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hw3sol - CSE 20 Discrete Mathematics Fall 2010 Problem Set...

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CSE 20: Discrete Mathematics Fall 2010 Problem Set 3 Instructor: Daniele Micciancio Due on: Wed. Oct. 20, 2010 Problem 1 (10 points) (a) Use the well-ordering principle to prove that every positive integer number is either even or odd , where “even(n)= x Z .n = 2 x ” and “odd(n)= x Z .n = 2 x + 1 ”. Proof: Let S = { n Z + | ¬ even ( n ) ¬ odd ( n ) } be the set of all positive integers that are neither even nor odd. We want to prove that S is empty. Assume for contradiction that S is not empty. Since S is a nonempty set of positive integers, by the well-ordering principle it contains a minimal element m S . Since m is the smallest element of S and m - 1 is smaller than m , m - 1 cannot be in S . Notice that 1 is an odd integer because 1 = 2 · 0 + 1 . Therefore 1 / S . In particular, m 6 = 1 and m - 1 > 0 is also a positive integer. But (by definition of S ) a positive integer that is not in S is either even or odd. We consider the two cases separately. If m - 1 is even, then m - 1 = 2 x for some x Z , and m = 2 x + 1 is odd. If m - 1 is odd, then m - 1 = 2 x + 1 for some x Z and m = 2 x + 2 = 2( x + 1) is even. In both cases, we got a contradiction because m is neither even nor odd.
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