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Proof for the greatest integer function inequality

  1. Jul 19, 2014 #1
    Can anyone help me prove the greatest integer function inequality-
    n≤ x <n+1 for some x belongs to R and n is a unique integer

    this is how I tried to prove it-
    consider a set S of Real numbers which is bounded below
    say min(S)=inf(S)=n so n≤x

    by the property x<inf(S) + h we have x< n+1 for some h=1
    thus we get n≤ x <n+1

    Is this method correct? and can I use the archimedian property to prove the above,how?
     
  2. jcsd
  3. Jul 19, 2014 #2

    Erland

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    I suppose you mean the following theorem:

    To every real number x, there exists a unique integer n such that n ≤ x < n+1.

    I don't undertand your attemped proof. What is S, and why is its supremum an integer?

    It is a good idea to use the Archimedian property here: Given x, there is always an integer m such that m*1>x. Hence, there is a smallest such integer...
     
  4. Jul 19, 2014 #3
    Yes,I mean that theorem.
    I am not able to solve it using Archimedian property.Can you work out the entire proof please?
     
  5. Jul 19, 2014 #4

    micromass

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    We don't do that in this forum. We only give hints to allow you to solve the problem on your own.

    So listen to the hints and questions of Erland and see if you can come up with a proof.
     
  6. Jul 19, 2014 #5
    Okay so finally I came up with something sensible

    Consider a set S defined by n>x for some integer n and any real x
    This set is non empty since there exist such n and x by archimedian property.
    Since the set is non empty it will have a least element by wop,let that be l
    So l>x since it is in S
    l-1<l and it contradicts its minimality so, l-1 wont belong to S
    Hence we get l-1<=x<l
    = l<=x<l+1

    Now is it correct?
     
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