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## Indirect Argument: Contradiction and Contraposition

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**Method of Proof by Contradiction**• Suppose the statement to be proved is false. • Show that this supposition logically leads to a contradiction. • 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, xQ+ s.t. for yQ+, y≥x. (1) Consider the number y*=x/2. x>0 implies that y*=x/2>0. (2) xQimplies thaty*=x/2Q. (3) x>0 implies thaty*=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. ■**Method of Proof by Contraposition**1. Express the statement to be proved in the form: xD, if P(x) then Q(x) . 2. Rewrite in the contrapositive form: xD, 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.**Method of Proof by Contraposition (Example)**Proposition 1:For any integer n, if n2 is even then n is also even. Proof: The contrapositive is: For any integer n, if n is not even then n2 is not even. (1) Let’s show (1) by direct proof. Supposen isnot even. Then n is odd. So n=2k+1 for some kZ. Hence n2 =(2k+1)2=4k2+4k+1 Thus, n2 isnot even. ■**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” ).**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 m2=2n2 (by basic algebra) (2)**Method of Proof by Contradiction (Ex.)**Proof (cont.): (2) implies thatm2is 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): 2n2 = m2 =(2k)2 = 4k2. By dividing both sides by 2, n2 = 2k2. Thus, n2 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). ■**Infinity of Prime Numbers**Lemma 1: For any integer a and any prime number p, ifp|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.■**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 p1=2, p2=3, …, pn in ascending order. Consider M = p1· p2·…·pn+1. p|M for some prime number p (1) (based on the th-m from handout 9/23). p is one of p1, p2, …, pn.. Thus, p | p1· p2·…·pn.. (2) By (2) and Lemma 1, p is not a divisor of M. (3) (3) contradicts (1).■