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Math Help - [SOLVED] Stumped - Help get me started

  1. #1
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    [SOLVED] Stumped - Help get me started

    Ok, so I have the following to prove:

    If a and b are real numbers s.t. 0<a<b, then there is a natural number n s.t. a<b-1/n<b.

    I am confused as to what I should do to prove this. Is this a two part proof? If so, what parts do I prove? Thanks.
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  2. #2
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    Quote Originally Posted by spearfish View Post
    Ok, so I have the following to prove:

    If a and b are real numbers s.t. 0<a<b, then there is a natural number n s.t. a<b-1/n<b.

    I am confused as to what I should do to prove this. Is this a two part proof? If so, what parts do I prove? Thanks.
    Consider the set of natural numbers

    \left\{n \in \mathbb{N}: n > \frac{1}{b-a} \right\} This is a non-empty set of natural numbers that is bounded from below.

    So by the well ordering of the postive integers this set has a least element n_0

    Obviously b > b-\frac{1}{n_0}

    Now we just need to show that b-\frac{1}{n_0}> a

    we know that

    n_0> \frac{1}{b-a} \iff b-a > \frac{1}{n_0} \iff b-\frac{1}{n_0}> a

    Putting the above two inequalities together we get

    a < b-\frac{1}{n_0}< b
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  3. #3
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    This makes sense. Thanks a bunch!
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