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Math Help - Proof of an impossible question.

  1. #1
    stokes
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    Proof of an impossible question.

    Hello all. I am very stuck on this question. I dont know where to start on how to tackle it.

    Please help:

    Show that berween any two different real numbers there is a reational number. (hint: If a<b, then b-a>0 so there is a natural number n such that
    1/n < b-a. Consider the set {K: k/n >b } and use the fact that a set of integers that is bounded from below contains a least element.) Show that between any two different real numbers there are ultimately many rational numbers.

    I am really having a real hard time with proof. Does anyone have any tips, advice or any suggestions of books that would help with proof of amathematical question?

    Thanks in advance for all your help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by stokes View Post
    Hello all. I am very stuck on this question. I dont know where to start on how to tackle it.

    Please help:

    Show that berween any two different real numbers there is a reational number. (hint: If a<b, then b-a>0 so there is a natural number n such that
    1/n < b-a. Consider the set {K: k/n >b } and use the fact that a set of integers that is bounded from below contains a least element.) Show that between any two different real numbers there are ultimately many rational numbers.

    I am really having a real hard time with proof. Does anyone have any tips, advice or any suggestions of books that would help with proof of amathematical question?

    Thanks in advance for all your help.
    Let a and b in R, a!=b. For simplicity we may assume 0<=a<b, the other
    cases can then be swept up without much trouble.

    Now follow the hint, there exists an integer n, such that:

    1/n<b-a,

    and consider the set {k: k/n>=b}, clearly this set is non-empty, and

    1/n<b,

    so it's bounded from below, and so has a least element k1>1, now
    consider r=(k1-1)/n. This is rational, and a<r<b (you may want to
    provide some amplification of this last point) which proves the
    result for 0<a<b.

    RonL

    (The other cases:

    1. a<b<=0, proven by the above for a'=-b, b'=-a.

    2. a<0<b, well 0 is rational and so result holds trivially.)
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  3. #3
    Global Moderator

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    Nobody seems to have mentioned Archimedean Ordering.
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