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Let be $\alpha$ an irrational positive number.

Prove that there are infinite many $n$ positive integers such that $$\{2^n\alpha\}\in (0,\frac{1}{4}) $$

{$x$} denotes the fractional part of $x$.

First I obtained that: for a given $\varepsilon>0$ there are $i$, $j$ such that: $|\{2^i\alpha\}-\{2^j\alpha\}| <\varepsilon$, using pigeonhole principle.

I wanted to prove that $A=\{\{2^n\alpha\} \mid \text{where }n\text{ is an positive-integered number}\}$ is dense in $[0,1]$

However, lately I realised that this is not true since we can construct irrational positive numbers $\alpha$ for which the set $A$ has no elements in a specific nondegenerate interval of numbers which is included in $[0,1]$.

How should I proceed?

Asaf Karagila
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shangq_tou
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  • What {$2^n \alpha$} means? – MH.Lee Oct 20 '21 at 18:11
  • It is the fractional part of $2^n\alpha$ – shangq_tou Oct 20 '21 at 18:13
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    Write the irrational in base $2$. Note that the fractional part must contain infinitely many zeroes. Though unless I am missing something this only gets you $\left( 0,\frac 12\right)$ – lulu Oct 20 '21 at 18:19
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    Why do you think the statement is true? What if you take $\alpha = 1 - \sum_{j=1}^\infty 2^{-j^2} = \sum_{k >= 1\text{ and }k\text{ is not a square}} 2^{-k}?$ If you write this in base $2,$ you can see that $\alpha$ is irrational (since the expansion doesn't repeat) and that the expansion doesn't contain two or more $0$ digits in a row (since there aren't any consecutive positive squares). – Mitchell Spector Oct 21 '21 at 00:03

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