I have recently recalled a problem from my school days, where I was asked if for every $\varepsilon > 0$ there exist positive integers $m, n$ such that $|m\sqrt 2 - n|< \varepsilon$. I have just realized, that I do not really know of a simple way to solve this problem. Furthermore, would be good if that solution could be applied also to any $q\in \Bbb R_+$ instead of $\sqrt2$.
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Like ${m\sqrt{2}}$ is dense in $[0,1)$? – Alexey Burdin Jul 21 '20 at 19:52
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Well, the equivalent to the OP is that it is dense at ${0}$ @AlexeyBurdin – SBF Jul 21 '20 at 19:56
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Well, $m=n=0$ is always a solution. – Vincenzo Tibullo Jul 21 '20 at 20:11
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thanks @enzotib, I've added the (strict) positivity condition explicilty – SBF Jul 21 '20 at 20:12
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1https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) – saulspatz Jul 21 '20 at 20:33
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Even easier with Dirichlet's approximation theorem. – rtybase Jul 21 '20 at 21:42
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There is also Liouville's lemma for lower bounds for algebraic numbers. One example here. – rtybase Jul 21 '20 at 21:45
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Let $0<q\in\Bbb R\setminus\Bbb Q$, $M=\{mq-n:m,n\in\Bbb Z\}$ and $a=\inf\{x:x\in M\land x>0\}$. Assume on contrary $a>0$.
If $a\in M$, then for every $0<x\in M$ we have $0\leq x-\lfloor x/a\rfloor a<a$ and since $x-\lfloor x/a\rfloor a\in M$ this implies $x=\lfloor x/a\rfloor a\in a\Bbb Z$. Then $q=au$ and $1=av$ for some $u,v\in\Bbb Z$, and we get the contradiction $q=u/v\in\Bbb Q$.
If $a\notin M$, by definition of infimum, then there exists $m',n'\in\Bbb Z$ such that $a<m'q-n'\lt 2a$ and, again by definition of infimum, there exists $m,n\in\Bbb Z$ such that $a<mq-n\lt m'q-n'$. Then $0\lt(m'-m)q-(n'-n)\lt a$ - a contradiction.
Fabio Lucchini
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It might be worth explaining why there exist such $m,m'.n.n'$ in the last paragraph? (The pigeonhole principle argument is certainly the standard way of handling this; I just haven't seen it phrased quite this way before.) – Steven Stadnicki Jul 21 '20 at 21:25
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