Density of a subset of $\mathbb{R}$ Let $\alpha$ be an irrational number.
Is it true that the set $\{\frac{m}{\alpha} + n | m,n \in \mathbb{Z} \}$ dense in $\mathbb{R}$?
If it is, how do we prove it?

(This is almost the same as yanko's argument, but from a different perspective)
Lemma. If $H$ is a  subgroup of $\Bbb R$, then $H=c\Bbb Z$ for some $c$, or $H$ is dense in $\Bbb R$.
Proof. Let $H$ be a non-cyclic subgroup of $\Bbb R$.
In particular, $H$ contains non-zero elements and, as $x\in H$ implies $-x\in H$, it contains some positive elements.
Let $s=\inf(H\cap \left]0,\infty\right[)$, which is a non-negative real.
Let $U\subseteq \Bbb R$ be a non-empty open set. Then it contains some open interval $\left]a,a+\delta\right[$ with $\delta>0$.
If $s<\delta$, pick $h\in H$ with $h>0$ and $s\le h<\delta$.
For $n\in\Bbb N$ large enough, we have $-nh<a$. Fix such $n$. Then for $m\in\Bbb N$ large enough, we have $(m-n)h>a$. For the minimal such $m$, we have $(m-n-1)h\le a$, hence $a<(m-n)h\le a+h<a+\delta$, i.e., $h\in H\cap U$, contradiction.
Thus $s\ge \delta>0$. Assume $s\notin H$. Then we can pick $h\in H$ with $s< h<2s$ and $h'\in H$ with $s<h'<h$. Then $h-h'\in H$ contradicting $0<h'-h<s$.
Therefore $s\in H$ and $s\Bbb Z\subseteq H$. By assumption, $H\ne s\Bbb Z$. So pick $h''\in H\setminus s\Bbb Z$. Wlog. $h''>0$.
For $n\in \Bbb N$ large enough, we have $ns>h''$. Then $(n-1)s<h''<ns$ and so $0<h''-(n-1)s<s$, contradicting the definition of $s$ as infimum. $\square$
Now note that all non-zero elements of $c\Bbb Z$ are rational if $c$ is rational, and all non-zero elements of $c\Bbb Z$ are irrational if $c$ is irrational. As the group $H:=\{\frac m\alpha+n\min m,n\in\Bbb Z\,\}$ contains the non-zero reational $1$ and the non-zero irrational $\frac 1\alpha$, hence $H$ is not cyclic, and by the lemma, $H$ is dense.
