# Is $\{n\pmod \pi: n \in \Bbb N\}$ dense in $[0,\pi]$?

Is $$\{n\pmod\pi: n\in\Bbb N\}$$ dense in $$[0,\pi]$$ ? Is there a proof or well known theorem for this result?

My intuition would say that it is dense.

• By $n \mod \pi$, do you mean $\frac{n}{\pi} - floor(\frac{n}{\pi})$? Sep 6, 2016 at 12:02
• Yes, n divided by pi and we take the remainder as the result. Sep 6, 2016 at 12:03
• Do you know whether $\bigl\{n\cdot \frac{1}{\pi} \bmod 1 : n \in \mathbb{N}\bigr\}$ is dense in $[0,1]$? Sep 6, 2016 at 12:04
• Not only for $\pi$, but for any irrational number. (I'm trying to remember the proof). Sep 6, 2016 at 12:04
• ZirconCode, a remotely related question. Jan 15, 2021 at 0:55

Let $\{\cdot\}$ be the fractional part. $\Bbb N$ is the set of positive integers. As a preliminary property, have in mind that if $x$ is real and $m,n$ are integers, then $\{m\{nx\}\}=\{mnx\}$.

Let $x$ be an irrational number. Then $\{nx\}\neq 0$ for every $n\in\Bbb N$. For the moment, let's see the following:

For every $\epsilon>0$, there exists some $n\in\Bbb N$ such that $\{nx\}<\epsilon$.

Suppose not. Then $\delta=\inf\big\{\{nx\}:n\in\Bbb N\big\}$ is not $0$. Obviously, $\delta<1$. Let $m$ be the natural such that $\delta m\ge 1$ and $\delta(m-1)<1$. We have then $1\le m\delta<1+\delta$.

Take any $0<\epsilon<1+\delta-m\delta$. There exists some $k\in\Bbb N$ such that $0<\{kx\}-\delta<\epsilon/m$, so $$1\le\delta m<m\{kx\}<m\delta+\epsilon<1+\delta<2$$ Therefore $$0<\{kmx\}<m\delta+\epsilon-1<\delta,$$ a contradiction.

Now the proof:

For every interval $(a,b)\subset[0,1]$, there is some $n$ such that $\{nx\}\in(a,b)$. That is, $\big\{\{nx\}:n\in\Bbb N\big\}$ is dense in $[0,1]$.

We know that there is $m\in\Bbb N$ such that $0<\{mx\}<b-a<1$. Now take the minimal integer $k$ such that $k\{mx\}>a$. Then $a<\{kmx\}<b$.

• This is great, thanks! Sep 6, 2016 at 13:31
• Jan 15, 2021 at 0:41

Hint By the Dirichlet Approximation Theorem $\frac{n}{\pi} \pmod{1}$ is dense in $[0,1)$.

You can easily show that this is equivalent to your statement.

• Thank you! Just to make sure, $\pi$ in this case could be any irrational number without any other requirement? Sep 6, 2016 at 12:42
• @ZirconCode Yes, the claim and proof are the same for any irrational number. Sep 6, 2016 at 14:26