induction

# Same argument 13 maximally weird fallacy in the proof

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Unformatted text preview: a−1=b−1. Hence a=b. Case 2: b − 1 ≥ a − 1. Same argument. 13 Maximally Weird!! Fallacy: In the proof we used the inductive hypothesis to conclude max {a − 1, b − 1} = n a − 1 = b − 1. However, we can only use the inductive hypothesis if a − 1 and b − 1 are positive integers. This does not have to be the case as the example b=1 shows. 14 More Examples 15 Factorials Theorem. n ￿ i=0 i(i!) = (n + 1)! − 1. By convention: 0! = 1 Induction basis: Since 0 = 1 − 1, the claim holds for n = 0. Induction Step: Suppose the claim is true for n.Then n+1 n ￿ ￿ i(i!) = (n + 1)(n + 1)! + i(i!) i=0 i=0 = = = (n + 1)(n + 1)! + (n + 1)! − 1 by ind. hyp. (n + 2)(n + 1)! − 1 (n + 2)! − 1 16 Divisibility Theorem: For all positive integers n, the number 7n-2n is divisible by 5. Proof: By induction. Induction basis. Since 7-2=5, the theorem holds for n=1. 17 Divisibility Inductive step: Suppose that 7n-2n is divisible by 5. Our goal is to show that this implies that 7n+1-2n+1 is divisible by 5. We note that 7n+1-2n+1 = 7x7n-2x2n= 5x7n+2x7n-2x2n = 5x7n +2(7n-2n). By induction hypothesis, (7n-2n) = 5k for some integer k. Hence, 7n+1-2n+1= 5x7n +2x5k = 5(7n +2k), so 7n+1-2n+1 =5 x some integer. Thus, the claim follows by induction on n. 18 Strong Induction 19 Strong Induction Suppose we wish to prove a certain assertion concerning positive integers. Let A(n) be the assertion concerning the integer n. To prove it for all n >= 1, we can do the following: 1) Prove that the assertion A(1) is true. 2) Assuming that the assertions A(k) are proved for all k<n, prove that the assertion A(n) is true. We can conclude that A(n) is true for all n>=1. 20 Strong Induction Induction basis: Show that A...
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## This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.

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