{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3sol

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}