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I think the answer should be $1$, but am having some difficulties proving it. I can't seem to show that, for any $\delta$ and $n > m$, $|n - k(2\pi)| &lt \delta$. Is there another approach to this or is there something I'm missing?

Asaf Karagila
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dhz
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  • I can understand that you have some difficulty proving your assessment. Maybe it is wrong and the limit does in fact not exist... – Fabian Apr 25 '12 at 18:18
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    @Fabian the lim sup of any bounded sequence of real numbers exists. –  Apr 25 '12 at 18:22
  • @countinghaus: i don't see a limsup... (maybe there is some problem with the display of the page on my computer) – Fabian Apr 25 '12 at 18:24
  • You might use the result here that given an irrational $x$ and a positive integer $n$, there exists at least one positive integer $j\le n$ for which $jx$ and the integer $m$ nearest to $jx$ differ in magnitude by less than $1/(n+1)$. – David Mitra Apr 25 '12 at 18:24
  • Interesting: the limsup is shown as lim on my computer???? Does somebody else with the mathjax setting svg have the same problem? – Fabian Apr 25 '12 at 18:25
  • Is it not true that $\limsup \cos(an) = 1$ for all $a$, rational or irrational? And isn't that easier by far to prove than $\pi$ is irrational? – GEdgar Apr 25 '12 at 18:41
  • @Fabian, yes, there is a bug with \limsup and \liminf in SVG mode. It will be fixed in the net release of MathJax. – Davide Cervone Apr 26 '12 at 10:42

2 Answers2

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No, that's exactly how you should show it. You can get what you want by using this question:

For every irrational $\alpha$, the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

Community
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You are on the right track. If $|n-2\pi k|&lt\delta$ then $|\frac{n}{k}-2\pi|&lt\frac \delta k$. So $\frac{n}{k}$ must be a "good" approximation for $2\pi$ to even have a chance.

Then it depends on what you know about rational approximations of irrational numbers. Do you know about continued fractions?

Thomas Andrews
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