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Proving that if $\omega \in [0, \frac{n}{n+1} ) \implies \omega \in [0, 1)$ is easy but I'm struggling with the other way round. How do we know that $\exists n: \omega < \frac{n}{n+1}$? I think I should use the archimedean principle but not sure how?

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    $\begingroup$ Do you know that $\displaystyle \lim_{n \to \infty} \frac{n}{n+1} = 1$? If so, use that. $\endgroup$
    – user169852
    Commented Jun 13, 2021 at 20:25
  • $\begingroup$ @Bungo How is this not an answer? Upvoting comments is not quite as satisfying as answers :) $\endgroup$ Commented Jun 13, 2021 at 20:28
  • $\begingroup$ @SeverinSchraven It's a judgment call, but for me a one-line hint doesn't really qualify as an answer unless the OP asked for a one-line hint. Also, because there's no context, it's not clear if the OP knows about limits yet, which is why I posed the comment as a question. $\endgroup$
    – user169852
    Commented Jun 13, 2021 at 20:29
  • $\begingroup$ @Bungo I would disagree. If the one-line contains all the info needed, I would even prefer that. This really is the essence. But of course that is a matter of one's personal philosophy. $\endgroup$ Commented Jun 13, 2021 at 20:31
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    $\begingroup$ @Bungo I do like the hint. Since we know that the limit of $\frac{n}{n+1}$ is 1, we can set $\epsilon = 1 - \omega$ and thus $\exists N$ such that $\forall n > N,$ $\omega < \frac{n}{n+1}$ $\endgroup$
    – Darby Bond
    Commented Jun 13, 2021 at 20:45

2 Answers 2

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$$\omega < \frac{n}{n+1} \iff n \omega + \omega < n \iff \omega < n(1 - \omega) \iff \frac{\omega}{1-\omega}<n$$ Take for instance $n = \left\lfloor\frac{\omega}{1-\omega}\right\rfloor+1$.

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The Archimedean property of real numbers says:

given $\ x>0,\ y\in\mathbb{R},\ \exists n\in\mathbb{N}\cup\{0\}\ $ such that $(n+1)x>y.$

Choosing $\ y=1\ $ and $\ x = 1-\omega\ (>0),\ $ it follows that $\ \exists\ n+1\in\mathbb{N}\ $ such that $\ \frac{1}{n+1}<1-\omega,\ $ which implies that $\ 1-\frac{1}{n+1} > 1 - (1-\omega) = \omega,\ $ i.e. $\ \frac{n}{n+1}>\omega.$

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