Multiples of an irrational number forming a dense subset Say you picked your favorite irrational number $q$ and looking at $S = \{nq: n\in \mathbb{Z} \}$ in $\mathbb{R}$, you chopped off everything but the decimal of $nq$, leaving you with a number in $[0,1]$.  Is this new set dense in $[0,1]$? If so, why? (Basically looking at the $\mathbb{Z}$-orbit of a fixed irrational number in $\mathbb{R}/\mathbb{Z}$ where we mean the quotient by the group action of $\mathbb{Z}$.)
Thanks! 

A bit of a late comer to this question, but here's another proof:
Lemma: The set of points $\{x\}$ where $x\in S$, (here $\{\cdot\}$ denotes the fractional part function), has $0$ as a limit point.
Proof: Given $x\in S$, Select $n$ so that $\frac{1}{n+1}\lt\{x\}\lt\frac{1}{n}$. We'll show that by selecting an appropriate $m$, we'll get: $\{mx\}\lt\frac{1}{n+1}$, and that would conclude the lemma's proof.
Select $k$ so that $\frac{1}{n}-\{x\}\gt\frac{1}{n(n+1)^k}$. Then:
$$
\begin{array}{ccc}
\frac{1}{n+1}  &\lt&  \{x\}      &\lt&  \frac{1}{n} - \frac{1}{n(n+1)^k} \\
1              &\lt&  (n+1)\{x\} &\lt&  1+\frac{1}{n} - \frac{1}{n(n+1)^{k-1}} \\
  & &  \{(n+1)x\} &\lt&\frac{1}{n} - \frac{1}{n(n+1)^{k-1}}
\end{array}
$$
If $\{(n+1)x\}\lt\frac{1}{n+1}$, we are done. Otherwise, we repeat the above procedure, replacing $x$ and $k$ with $(n+1)x$ and $k-1$ respectively. The procedure would be repeated at most $k-1$ times, at which point we'll get: 
$$
\{(n+1)^{k-1}x\}\lt\frac{1}{n} - \frac{1}{n(n+1)}=\frac{1}{n+1}.
$$
Proposition: The set described in the lemma is dense in $[0,1]$.
Proof: Let $y\in[0,1]$, and let $\epsilon\gt0$. Then by selecting $x\in S$ such that $\{x\}\lt\epsilon$, and $N$ such that $N\cdot\{x\}\le y\lt (N+1)\cdot\{x\}$, we get: $\left|\,y-\{Nx\}\,\right|\lt\epsilon$.
