Prove that exists an irrational number $a$ and $\epsilon>0$ such that $\epsilon \leq \{2^na\} \leq 1-\epsilon,\forall n \in \Bbb{N}$.
I have a difficulty solving this.
My first thoughts are to work with binary representation of numbers using the fact that when we multiply a number in its binary form with $2^n$ ,we shift its digits $n-$ positions to the left.
And also we thow out some digits because we care about fractional parts.
Can someone help me?
I just want a hint not a full answer.
Thank you in advance.
Think of the irrational expressed in the binary system $$ \alpha=0.a_1a_2\ldots a_na_{n+1}\ldots, \quad a_i\in\{0,1\}. $$ Then $$ 2^n\alpha=a_1a_2\ldots a_n.a_{n+1}\ldots $$ and $$ \{2^n\alpha\}=0.a_{n+1}a_{n+2}\ldots\,\,\text{or}\,\, 1-0.a_{n+1}a_{n+2}\ldots $$ You need to find $\varepsilon$ and $a_i$'s such that $$ \varepsilon<0.a_{n+1}a_{n+2}\ldots, 1-0.a_{n+1}a_{n+2}\ldots<1-\varepsilon $$ Define $\hat a=1-a$. Then $1-0.a_{n+1}a_{n+2}\ldots=.\hat{a}_{n+1}\hat{a}_{n+2}\ldots$.
Answer. Consider for example, $\varepsilon=1/8$, and a sequence of pairs $$ (a_1,a_2), (a_3,a_4),\ldots,(a_{2n-1},a_{2n}),\ldots, $$ where $a_2=1-a_1$ and this sequence is not periodic.