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jan28 - Illinois Institute of Technology Department of...

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Unformatted text preview: Illinois Institute of Technology Department of Computer Science Lectures 2–3: January 28–30, 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 330—Spring, 2008 2 Lectures 2–3: January 28–30, 2008 1.3 Bertrand Russell’s 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....
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