Does $\pi$ consist of $\pi$ in it?

Question

We do not know whether $\pi$ consists every real number in its decimals, or not. However, If we assume that $\pi$ is consisted in $\pi$ then (I think) we reach a contradiction.

$\pi=\underbrace{3.14159265359...}_{n\; terms}\underbrace{314159265359...}_{\pi's\; decimals}$

Then we have:

$\pi 10^{n-1}=\underbrace{314159265359.....}_{an\;integer}\underbrace{.}_{dot}314159265359....$

$\lfloor x\rfloor$ : greater integer function

so

$$\pi 10^{n-1}=\lfloor\pi10^{n-1}\rfloor+\dfrac{\pi}{10}\\ \Rightarrow \pi\underbrace{(10^n-1)}_{\in \mathbb Z}=\underbrace{10\lfloor\pi10^{n-1}\rfloor}_{\in \mathbb Z}$$

However, it is contradiction.

Result: If the calculation is true, then can we exactly say that $\pi$ does not consist every number? Or can we just say that $\pi$ doesnot consist of $\pi$?

Edit:

In comment section, as @fleablood has mentioned, I wanted to draw attention to what if $\pi$ has interesting irrationals in it (I mean at its tail), how can we verify.

Answer 1

Your argument applies in greater generality. In fact, it is a proof that any real number with decimal expansion that eventually repeats is rational.

In your case, you assumed that $\pi$ had repeating decimal expansion, and then proved it was rational.

Answer 2

Note, that if we construct a number in a way you've mentioned, we will obtain a repeater fraction. The numbers, that can be represented as a repeter fraction are rationals, while $\pi$ is not.

$$x10^n = 10 \lfloor10^{n-1}x\rfloor+x$$

$$x= \frac{10 \lfloor10^{n-1}x\rfloor}{10^n-1}\in\mathbb{Q}$$

Thus we could say, that for every non-rational number $x$ (eg. $\pi$) there is no $n\in\mathbb{N}$ , such that $x$ satisfies the first equation.