# Density of positive multiples of an irrational number

Let $x$ be irrational. Use $\{r\}$ to denote the fractional part of $r$: $\{r\} = r - \lfloor r \rfloor$. I know how to prove that the following set is dense in $[0,1]$: $$\{\{nx\} : n \in \mathbb{Z}\}.$$ But what about $$\{\{nx\} : n \in \mathbb{N}\}?$$ Any proof that I’ve seen of the first one fails for the second one.

Really? I thought exactly the same proof worked for $\Bbb N$.

Let $k\in\Bbb Z$ with $k\ne 0$ and define $$f(t)=e^{2\pi ikt}.$$

Then $f$ has period $1$, and $$\frac1N\sum_{n=0}^{N-1}f(nx) =\frac1N\sum_{n=0}^{N-1}\left(e^{2\pi ikx}\right)^n=\frac1N \frac{e^{2\pi ikxN}-1}{e^{2\pi ikx}-1}\to0\quad(N\to\infty).$$ So the usual approximation shows that $$\frac1N\sum_{n=0}^{N-1}f(nx)\to\int_0^1 f(t)\,dt$$for $f\in C(\Bbb T)$ and you're done, as usual.

How is this any different from the case $n\in\Bbb Z$?

• How does that imply that $\{ \exp (2\pi i {n x}) :n\in N\}$ is dense in the unit circle in $C$ ? Anyway, a proof from the most elementary method works for N just as well as for Z. – DanielWainfleet Jan 24 '16 at 5:33
• @user254665 Say those points are not dense. There is then an interval, or arc, $I$ on the circle that contains none of the points. Take $f\in C(\Bbb T)$ such that $f=0$ on the complement of $I$ but $\int f > 0$. Then $\frac1N\sum_0^{N-1}f(nx)=0$ for every $N$, so it does not converge to $\int f$. (Yes, the pigeonhole argument works as well. I really can't think of an argument that works for $n\in\Bbb Z$ but not $\Bbb N$. In any case, ignoring the question of which is simpler, this argument shows more than just that the points are dense, it shows they are asymptotically uniformly distributed.) – David C. Ullrich Jan 24 '16 at 14:21
• very nice integral proof, which yields extra info (distribution) too. – DanielWainfleet Jan 24 '16 at 20:10
• @user254665 Yes it is very nice. The equidistribution is Weyl's theorem - I think it's his proof, not sure, in any case it's a standard thing. – David C. Ullrich Jan 24 '16 at 21:27

A minor modification of the pigeonhole argument works. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\in\{1,\dots,m+1\}$ and $k\in\{0,\dots,m-1\}$ such that $\frac{k}m\le\{ix\},\{jx\}<\frac{k+1}m$; clearly $\{|(j-i)x|\}<\frac1m$. Let $\ell$ be the largest positive integer such that $\ell\{|(j-i)x|\}<1$, let $A_m=\{n|j-i|x:0\le n\le\ell\}$, and let $D_m=\big\{\{y\}:y\in A_m\big\}$.

If $x>0$, every point of $[0,1)$ is clearly within $\frac1m$ of $A_m=D_m$. If $x<0$, then

$$D_m=\{1-|y|:y\in A_m\}\;,$$

so every point of $[0,1)$ is again within $\frac1m$ of the set $D_m$. Since $D_m\subseteq\big\{\{nx\}:n\in\Bbb N\big\}$, we’re done.