Obviously, for any $\epsilon >0$, there exist $m,n\in \mathbb{N}$ such that$$\sqrt{2}\frac{n}{m}<\epsilon \; \textrm{.}$$ Is it also true that for all $\epsilon >0$, there exist $m,n\in \mathbb{N}$ such that$$\sqrt{2}mn<\epsilon \; \textrm{?}$$ If so, does it also hold for transcendental numbers?
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1In some sense, transcendental numbers can be approximated BETTER by rational numbers than irrational numbers (see Liouville's approxmation therem). – Peter Aug 06 '16 at 22:11

@Peter you mean algebraic? – Cameron Williams Aug 06 '16 at 22:18

@CameronWilliams Of course ... – Peter Aug 06 '16 at 22:19

1For $\sqrt2$ specifically you can also look at [this question](http://math.stackexchange.com/q/73262/11619). Some discussion about the general case is also there, Mariano's answer in particular. – Jyrki Lahtonen Aug 06 '16 at 22:19
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Here
https://en.wikipedia.org/wiki/Diophantine_approximation
it is shown that every irrational number $\alpha$ satisfies
$$\alpha\frac{p}{q}<\frac{1}{q^2}$$
for infinite many pairs $(p,q)$. If you multiply with $q$, you see that the answer to your question is "yes".
Peter
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