Is the set $\{\textrm{fractional part of } n\sqrt{2}|n \in \mathbb{N}\}$ dense in $[0,1]$

Question

Let $m,n \in \mathbb{N}$ such that $m$ is unequal to $n$ then the fractional part of $m\sqrt{2}$ is unequal to the fractional part of $n\sqrt{2}$. So my questions is: is it true that the set $\{\textrm{fractional part of } n\sqrt{2}|n \in \mathbb{N}\}$ is dense in $[0,1]$? My intuition is yes, however, I have no idea how to even begin.

Answer 1

It is true.

Let $S:=\{$fractional part of $n\sqrt{2}~|~ n\in \mathbb{N}\}$. And we denote the fractional part of $r\in \mathbb{R}$ as $0\le F(r)<1$.

In fact for $\epsilon>0$, we take an integer $N$ s.t. $\frac{1}{N}<\epsilon$ and by pigeonhole principle and your argument there is a $1\le k< N$ s.t. there are two or more elemtnts in $S\cap [\frac{k}{N}, \frac{k+1}{N}]$. Now there elements denote $F(n\sqrt{2}), F(m\sqrt{2})$ ($n<m$). Then $0<a:=F((m-n)\sqrt{2})<\frac{1}{N}<\epsilon$.

Now for all $0\le x\le 1$, we write $x$ as $x=Ma+y$ where $M\in \mathbb{N}, 0\le y<a$. Then $x-F(M(m-n)\sqrt{2})=x-Ma=y<\epsilon$. So $S$ is dense in [0,1].