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?

Yes it is
Lemma: Let $H$ be a subgroup of $\mathbb{R}$, then $H$ is dense if and only if there exists a convergence sequence $h_n\rightarrow \xi$ for some $\xi\in \mathbb{R}$ and $h_n\in H$ with the property that $h_n$ is not eventually constant.
Proof: The first direction is obvious, For the other direction, let $h_n\rightarrow \xi$. Let $\varepsilon>0$, as $h_n$ is a Cauchy sequence there exists $h\in H$ for which $|h|<\varepsilon$, as $H$ is a group you have that $kh\in H$ for every $k\in\mathbb{Z}$, it follows that $\{kh:k\in\mathbb{Z}\}$ is $\varepsilon$-dense in $\mathbb{R}$, as $\varepsilon$ is arbitrary, we have that $H$ is dense in $\mathbb{R}$.
Now, the set you mentioned is a group, hence it is enough to find a convergence sequence which is not a eventually constant.
For every $m$, let $n_{m} = \lfloor m/\alpha\rfloor$, and consider the set $\frac{m}{\alpha}-n_m$ this is a set of infinitely many different numbers in $[0,1]$ hence there is a converging sub sequence $\frac{m_k}{\alpha}-n_{m_k}$, using the Lemma this completes the proof.
Note: another way to do this is to use the fact that an irrational rotation of the circle has a dense orbit (with rotation $1/\alpha$)
