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it might be a simple question.

I need to prove that $1$ is the supremum of the following set: $A=\left\{\frac{m}{n}\mathrel{}\middle|\mathrel{} m So, actually I need to prove 2 things:

  1. $\forall x\in A .x\le 1$
  2. $\forall \varepsilon>0 \ \exists x\in A .x>1-\varepsilon$

So, the the first requirement is easy to prove, from the defeinition of $A$. but the second is not so simple for me. I've tried to show that this $x$ exists, but I can't show how it looks like.. I guess I should express it with $\varepsilon$, but I have no success so far.

If there are other ways to prove it, it'll fine too, and I'll be happy to hear about them.

Thanks!

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    Can you find an integer $N$ with N>\frac 1 {\epsilon}? (The Axiom of Archimedes, if you have encountered it)2012-05-06

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Hint: ${n\over n+1} =1-{1\over n+1}$.

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    there supposed to be a continuation to that comment, but I accidently pressed enter, and my chrome browser did not allow me to delete that comment. any way, I get it now, I should select $x$ to be $\frac{n}{n+1}$ where n > \frac{1}{\epsilon}. that is possible from the reason Mark mentioned in his comment - the axiom of archimedes. thanks guys!2012-05-06