For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$

Ok, since you've asked and it doesn't fit into a comment, there you go. I'll do it on a circle since it's slightly easier to explain and I'll leave it to you to complete it in the case of an interval. let's say you have a circle of length $1$. you take 'steps' along the circle of an irrational length, let's say counter-clockwise. you'll never hit the same spot twice so for any fixed $\epsilon > 0$ you'll eventually find two 'steps' $a_n$ and $a_m$ such that $0 < |a_n - a_m| < \epsilon$. the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$ and so on. therefore if you let $k:= n-m$ and you only consider each $k$-th step you'll be going around the circle travelling a distance smaller than $\epsilon$ hence if you divide your circle into arcs of equal lengths greater than $\epsilon$ (but just slightly, say smaller than $2 \epsilon$) you'll have to land in each one of those in order to make your way all around the circle (because your steps are to small to jump over them). Every point of the circle is in at least one of those intervals which means that for each point of the circle you can find a number $a_j$ in your sequence that is closer than $2 \epsilon$ to it. Now conclude taking smaller and smaller $\epsilon$'s.

edit: oh, just note that I'm taking the distance along the circle, not the euclidean one