For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$ 
Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$. 

Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.
I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense 
and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer. 
This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$. 
Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.
I think this problem has not been solved.
Thanks!
 A: A pictoral representation of mm-aops's proof. I also avoid use of: "the distance from $a_n$ to $a_m$ is the same as between $a_{n-m}$ and $a_0 = 0$."
Let $\ \varepsilon = \frac{1}{200}.\ $ Then, we split the circle up into $201$ arcs of equal length. Think of these arcs as "boxes" in the sense of the pigeonhole principle: then there must be $\ n,m \leq 201\ $ such that $\ \mid a_{n} - a_m\mid \leq \frac{1}{201} <\varepsilon.$
Suppose the first two such $\ n,m\ $ are $\ a_{45}\ $ and $\ a_{162}\ .$
Any point on the circle must lie within $\ \varepsilon=\frac{1}{200}\ $ distance from $\ a_{45}\ $ or $\ a_{162}\ $ or $\ a_{279}\ $ or $\ a_{396}\ $ or ...

A: 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
A: $f \colon x \rightarrow x-\lfloor x \rfloor$ 
$F=\{f(nx),n \in \Bbb{Z}\}$
$F \neq \emptyset$ and $x \ge 0$ $\exists d=\inf(F)$
We can prove that $d=0$ (suppose a $\gt$ 0 and ...)
hence $\forall$ y $\gt$ 0 $\exists$ n $\in$ $\Bbb{N}$ , $y \gt f(nx) \gt 0$ 
$a,b \in [0;1[$
if $b-a \gt 0$ $\exists n \in \Bbb{N}$, $b-a \gt f(nx) \gt 0$
$\exists p \in \Bbb{N}$ , pf(nx) $\gt$ a (1) , if we suppose that p is the smallest number who verifies (1) we have that (p-1)fn(x) $\lt$ a therefore
 pf(nx) $\lt$ a+f(nx) $\lt$ b 
therefore pf(nx) $\in$ [a;b] and pf(nx)=f(pnx) 
we can conclude about the density 
First time I use this way of writing I hope I have been clear
