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Proof involving the Archimedean Property

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    If x is a real number, show that there is an integer m such that:
    m≤x<m+1
    Show that m is unique


    2. Relevant equations
    Archimedean Property: The set of natural numbers has no upper bound


    3. The attempt at a solution
    I'm having trouble with showing that m is unique. If x is a real number, I can find integers that are smaller and bigger than it. If m=x, then m≤x. By the Archimedean property, m+1>x and m+1>m, so m≤x<m+1
     
  2. jcsd
  3. Mar 11, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that's fine if x happens to be an integer (that's what you imply when you say "if m= x". What if it isn't?

    Whatever x is, you are correct when you say that the Archimedean property says that there exist integers larger than or equal to x. What can you say about the set of integers larger than or equal to x?
     
  4. Mar 11, 2012 #3
    Thanks for the advice.

    So if I understand correctly, we have three cases:
    1) x is an integer. Then, we can say m=x
    2) x is rational. Then by the Archimedean property, we an find integers that are strictly greater and less than x, so we can let m be an integer such that m=x+1/2, then m<x<m+1 and x is halfway point here
    3) x is irrational, then suppose we have the interval (m, m+1) where m is an integer, we let x be an irrational number somewhere in that interval
     
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