Proving for all $n\in \mathbb{N}$, $(2\pi k, 2\pi k+1/n) \cap\mathbb{N} \ne\emptyset $ for some $k \in \mathbb{N}$

Question

I have been trying to prove the following for quite a while but have been unable to do so:

For all $n\in \mathbb{N}$, $(2\pi k, 2\pi k+1/n) \cap\mathbb{N} \ne\emptyset $ for some $k \in \mathbb{N}$.

I believe $2\pi$ is not something special in the claim; if $2\pi$ is replaced by some irrational number greater than $1$, it still holds.

I tried using the density of rationals in the reals but clearly it won't help (unless used in a very clever way!). Hints in the direction to prove this would be appreciated.

score 4 · Accepted Answer · answered Jan 06 '20 at 05:48

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By the pigeonhole principle, if you are given more than $n$ points in $[0,1]$ then there will be at least two of them within distance $\tfrac{1}{n}$ of each other.

You can apply this to $2\pi k\mod 1$ to find two integers $k,m$ such that $2\pi k$ and $2\pi m$ are within $\tfrac{1}{n}$ modulo $1$. Then consider $2\pi(k-m)$...

answered Jan 06 '20 at 05:48

pre-kidney

30,223

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What is the meaning of an irrational mod n? – Dhanvi Sreenivasan Jan 06 '20 at 05:49
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Given any real number $r$, the meaning of $r\mod 1$ is the fractional part of $r$, i.e., $r-\lfloor r\rfloor$ where $\lfloor r\rfloor$ is the greater integer less than or equal to $r$. – pre-kidney Jan 06 '20 at 05:51
Oh got it, quite clever. – Dhanvi Sreenivasan Jan 06 '20 at 05:52
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2009&task=show_msg&msg=3451.0001 – Dhanvi Sreenivasan Jan 06 '20 at 05:52

Proving for all $n\in \mathbb{N}$, $(2\pi k, 2\pi k+1/n) \cap\mathbb{N} \ne\emptyset $ for some $k \in \mathbb{N}$

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