Given an irrational number $\alpha$ and a real number $x$, show that for all $\epsilon > 0$ there exist integers $N$ and $M$ such that $|N\alpha-M-x|<\epsilon$.
I am stuck on this. I have tried approximating $\alpha$ with rational numbers but that doesn't seem to be getting me anywhere. Any hints? Is there a theorem that covers this?
The set $G:=\alpha\Bbb Z+\Bbb Z$ is a subgroup of $\Bbb R$. Let $a=\inf\bigl(G\cap(0,\infty)\bigr)$. Clearly, $0\le a\le 1$ . If $a>0$, then $G=a\Bbb Z$ and that would imply $\alpha\in\Bbb Q$. We conclude that $a=0$. Hence there exist $n,m\in\Bbb Z$ such that $0<n\alpha-m<2\epsilon$. Then a suitable multiple $kn\alpha-km$ will hit the interval $(x-\epsilon,x+\epsilon)$.