Proof by Contradiction 1.Suppose the statement to be proved is false. 2.Show that this supposition logically leads to a contradiction. 3.Conclude that the statement to be proved is true.

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Method of Proof by Contradiction (Ex.) Theorem: There is no least positive rational number. Proof: Suppose the opposite: least positive rational number x. That is, x Q + s.t. for y Q +, y≥x. (1) Consider the number y*=x/2. x>0 implies that y*=x/2>0.(2) x Q implies that y*=x/2 Q.(3) x>0 implies that y*=x/2<x.(4) Based on (2),(3),(4), y* Q + and y<x. This contradicts (1). Thus, the supposition is false and there is no least positive rational number. ■

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4 Method of Proof by Contraposition 1. Express the statement to be proved in the form: x D, if P(x) then Q(x). 2. Rewrite in the contrapositive form: x D, if Q(x) is false then P(x) is false. 3. Prove the contrapositive by a direct proof: (a) Suppose x is an element of D such that Q(x) is false. (b) Show that P(x) is false.

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5 Method of Proof by Contraposition (Example) Proposition 1: For any integer n, if n 2 is even then n is also even. Proof: The contrapositive is: For any integer n, if n is not even then n 2 is not even. (1) Let’s show (1) by direct proof. Suppose n is not even. Then n is odd. So n=2k+1 for some k Z. Hence n 2 =(2k+1) 2 =4k 2 +4k+1 Thus, n 2 is not even. ■

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6 Comparison of Contradiction and Contrapositive methods Advantage of contradiction method: –Contrapositive method only for universal conditional statements. –Contradiction method is more general. Advantage of contrapositive method: –Easier structure: after the first step, Contrapositive method requires a direct proof. –Contradiction method normally has more complicated structure.

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When to use indirect proof Statements starting with “There is no”. (E.g., “There is no greatest integer” ). If the negation of the statement deals with sets which are easier to handle with. (E.g., “ is irrational”; rational numbers are more structured and easier to handle with than irrational numbers). If the infinity of some set to be shown. (E.g., “The set of prime numbers is infinite” ).

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8 Method of Proof by Contradiction (Ex.) Theorem: is irrational. Proof: Assume the opposite:is rational. Then by definition of rational numbers, (1) where m and n are integers with no common factors. ( by dividing m and n by any common factors if necessary) Squaring both sides of (1), Then m 2 =2n 2 (by basic algebra) (2)

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Method of Proof by Contradiction (Ex.) Proof (cont.): (2) implies that m 2 is even. (by definition) Then m is even. (by Proposition 1)(3) So m=2k for some integer k. (by definition)(4) By substituting (4) into (2): 2n 2 = m 2 =(2k) 2 = 4k 2. By dividing both sides by 2, n 2 = 2k 2. Thus, n 2 is even (by definition) and n is even (by Prop. 1).(5) Based on (3) and (5), m and n have a common factor of 2. This contradicts (1). ■

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10 Infinity of Prime Numbers Infinity of Prime Numbers Lemma 1: For any integer a and any prime number p, if p|a then p doesn’t divide a+1. Proof (by contradiction): Assume the opposite: p|a and p|(a+1). Then a=p·n and a+1=p·m for some n,m Z. So 1=(a+1)-a=p·(m-n) which implies that p|1. But the only integer divisors of 1 are 1 and -1. Contradiction. ■

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11 Infinity of Prime Numbers Theorem: The set of prime numbers is infinite. Proof (by contradiction): Assume the opposite: The set of prime numbers is finite. Then they can be listed as p 1 =2, p 2 =3, …, p n in ascending order. Consider M = p 1 · p 2 ·…·p n +1. p|M for some prime number p (1) (based on the th-m from handout 9/23). p is one of p 1, p 2, …, p n.. Thus, p | p 1 · p 2 ·…·p n..(2) By (2) and Lemma 1, p is not a divisor of M. (3) (3) contradicts (1). ■

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