# About fractional part of $nx$

Demonstrate that if a number $x$ is irrational and $c(n):=nx-\lfloor nx \rfloor$, then for any $a$ and $b$ belonging to $[0,1]$ infinitely many members of the sequence $c(n)$ lie in the interval $(a,b)$.

(Does this theorem have a name?)

-

You are asking if the collection of the points $c(n)$ is dense in $[0,1]$, which I believe is a theorem of Dirichlet. This can be shown using the pigeonhole principle. It is important to note that since $x$ is irrational, $c(n)$ is injective, in the sense that if $c(n)=c(n')$, then $n=n'$.
We will show that $0$ is a limit point, and leave the generalization to you. For any positive integer $m$, cut $[0,1]$ into $m$ pieces of equal size $[0,\frac{1}{m}]$,$(\frac{1}{m},\frac{2}{m}],\ldots,(\frac{m-1}{m},1]$. By the pigeonhole principle, there must exist two points $c(n)$ and $c(n')$ in the same piece, with $n>n'$. It follows that $c(n-n')\in[0,\frac{1}{m}]$, completing the proof.