Approximation of integer by multiple of irrational number

Question

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$$|\sqrt{2}m-n|<\epsilon \; \textrm{?}$$ If so, does it also hold for transcendental numbers?

In 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
For $\sqrt2$ specifically you can also look at this question. Some discussion about the general case is also there, Mariano's answer in particular. — Jyrki Lahtonen, Aug 06 '16 at 22:19

score 3 · Accepted Answer · answered Aug 06 '16 at 22:15

3

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".

answered Aug 06 '16 at 22:15

Peter

84,454

1

Thank you, that's what I was looking for – Philmore Aug 06 '16 at 22:20

Approximation of integer by multiple of irrational number

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