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230hw5sol - Math 230 E Homework 5 Solutions Fall 2011 Drew...

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Math 230 E Fall 2011 Homework 5 Solutions Drew Armstrong Problem 1. Let p, a 1 , a 2 , . . . , a n be integers, with p prime. For all n 2 prove that if p | a 1 a 2 · · · a n then p | a i for some 1 i n . Proof. Let p Z be prime and let P ( n ) =“for all a 1 , . . . , a n Z with p | a 1 · · · a n we have p | a i for some 1 i n ”. We wish to prove P ( n ) = T for all n 1. First note that P (1) = T because certainly, “for all a Z with p | a we have p | a ”! Now fix some arbitrary k 1 and (OPEN MENTAL PARENTHESIS. assume that P ( k ) = T . In this case we wish to show that P ( k + 1) = T . So consider any a 1 , . . . , a k +1 Z and (OPEN MENTAL PARENTHESIS. suppose that p | a 1 · · · a k +1 . In particular, this means that p | ( a 1 · · · a k ) a k +1 . If p | a k +1 we are done, and otherwise Euclid’s Lemma implies that p | a 1 · · · a k . But since P ( k ) = T by assumption, this implies that p | a i for some 1 i k . CLOSE MENTAL PARENTHESIS.) Hence p | a i for some 1 i k + 1, and we conclude that P ( k + 1) = T . CLOSE MENTAL PARENTHESIS.) We have show that P (1) = T and P ( k ) P ( k + 1) for all k 1. By induction we condlude that P ( n ) = T for all n 1. [A real world proof would be much shorter than this, but I want to emphasize the logical structure.] Problem 2. Let a, q be real numbers, with q 6 = 0. Use induction to prove that for all integers n 1 we have a + aq + aq 2 + · · · + aq n - 1 = a (1 - q n ) 1 - q . Proof. Let a, r R with q 6 = 0 and say P ( n ) =“ a + aq + · · · + aq n - 1 = a (1 - q n ) 1 - q ”. We wish to show that P ( n ) = T for all n 1. First note that P (1) = T since a = a (1 - q ) / (1 - q ). Now fix some arbitrary k 1 and (OPEN MENTAL PARENTHESIS. assume that P ( k ) = T ; i.e.
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