Recurring dreams with floor function?

In general, we use base 10 to represent any real number. For example $652.333333...=6*10^{2}+5*10^{1}+2+\frac{3}{10}+\frac{3}{10^{2}}+\frac{3}{10^{3}}+....=652\frac{1}{3}.$ In a similar way any number can be represented in base 2 by a series in the following way: if x is any real number then there is sequence $a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n}, ...,$ such that any $a_{i}= 0$ or $1$ and $x=\sum_{k=0}^{\infty}a_{k}*2^{m-k}$ for a certain integer $m$ that depends on $x$ . In particular, the number $x=2^{\frac{1}{2}}=\sqrt{2}$ can be represented by a series $x=\sum_{k=0}^{\infty}b_{k}*2^{-k}.$ Now it is easy to see that for any integer $n\geq 0$ $f(n)=b_{n+1}$ . So if $f(n)$ were periodic, it would be easy to prove, using this representation, that $\sqrt{2}$ would be a rational number that would be a contradiction. Therefore the answer is non-periodic.

f (n) can be simplified as Determine( $Frac(\sqrt2 \times 2^n)$ ) of 1 when fractional part of $\sqrt2 \times 2^n$ greater than or equals to 0.5, OR 0 when less than 0.5!!!

$\sqrt2 \times 2 \times 2 \times 2 \times 2...$ is a change of 1.4142135623730950488016887242097+ into

2.8284271247461900976033774484194+ for 1

5.6568542494923801952067548968388+ for 1

11.313708498984760390413509793678+ for 0

etc.

Since $\sqrt2$ is an irrational number, a dragged multiples being pulled from far decimals and also change of times simultaneously shall not be regular, such that not predictable and it is also not zero but of infinitely endless terms. Therefore, the simplified equivalent of seeing infinite decimals of irrational number $\sqrt2$ implies for an answer of Not eventually periodic.

All these conclusions are drawn from observations and thinking. f (n) is not eventually periodic and it also remained as either 1 or 0 only when n tends to $\infty$ .

The question can change 1.5 and 0.5 into 1.1 and 0.1 and etc. for consideration onto other cases.

http://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil