This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Illinois Institute of Technology Department of Computer Science Lectures 23: January 2830, 2008 CS 330 Discrete Structures Spring Semester, 2008 1 Proof by contradiction We will now look at an important but easily misunderstood proof technique. The essence of this technique is that to prove A B , we will instead show B A . 1.1 Why does a proof by contradiction work? It can be shown that these two forms are equivalent by examining the truth table of both functions. Since A B B A is a tautology , proving one implication proves the other implication, and disproving one implication disproves the other. A B A B B A B A T T T F F T T F F T F F F T T F T T F F T T T T 1.2 A sample proof by contradiction Theorem: 2 is irrational. That is, 2 cannot be written as a b , where a and b are integers with no common factors. We first need to convert this to the A B form as above. One simple conversion is T 2 is irrational. (Convince yourself by examining the truth table above that this is indeed a valid conversion.) Proof by contradiction: We will show that 2 is rational implies F that is, that if we assume that 2 is rational, we can derive a contradiction. By the definition of rationality, 2 = a b , for two relatively prime integers a and b . Thus 2 b = a . It follows that 2 b 2 = a 2 and, by the definition of an even number, that a 2 is even. We now take a small diversion to help us arrive at our goal. Lemma: a 2 is even a is even Proof by contradiction: We show a is odd a 2 is odd. Since a is odd, it has the form 2 k + 1, for some integer k . Thus a 2 = (2 n + 1) 2 = 4 n 2 + 4 n + 1 = 2(2 n 2 + 2 n ) + 1. Thus a 2 is odd. Now that we have concluded that a 2 is even a is even, we can resume our original proof. Since a is even, it can be written as 2 c , where c is an integer. Thus 2 b 2 = (2 c ) 2 , so 2 b 2 = 4 c 2 , and b 2 = 2 c 2 . Thus b 2 is even, and by the lemma we know that b is even. So both a and b are even. They share the common factor 2. But we originally assumed that a and b had no common factors! Thus we have arrived at a contradiction and proved the original theorem, that 2 is irrational. CS 330Spring, 2008 2 Lectures 23: January 2830, 2008 1.3 Bertrand Russells Proof Bertrand Russell, the famous philosopher/mathematician, was challenged that because a false proposition implies any proposition, could he prove that if 2 + 2 = 5, then he is the pope. Yes, he responded:implies any proposition, could he prove that if 2 + 2 = 5, then he is the pope....
View
Full
Document
This note was uploaded on 04/15/2008 for the course CS 330 taught by Professor Reingold,edwardm. during the Spring '08 term at Illinois Tech.
 Spring '08
 Reingold,EdwardM.
 Computer Science

Click to edit the document details