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Math Help - supremum and infimum

  1. #1
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    supremum and infimum

    I need help on how to prove this:

    If x is an arbitrary real number, prove that there are integers m and n such that m < x < n.

    Thank you
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  2. #2
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    Quote Originally Posted by PersaGell View Post
    I need help on how to prove this:

    If x is an arbitrary real number, prove that there are integers m and n such that m < x < n.

    Thank you
    WLOG for any x>0 consider k=1. By the Archimedean Ordering on \mathbb{R} there shall exist n such that n\cdot 1 > x. Hence we can find n>x. That completes the second inequality.

    For any x>0 consider x-1. I claim we can choose an integer m such that x-1\leq m <x.

    Above we can find an integer n>x. Hence consider the interval [-n,n]. Define the set of integers S=\{ k \in \mathbb{Z} | x-1 \leq k \leq n\}. This is a finite non-empty set. So choose m=\min S. I leave it to you to prove that m is the desired integer.
    Last edited by ThePerfectHacker; May 24th 2007 at 06:28 PM.
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