Is $ \{ m+rn\mid m,n \in \mathbb{Z}\}$ dense in $\mathbb{R}$ if $r\in\mathbb{Q}$? A classic exercise is in real analysis is to prove that for a fixed $\alpha\notin \mathbb{Q}$, 
$$\{ m+ \alpha  n\mid m,n \in \mathbb{Z} \}$$ is dense in $\mathbb{R}$.
Will this statement still be true if the parameter $\alpha$ is a rational number?
I need to stress here that r is a fixed rational number.
 A: For a fixed $\alpha\in\mathbb{Q}$, write $\alpha = \frac{a}{b}$ with $\gcd(a,b)=1$, $b\gt 0$. Then
$$\{m+\alpha n\mid m,n\in\mathbb{Z}\} = \left\{r\in\mathbb{Q}\;\left|\; r=\frac{t}{v},\ t,v\in\mathbb{Z}, \gcd(t,v)=1, v\gt 0, v|b\right\}\right..$$
To prove the right hand side is contained on the left hand side, 
given any $\frac{t}{v}$ as given, write $\frac{t}{v} = m + \frac{q}{v}$, with $m\in\mathbb{Z}$, $0\leq q\lt v$; then $\gcd(q,v)=1$ (since $t = vm + q$ is relatively prime to $v$). Since $v|b$, we can write $b=vs$. So $\frac{t}{v} = m + \frac{qs}{b}$. Since $\gcd(a,b)=1$, there exist $k,\ell$ such that $ka+\ell b = 1$. Hence $kqsa + \ell qsb = qs$. Hence
$$\frac{t}{v} = m+ \frac{qs}{b} = m + \frac{kqsa + \ell qsb}{b} = m + \ell qs + kqs\alpha.$$
For the converse inclusion, any element on the left hand side can be written as 
$$ m + n\alpha = \frac{bm+an}{b},$$
which when written in lowest terms will be of the form $\frac{t}{v}$ with $v|b$. This gives equality.
However, the right hand side describes a discrete subset of $\mathbb{R}$ (since any two elements of the set are at least $\frac{1}{b}$ apart), hence it is not dense. It is none other than the group $\mathbb{Z}[\frac{1}{b}]$.
