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How do i prove the following:

For every irrational number q, given e>0, there exists natural numbers N and M such that |Nq-M|< e

Thank you.

3 Answers 3

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Hint: Look at the fractional parts of $q,2q, 3q, \dots $. There are an infinite of them squeezed into $(0,1)$.

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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 infinitess.

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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).