Prove that $2^n\alpha-[2^n\alpha]$ is dense in [0,1]

Solution 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?