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I'm denoting the fractional part of a real number by $\langle \cdot \rangle$. I've already proven that if $\theta \in \mathbb{R}\setminus \mathbb{Q}$, then the set $\{\langle k\theta \rangle \colon k \in \mathbb{N}\}$ is dense in $[0,1]$. Is this still true if we take the fractional parts of odd multiples of $\theta$?

Edit: My proof of the density of $\{\langle k\theta \rangle \colon k \in \mathbb{N}\}$ is similar to the one found in the top answer of this post, but I don't see how could I tweak that argument to show that the set $\{\langle (2k-1)\theta \rangle \colon k \in \mathbb{N}\}$ is also dense in $[0,1]$. As Sayan Dutta has replied, this would be easy to prove using Weyl's criterion, but I'm trying to find a more elementary proof that doesn't require any knowledge of equidistribution theory.

David
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  • Yes, it is true. You can use Weyl's criteria to show this. – Mittens May 05 '25 at 14:31
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    If you are familiar with uniform distribution mod 1, you can also use a more pedestrian argument: $2\theta$ is irrational whenever $\theta$ is. Then $2n\theta$ is u.d mod 1. Then you can show that is $x_n$ is a sequence that is u.d. mod 1, then for $a$ fixed, so is $x_n+a$. Now if density is you concern, perhaps a small teak to your proof os density of $n\theta$ mod 1 will work. But you are not showing your arguments. – Mittens May 05 '25 at 14:57
  • More generally, if $r\in \mathbb{R}$ and $\theta \in \mathbb{R}\setminus \mathbb{Q}$, then ${\langle k\theta + r\rangle : k \in \mathbb{N}}$ is dense in $[0,1]$. The argument is essentially the same as the case of $r=0$. (note this is just rephrasing Mittens's comment closer to your notation, and without caring about uniform distribution) – Brian Moehring May 05 '25 at 15:52

2 Answers2

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Note that for any real numbers $x,y$ we have $$\langle x+y \rangle = \langle \langle x \rangle + \langle y \rangle \rangle$$ (where $\langle . \rangle$ means the fractional part.) Then one can prove that
$$ \{ \langle (2k+1)\theta \rangle \;:\;k\in\mathbb Z\} = \{ \langle 2k\theta \rangle \;:\;k\in\mathbb Z\} + \langle \theta \rangle$$ But $2\theta$ is irrational and therefore $\{ \langle 2k\theta \rangle \;:\;k\in\mathbb Z\}$ is dense (by what you already have proved) and so its translation by $<\theta>$ is also dense.

Smiley1000
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Sara
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Notice $$\sum_{n=1}^N e^{2\pi i k (2n+1)\theta} = \frac{e^{6\pi i k \theta} \left(1 - e^{4\pi i k \theta N} \right)}{1 - e^{4\pi i k \theta}}$$ which implies $$\lim_{N\to \infty} \frac 1N \sum_{n=1}^N e^{2\pi i k (2n+1)\theta} = 0$$ since $$\left| \frac{1}{N} \cdot \frac{1 - e^{4\pi i k\theta N}}{1 - e^{4\pi i k\theta}} \right| \le \frac{2}{N |1 - e^{4\pi i k\theta}|} \xrightarrow{N\to \infty} 0$$ hence completing the proof (by Weyl Criterion).

Sayan Dutta
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