How do I prove the following:
For every irrational number $q$, given $\varepsilon>0$, there exist natural numbers $N$ and $M$ such that $|Nq-M|<\varepsilon$.
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Hint: Look at the fractional parts of $q,2q, 3q, \dots $. There are an infinite of them squeezed into $(0,1)$.
Aryabhata
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1How does this prove the question? For example: all of the fractional parts might lie in (0.1,0.9) – Agile_Eagle Feb 26 '21 at 18:11
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@Agile_Eagle: $nq=n(\lfloor q\rfloor+{q})$, ergo ${nq}={n{q}}$. If ${q}=\overline{0.a_1a_2a_3\ldots}$ is irrational, you can get any possible value in $(0,1)$ as a fractional part of $nq$ for some $n=10^k$. They cannot all lie in $(0.1,0.9)$, because, say a multiple has fractional part $0.10001$, then multiplying it by $9$, we get a multiple with fractional part $0.90009\gt 0.9$ – Prasun Biswas Jul 25 '21 at 20:24
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Here's a small variant:
Consider the quotient ${\Bbb R}/{\Bbb Z}$ which can be identified to the circle $S^1$. The irrationality of $q$ is equivalent to the injectivity of the quotient map $\pi:{\Bbb R}\rightarrow S^1$ restricted to ${\Bbb Z}q$. The assertion amounts to prove that for any open neighborhood $V\ni0$ in $S^1$ there is always a $n\bar{q}\in V$ with $n\neq0$, i.e. $\pi({\Bbb Z}q)\cap(V\setminus(0))\neq\emptyset$
Consider the translates $V_n=n\bar{q}+V\subset S^1$ for $n\in{\Bbb Z}$ (still open, of course). If you can find an $m\in{\Bbb Z}$ such that there exists $n\bar{q}\in V_m\setminus(m\bar{q})$ then $(n-m)\bar{q}\in V\setminus(0)$ and you are done.
If not, it means that for all $m\in{\Bbb Z}$, $\pi({\Bbb Z}q)\cap V_m=\{m\bar{q}\}$. But this says that $\pi({\Bbb Z}q)$ is discrete in $S^1$ and since $S^1$ is compact, this is in contradiction with its infinitness.
Andrea Mori
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Not to toot my own horn or anything, but my answer to this question
Can we make $\tan(x)$ arbitrarily close to an integer when $x\in \mathbb{Z}$?
includes an outline of how you would prove this (similar to Moron's hint).
