2
$\begingroup$

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...

$\endgroup$
3
$\begingroup$

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 $.

$\endgroup$
  • $\begingroup$ oh yeah, I see... Damn it's annoying being stuck on a question because you accidently turned a sign around... Thanks! $\endgroup$ – Nescio Nov 14 '12 at 11:18
2
$\begingroup$

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 $$\frac{n-1}{n+1}=\frac{(n+1)-2}{n+1}=1-\frac2{n+1}\;.$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Nescio Nov 14 '12 at 11:19
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Nov 14 '12 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.