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$\{$ fractional part of $n\alpha\mid n \in \mathbb{N}$ $\}$ is a dense subset of $[0,1]$ for an irrational number $\alpha$. - this is a known theorem.

Wouldn't it be true with $\mathbb{N}$ replaced with the set of primes?

For the set of powers of 2 it doesn't hold, I found an interesting topic: The density --- or otherwise --- of $\{\{2^N\,\alpha\}:N\in\mathbb{N}\}$ for ALL irrational $\alpha$.

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tong_nor
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1 Answers1

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The answer is YES. A stronger result is known that $2\alpha, 3\alpha, 5\alpha, \ldots \pmod 1$ is equidistributed on $[0,1)$.

Reference: https://en.wikipedia.org/wiki/Equidistributed_sequence#Equidistribution_modulo_1

Zilin J.
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  • Thanks, but this is a respectable theorem, the density is only an easy corollary from it. So now I ask if it's possible to give an elementary proof of density without using Vinogradov's theorem. – tong_nor Mar 10 '15 at 18:30