TITLE: Existence of a sequence of integers $\lbrace a_k\rbrace_{k\geq 1}$ so that the first $k$ digits of $a_k\alpha$ are $0$ where $\alpha$ is irrational.
QUESTION [2 upvotes]: Let $\alpha$ be an irrational number. 
Is there a sequence  $\lbrace a_k\rbrace_{k\geq 1}$ of integers so that the first $k$ digits of the fractional part of $a_k·\alpha$ are $0$?
(in base $2$, for instance, but it doesn't really matter)
How would one find such sequence if all the digits of $\alpha$ are known?

REPLY [1 votes]: I will assume that you are talking about the first $k$ digits of $a_k \alpha$ after the decimal point. 
Let $\alpha\gt 0$. We use the fact that if $\alpha$ is irrational, then the set of fractional parts of the numbers $n\alpha$, where $n$ ranges over the positive integers, is dense in the interval $[0,1]$. This result has been proved many times on MSE.
For $\alpha\lt 0$, use negative integers as multipliers. 
Remark: The proof of denseness is constructive, so we can actually construct an $a_k$ by following the proof. This is very inefficient, and usually produces unnecessarily large $a_k$. The continued fraction expansion of $\alpha$ should product substantially cheaper $a_k$.
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