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it might be a simple question.

I need to prove that $1$ is the supremum of the following set: $$A=\left\{\frac{m}{n}\mathrel{}\middle|\mathrel{} m<n\wedge m,n\in \mathbb{N}\right\}$$ 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) –  Mark Bennet May 6 '12 at 15:24
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1 Answer

up vote 1 down vote accepted

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

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I'm not sure how to use it. I guess I should look at $\frac{1}{n+1}$ as $\varepsilon$, but still I don't understand how to build $x$ from it :( –  RB14 May 6 '12 at 15:41
    
@RB14 Given $\epsilon>0$, select $n$ such that ${1\over n+1}<\epsilon$ (why can you do this?). Then show that $1-{1\over{n+1}} >1-\epsilon$. –  David Mitra May 6 '12 at 15:45
    
yeah, apprently I'm a bit idiot :) –  RB14 May 6 '12 at 15:48
    
@RB14 Not at all; this stuff is hard unless you've been doing it for 15+ years... –  David Mitra May 6 '12 at 15:49
    
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! –  RB14 May 6 '12 at 15:54
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