practice - ZZ has the least upper bound M . Now M-1 is not...

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Math 131A practice exam Problem 3: Solution We proceed by contradiction: Suppose not. Then - A n for all integers n . Put minus on both side of the inequality to obtain A m for all integers m (= - n ). In other words, the set of all integers ZZ has an upper bound. The rest of the proof is the same as what we did in class to prove that natural numbers does not have an upper bound. Due to the completeness Axiom, the set of integers
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Unformatted text preview: ZZ has the least upper bound M . Now M-1 is not an upper bound, and so there exists some integer m which is bigger than M-1, i.e., m > M-1. But then m + 1 is also an integer and m + 1 > M , which contradicts the fact that M is an upper bound for ZZ . 1...
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This note was uploaded on 11/15/2010 for the course MATH math 131 taught by Professor Kim during the Spring '10 term at UCLA.

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