Show that for any $\epsilon>0$, there are positive integers, $m,n$ such that $|m-n\pi|<\epsilon$. Show that for any $\epsilon>0$, there are positive integers, $m,n$ such that $|m-n\pi|<\epsilon$. I suppose there is some theorem in real analysis can be used here? Can anyone point out what it might be?
Is there such result if $\pi$ is replaced with other irrational numbers, like $\sqrt{2}$?

A simple proof: Let $\alpha$ be irrational. The set 
$$G:=\{\,n+m\alpha\mid n,m\in\Bbb Z\,\} $$
is a non-trivial subgroup of $\Bbb R$.
Let $G^+=\{\,x\in G\mid x>0\,\}\ne\emptyset$ and $h=\inf G^+$. The desired result is equivalent to $h=0$.
So assume $h>0$.
Then we can $x\in G$ with $h\le x<2h$.
If $G\ne x\Bbb Z$, we can also pick $y\in G^+\setminus x\Bbb Z$. For suitable $n\in \Bbb Z$, $z:=y-nx\in G$ is in $(0,x)$ and because $x<2h$, one of $z,x-z$ is in $(0,h)$, contradiction. Therefore $G=x\Bbb Z$.
In particular, $1\in G$ and $\alpha\in G$ implies that $1=nx$, $\alpha=mx$ for suitable $n,m\in\Bbb Z$, and thus $\alpha=\frac nm\in\Bbb Q$.
