Prove that $2^n\alpha-[2^n\alpha]$ is dense in $[0,1]$, if $\alpha$ is a positive irrational number.
$[x]$ represents the largest integer smaller than $x$.
I only know how to prove $n\alpha-[n\alpha]$ is dense in $[0,1]$, using pigeon hole principle.
I am looking for similar method to solve this problem.

  • 1
    $\begingroup$ Hi, and welcome! Can you please share your thoughts on the problem and explain what tools you have available, and what you've tried? Without context or details, it's difficult to answer the question and your question might be closed; so please edit your post. $\endgroup$
    – user61527
    Dec 16, 2013 at 5:46
  • $\begingroup$ As a note, from wikipedia, we have this problem is true for almost $\alpha$. $\endgroup$
    – Du Phan
    Dec 16, 2013 at 7:06

1 Answer 1


The statement is not true.

Recall that a number is rational if and only if the decimal representation (in any base) terminates or is eventually repeating.

Let $0< \beta < 1 $ be any irrational number. Let it's base 3 representation be $0. b_1 b_2 b_3 \ldots$. (I'm using base 3 to reduce confusion with the latter part. If you want, you can use base 2 here.)

Henceforth, we are working in base 2.

Construct the following irrational number $\alpha$ (in base 2).
Start with 0.1,
then add $b_1+1$ 0's and then a 1,
then add $b_2+1$ 0's and then a 1,
then add $b_3+1$ 0's and then a 1, etc.
For example, if $b_1 = 0, b_2 = 1, b_3 = 2, b_4 = 0, b_5=2 \ldots$, we get the number $ \alpha = 0.1010010001010001\ldots$.
This number is irrational because the binary representation doesn't terminate nor eventually repeat (from the construction).

Claim: The sequence $2^n \alpha - \lfloor 2^n \alpha \rfloor $ does not have $\frac{1}{2} = 0.1_2$ as a limit point.

This is obvious since it never takes on any value in the range $(0.\overline{01}_2, 0.\overline{1000}_2)$, which has a length of $\frac{1}{5}$! Pretty amazing eh?

  • $\begingroup$ Sorry if I missunderstand something. Is $4\alpha$ mod $1$ in base $2$ is: $0.100\ldots$? Then it is in your interval. $\endgroup$
    – Du Phan
    Dec 16, 2013 at 6:59
  • $\begingroup$ @DuPhan That is true. I've changed the upper limit, and it's open on that side. $\endgroup$
    – Calvin Lin
    Dec 16, 2013 at 7:09
  • $\begingroup$ Excellent! I understand your approach now. $0.10001$ is enough. Thanks! $\endgroup$
    – Du Phan
    Dec 16, 2013 at 7:18
  • $\begingroup$ I think $[0.011,0.10001]$ is better bound. $\endgroup$
    – Du Phan
    Dec 16, 2013 at 7:27
  • $\begingroup$ That's really nice! Yep, really nice. $\endgroup$
    – davidlowryduda
    Dec 16, 2013 at 7:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .