# Proving the sup of $\{\frac{n-m}{n+m}|n,m\in \mathbb{N}, m<n\}$

The question was to prove $\{\frac{n-m}{n+m}|n,m\in \mathbb{N}, m<n\}$ is bounded, and have a supremum and an infimum.
Proving it's bounded was easy, But I got stuck on proving 1 is it's supremum.

This is what I have:
i. We'll hypothesize that $\sup\left(A\right)=1$
This means that $\forall\epsilon>0,\exists x\in A,x>1-\epsilon$

$$\frac{n-m}{n+m}>1-\epsilon\implies n-m<\left(1-\epsilon\right)\left(n+m\right)\implies$$ $$\implies n-m<n+m-\epsilon n-\epsilon m\implies\epsilon n<\left(2-\epsilon\right)m$$

But I'm not sure how I continue from here. If epsilon is $\geq1$ then this can never be right, and I'm actually confused as to why (or even if) the steps I did untill now are true. Can I just find an example for when $\epsilon\geq1$ and then expain why a solution exist for when $\epsilon<1$ (Though I'm not sure how to do that either...)?

I'm assuming proving supremum is the same, so I guess this is a double question :p

p.s. I don't know how to tag this, but it's homeworks in infi so I guess calculus it is...

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From $n-m<n+m$ (and $n+m>0$) we find immediately that $\frac{n-m}{n+m}<1$, so $1$ is an upper bound. You want to show that $1-\epsilon$ is not an upper bound if $\epsilon>0$. That is, you want to exhibit $n,m$ with $\frac{n-m}{n+m}>1-\epsilon$ or equivalently with $\epsilon n >(2-\epsilon)m$ (you mixed $>$ and $<$, otherwise all your $\Rightarrow$'s are in fact $\Leftrightarrow$'s). Dividing by $\epsilon$ then gives $n>\frac{2-\epsilon}\epsilon m$ as condition, which is easily fulfilled by choosing $m=1$ and $n>\frac{2-\epsilon}\epsilon$.

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oh yeah, I see... Damn it's annoying being stuck on a question because you accidently turned a sign around... Thanks! –  Nescio Nov 14 '12 at 11:18

HINT: For a given $\epsilon>0$ how big does $n$ have to be to ensure that $$\frac{n-1}{n+1}>1-\epsilon\;?$$ It may help to notice that $$\frac{n-1}{n+1}=\frac{(n+1)-2}{n+1}=1-\frac2{n+1}\;.$$

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Though I see how I can use it in this question, appereantly it wasn't necessery. Thanks for this neat idea though, I'm sure it will be useful for me in the future. –  Nescio Nov 14 '12 at 11:19
@Nescio: No, it’s not necessary; it’s just the easiest way that occurred to me. You’re welcome; this kind of dividing out is often useful. –  Brian M. Scott Nov 14 '12 at 11:21