1 Indirect Argument: Contradiction and Contraposition

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.

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. ■

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.

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. ■

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.

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” ).

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)

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). ■

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. ■

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). ■