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280wk3_x4 - Faulty Inductions Part of why I want you to...

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Unformatted text preview: Faulty Inductions Part of why I want you to write out your assumptions carefully is so that you don’t get led into some standard errors. Theorem: All women are blondes. Proof by induction: Let P ( n ) be the statement: For any set of n women, if at least one of them is a blonde, then all of them are. Basis: Clearly OK. Inductive step: Assume P ( n ). Let’s prove P ( n + 1). Given a set W of n + 1 women, one of which is blonde. Let A and B be two subsets of W of size n , each of which contains the known blonde, whose union is W . By the induction hypothesis, each of A and B consists of all blondes. Thus, so does W . This proves P ( n ) ⇒ P ( n + 1). 1 Take W to be the set of women in the world, and let n = | W | . Since there is clearly at least one blonde in the world, it follows that all women are blonde! Where’s the bug? 2 Theorem: Every integer > 1 has a unique prime fac- torization. [The result is true, but the following proof is not:] Proof: By strong induction. Let P ( n ) be the statement that n has a unique factorization. We prove P ( n ) for n > 1. Basis: P (2) is clearly true. Induction step: Assume P (2) , . . . , P ( n ). We prove P ( n + 1). If n + 1 is prime, we are done. If not, it factors somehow. Suppose n + 1 = rs r, s > 1. By the induction hypothesis, r has a unique factorization Π i p i and s has a unique prime factorization Π j q j . Thus, Π i p i Π j q j is a prime factorization of n + 1, and since none of the factors of either piece can be changed, it must be unique. What’s the flaw?? 3 Problem: Suppose n + 1 = 36. That is, you’ve proved that every number up to 36 has a unique factorization. Now you need to prove it for 36. 36 isn’t prime, but 36 = 3 × 12. By the induction hy- pothesis, 12 has a unique prime factorization, say p 1 p 2 p 3 . Thus, 36 = 3 p 1 p 2 p 3 . However, 36 is also 4 × 9. By the induction hypothesis, 4 = q 1 q 2 and 9 = r 1 r 2 . Thus, 36 = q 1 q 2 r 1 r 2 . How do you know that 3 p 1 p 2 p 3 = q 1 q 2 r 1 r 2 . (They do, but it doesn’t follow from the induction hy- pothesis.) This is a breakdown error . If you’re trying to show some- thing is unique, and you break it down (as we broke down n +1 into r and s ) you have to argue that nothing changes if we break it down a different way. What if n + 1 = tu ? • The actual proof of this result is quite subtle 4 Theorem: The sum of the internal angles of a regular n-gon is 180( n- 2) for n ≥ 3. Proof: By induction. Let P ( n ) be “the sum of the internal angles of a regular n-gon is 180( n- 2).” For n = 3, the result was shown in high school. Assume P ( n ); let’s prove P ( n + 1). Given a regular ( n + 1)-gon, we can lop off one of the corners: By the induction hypothesis, the sum of the internal an- gles of the regular n-gon is 180( n- 2) degrees; the sum of the internal angles of the triangle is 180 degrees. Thus, the internal angles of the original ( n +1)-gon is 180(...
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This note was uploaded on 10/25/2009 for the course CS 2800 taught by Professor Selman during the Fall '07 term at Cornell.

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280wk3_x4 - Faulty Inductions Part of why I want you to...

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