Denseness of the set $\{ m+n\alpha : m\in\mathbb{N},n\in\mathbb{Z}\}$ with $\alpha$ irrational [duplicate]

How to prove that the set $\{ m+n\alpha : m\in\mathbb{N},n\in\mathbb{Z}\}$, ($\alpha$ is an irrational number) is dense in $\mathbb{R}$?

Using the fact every additive subgroup of $\mathbb{R}$ is either discrete or dense in $\mathbb{R}$, we can prove that the set $\{ m+n\alpha : m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$. But how to prove the denseness of the above set?

In this answer I use the pigeonhole principle to give a short, easy proof that if $\alpha$ is irrational, then $\{n\alpha\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$, where $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$. It’s easy to check that the proof still works if we replace $\Bbb Z$ either by $\Bbb Z^+$ or by $\Bbb Z^-$.

Now let $(a,b)$ be an open interval of real numbers; we can certainly find an interval $(c,d)\subseteq(a,b)$ such that $(c,d)\cap\Bbb Z=\varnothing$. Let $k=\lfloor c\rfloor$; then $0<c-k<d-k<1$, so there is an $n\in\Bbb Z^+$ such that $n\alpha\bmod 1\in(c-k,d-k)$. If $\ell=\lfloor n\alpha\rfloor$, $n\alpha=\ell+(n\alpha\bmod 1)$, so

$$c-k+\ell<n\alpha<d-k+\ell\;.$$

Now let $m=k-\ell$ to find that $c<m+n\alpha<d$ and hence that $m+n\alpha\in(a,b)$.

This doesn’t quite work, because $m$ may be negative. To get around that problem, we pick infinitely many pairwise disjoint subintervals of $(c,d)$. Let $\langle c_k:k\in\Bbb N\rangle$ be a strictly increasing sequence lying entirely within the interval $(c,d)$, and for $k\in\Bbb N$ let $I_k=(c_k,c_{k+1})$. If $\alpha>0$, the argument that I applied to $(c,d)$ applies equally well to each of the intervals $I_k$, so for each $k\in\Bbb N$ there is an $n_k\in\Bbb Z^-$ such that $\lfloor c_k\rfloor+(n_k\alpha\bmod 1)\in I_k$; if $\alpha<0$, for each $k\in\Bbb N$ there is an $n_k\in\Bbb Z^+$ such that $\lfloor c_k\rfloor+(n_k\alpha\bmod 1)\in I_k$. In either case the numbers $n_k\alpha$ are negative, and because the intervals $I_k$ are pairwise disjoint, the numbers $n_k$ are distinct, and the set $\{n_k\alpha:k\in\Bbb N\}$ is unbounded. There must therefore be a $k\in\Bbb N$ such that $\lfloor c_k\rfloor-\lfloor n_k\alpha\rfloor>0$, and if we set $m=\lfloor c_k\rfloor-\lfloor n_k\alpha\rfloor$, then $m\in\Bbb N$, and we have $m+n_k\alpha\in I_k\subseteq(a,b)$.

Thus, $\{m+n\alpha:m\in\Bbb N\text{ and }n\in\Bbb Z\}$ meets every non-empty open interval $(a,b)$ and is therefore dense in $\Bbb R$.