Prove that $x,y$ are positive real, $x/y$ is irrational, then the set $\{ mx+ny:m,n\in \Bbb Z \}$ .is dense in $\Bbb R$ Anyone can help with the following problem,

Prove that if $x,y$ are positive real, $x/y$ is irrational, then the set $\{ mx+ny:m,n\in \Bbb Z \}$ is dense in $\Bbb R$.

Thanks a lot!
 A: Sketch of proof: Note that it is equivalent to show that the set 
$$
\left\{\frac{xm + yn}{y} :m,n \in \Bbb Z \right\} = \left\{m\frac{x}{y} + n :m,n \in \Bbb Z \right\} = 
\left\{\alpha m + n :m,n \in \Bbb Z \right\} \quad (\alpha = x/y)
$$
is dense. Note that it sufficient to show that this set is dense over the interval $(0,1)$.  For any $x \in \Bbb R$, let $\langle x \rangle$ denote the "fractional part" of $x$ (i.e. the value of $x$ modulo $1$), so that $\langle x \rangle = x - \lfloor x \rfloor$.  It is equivalent to show that
$\{\langle \alpha m\rangle : m \in \Bbb Z\}$ is dense in $(0,1)$.
Now, convince yourself that the following is true:

For any $p,q \in \Bbb R$, $\langle p+q \rangle = \langle\langle p \rangle + \langle q \rangle \rangle$

Then, use the pigeonhole principle to prove the following:

For any $N = 1,2,3,\dots$, there exists integers $n_1,n_2$ such that $0 < \langle \alpha n_1 \rangle - \langle\alpha n_2 \rangle < 1/N$

Note also that $\langle \alpha n_1 \rangle - \langle\alpha n_2 \rangle = \langle \alpha(n_1 - n_2) \rangle$. So, for any $N$, there is an $n$ such that $0 < \langle \alpha n\rangle < 1/N$.
It follows that we can take successively better approximations of any $r \in [0,1]$ by selecting $n_1,\dots,n_k$ such that
$$
r \geq \langle \alpha n_1 \rangle + \cdots + \langle \alpha n_k \rangle = 
\langle \alpha(n_1 + \cdots + n_k)\rangle
$$
A: Hint:
First let's write the set, say $A$ as $A=\{m\alpha +n\mid m,n\in \mathbb{Z}\}$ where $\alpha=\frac{x}{y}$ is irrational. We want to show that given any open interval $(a,b)$ in $\mathbb{R}$, $(a,b)\cap A\neq \varnothing$. I will show two things :
1) Given any $\epsilon >0$, $\exists$ $m,n \in \mathbb{Z}$ such that $0<m\alpha+n<\epsilon$. This relies on the Archimedian property. My suggestion is to take $\alpha=\sqrt{2}$ and $\epsilon=0.05$ and see how the argument works.
2) $\forall k\in \mathbb{Z}, \forall x\in A,\ kx\in A$.
Now take any $(a,b)$ and choose an element $x\in A$ in $(0,\epsilon)$ from $1)$ and just "translate" by the appropriate integer using $2)$.
