dense set on $[0,1]$ and more... I want to prove that $A=\{\{n\alpha\} \mid \text{where }n\text{ is an integer number}\}$ is dense in $[0,1]$ and  I do not know how to continue.
$\alpha$ is irrational.
I obtained that:
for a given $\varepsilon>0$ there are $i$, $j$ such that: $|\{i\alpha\}-\{j\alpha\}| <\varepsilon$.
How should I continue. I would really appreciate an answer in the easiest way possible and one that is complete as well.
 A: Hint
The subset $G = \{n\alpha + m \mid (m,n) \in \mathbb Z^2\}$ is an additive subgroup of the reals. You probably know (or should prove) that the additive subgroups of the reals are either discrete or dense.
You can prove that $G$ discrete implies $\alpha \in \mathbb Q$. In contradiction with the hypothesis. Hence $G$ is dense.
Finally prove that $G$ dense in $\mathbb R$ implies that $A$ is dense in $[0,1]$.
Note: to prove that the additive subgroups of the reals are either discrete or dense, consider $a = \inf \{g \in G \mid g \gt 0\}$ and the options $a=0$ or $a \gt 0$.
A: A relatively simple proof hinges on the so called Dirichlet lemma: 
If $\alpha$ is irrational then for any $n$ one can find $p$ and $q$ such that 
$|\alpha -\frac{p}{q}| < \frac{1}{nq}$ with $q<n$. (p,q,n - integers)
So, in particular, one can find $p_n->\infty$ and $q_n->\infty$ such that
$$|\alpha -\frac{p_n}{q_n}| < \frac{1}{q_n^2} $$
From this one easily deducts that ${\{\alpha n\}}$  has a cluster point either at $0$ or at $1$.
Say, it has a cluster point at $0$. Then, if ${\{\alpha n\}}$ is "small" then
$${\{\alpha n\cdot k\}} = k{\{\alpha n\}}$$ which means density.
 Lastly, using similar trick one shows that if $1$ is a cluster point of ${\{\alpha n\}}$ then so is $0$.
