# Prove: every irrational number q , given e &gt; 0 , there exists natural numbers

Prove: every irrational number $q$, given $e>0$, there exists natural numbers $N$ and $M$ such that $|Nq-M|
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SweallySnicles3
Consider the quotient $\mathbb{R}/\mathbb{Z}$ which can be identified to the circle ${S}^{1}$. The irrationality of $q$ is equivalent to the injectivity of the quotient map $\pi :\mathbb{R}\to {S}^{1}$ restricted to $\mathbb{Z}q$. The assertion amounts to prove that for any open neighborhood $V\ni 0$ in ${S}^{1}$ there is always a $n\overline{q}\in V$ with $n\ne 0$, i.e. $\pi \left(\mathbb{Z}q\right)\cap \left(V\setminus \left(0\right)\right)\ne \mathrm{\varnothing }$
Consider the translates ${V}_{n}=n\overline{q}+V\subset {S}^{1}$ for $n\in \mathbb{Z}$ (still open, of course). If you can find an $m\in \mathbb{Z}$ such that there exists $n\overline{q}\in {V}_{m}\setminus \left(m\overline{q}\right)$ then $\left(n-m\right)\overline{q}\in V\setminus \left(0\right)$ and you are done.
If not, it means that for all $m\in \mathbb{Z}$, $\pi \left(\mathbb{Z}q\right)\cap {V}_{m}=\left\{m\overline{q}\right\}$. But this says that $\pi \left(\mathbb{Z}q\right)$ is discrete in ${S}^{1}$ and since ${S}^{1}$ is compact, this is in contradiction with its infinitness.
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