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Given an irrational number $x \in \mathbb{R}\setminus \mathbb{Q}$, is it possible to find a map $T: \mathbb{N} \to \mathbb{N}$ strictly increasing such that $\left\{T(n) x\right\} \to 0$ as $n \to \infty$, where $\left\{\cdot\right\}$ is the fractional part?

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Apologies, I realized Dirichlet's theorem is not appropriate here, we must instead use (a one variable version of) Kronecker's theorem to get $qx$ within $1/2N$ of $1/2N$. So the result depends on a slightly stronger theorem but still much weaker than uniform distribution. –  user50336 Nov 24 '12 at 13:50
    
Another question: which one of Kronecker's theorems? –  Jaques Nov 24 '12 at 17:00

2 Answers 2

  • If $x$ is irrational, by Kronecker's theorem, for all $N$ there exists some $1 \le q$ such that $\{qx\}<\frac{1}{N}$.
  • If $x$ is irrational then $nx$ is irrational.

Let $T(1) = 1$ and suppose we have built $n$ terms of the sequence $T(n)$, to build the next pick $N$ large enough that $\frac{1}{N} < \{T(n)x\}$ and apply Kronecker's theorem on the irrational $T(n)x$ to get some $q > 1$ such that $\{qT(n)x\}$ and put $T(n+1) = T(n)q$.

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I would like to add a second proof of this theorem based on a deleted answer.

Since $x$ is irrational the continued fraction gives a sequence of convergents (best rational approximations) $\frac{p_n}{q_n} \to x$ with (at least) quadratic convergence, further from the theory of continued fractions $p_n x - q_n$ is alternating in sign so we may choose a subsequence $r$ from it such that $\{p_{r(n)} x\} \to 0$.

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