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Let $\theta \in \mathbb{R} \setminus \mathbb{Q}$. Is the set $\{ (2n+1) \theta \bmod 1: n \in \mathbb{N} \}$ dense in $[0,1]$?

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Yes. In fact, for any irrational number $\alpha$ and and real number $\beta$, the set $\{\alpha n+\beta \bmod 1\colon n\in\Bbb N\}$ is dense in $[0,1]$. (A proof follows from showing that every interval of the form $[0,\varepsilon)$ contains some multiple of $\alpha\bmod1$.) The answer to your question follows from taking $\alpha=2\theta$ and $\beta=\theta$.

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  • $\begingroup$ Why this answer get two downvotes? $\endgroup$ – Hanul Jeon Oct 26 '14 at 6:51
  • $\begingroup$ What downvotes? Are there issues with this answer? $\endgroup$ – user173897 Oct 27 '14 at 4:27

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