final_pract_key - COT 3100-01 April 18, 2001 Practice...

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COT 3100-01 April 18, 2001 Practice test#3 (Total 105pts) Name:_________________ SSN:__________________ 1)(10pts) Prove by induction that for any integers a and b ( a - b ) | ( a n - b n ), where n is a positive integer, n 1. (Hint: use induction on n and for the inductive step relate a k +1 - b k +1 = a ( a k - b k )+…?) Proof. We are going to prove by induction on n 1, that for any integers a and b , ( a n - b n ) is divisible by ( a - b ). By the definition of divisibility it is sufficient to prove that for any n 1, for any a and b there exists an integer m such that ( a n - b n )= m ( a - b ). Basis. n =1, ( a - b ) = 1 ( a - b ). IH . Assume that proposition is true for n = k , where k is some integer, k 1. In other words, we assume that for some k 1 and any a and b there exists an integer m such that ( a k - b k )= m ( a - b ). IS. We need to prove that the proposition holds for n = k +1, i.e. for any a , b there exists some integer s , such that ( a k +1 - b k +1 )= s ( a - b ) To reduce k +1 case to the case n = k , we consider the following identity: a k +1 - b k +1 = a ( a k - b k )+ a b k - b k +1 = a ( a k - b k )+ b k ( a - b ) By IH we have for these a and b some integer m such that ( a k - b k )= m
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final_pract_key - COT 3100-01 April 18, 2001 Practice...

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