Given $\alpha\in \mathbb{R},$ do there exist arbitrarily large $m,k$ such that $\vert \alpha - \frac{k}{m}\vert < \frac{1}{m^2}$?

Asked Oct 02 '22 at 19:06

Active Oct 02 '22 at 19:46

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Given $\alpha\in \mathbb{R},\ n\in\mathbb{N}$ does there exist $m\in\mathbb{N}$ with $m>n,\ $ and $\ k\in\mathbb{Z}, $ such that

$$ \alpha \in\left[\frac{k}{m}-\frac{1}{m^2}, \frac{k}{m} + \frac{1}{m^2} \right]\ ? $$

I suspect the answer is yes somehow, partly due to the fact that the sum of the measure (width) of all these intervals, $\displaystyle\sum_{m\in\mathbb{N}} m\frac{2}{m^2},\ $ diverges. However, since many of the intervals overlap, the result is not so clear...

edited Oct 02 '22 at 19:46

asked Oct 02 '22 at 19:06

Adam Rubinson

20,052

Yes, this is a classical application of pigeonhole (and afaik is the reason pigeonhole is sometimes called "Dirichlet's principle"): https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem – Qiaochu Yuan Oct 02 '22 at 19:09
Oh, brilliant! I was not aware of that theorem. Thanks! – Adam Rubinson Oct 02 '22 at 19:10
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In the title, I think there is the impression that $k$ is constant. – dmtri Oct 02 '22 at 19:26
@QiaochuYuan Please make that an answer! - it seems quite interesting – whoisit Oct 02 '22 at 19:30
I think that this answer deals with this question. – robjohn Oct 02 '22 at 21:51

Given $\alpha\in \mathbb{R},$ do there exist arbitrarily large $m,k$ such that $\vert \alpha - \frac{k}{m}\vert < \frac{1}{m^2}$?

0 Answers0