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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.

Micheal Brain Hurts
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  • $\pi$ has infinitely many digits! Where'd you get $n$ ? – Dzoooks Jun 12 '18 at 20:41
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    In your notation $3.14159...314159...$, what does the first '...' mean? The usual meaning of $...$ means 'and so on', but if your '...' means 'and so on', then the decimal expansion continues indefinitely, and hence cannot be followed by anything. – preferred_anon Jun 12 '18 at 20:41
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    the definition of "normal" number refers to finite strings of digits. Clearly, for instance, $\pi$ does not contain the string $111\cdots$ – lulu Jun 12 '18 at 20:42
  • What do you think you do along the $\Rightarrow$? – Jasper Jun 12 '18 at 20:42
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    The claim is that $\pi$ contains all finite strings of decimals. That is different from including the infinite string that is $\pi$. We do not know whether the claim is trrue. – Ross Millikan Jun 12 '18 at 20:43
  • If your claim was that $\pi$ has some digits and then it ends with a copy of the whole decimal expansion of $\pi$, then there is an even simpler absurd: the first $n$ digits of $\pi$ must actually be the period of $\pi$. –  Jun 12 '18 at 20:45
  • "..." doesnot mean infinity it can be seen because I opened unbrace "n terms"//

    I just assumed that if it has pi in it then it must be rational but pi isnot so every number combianiton isnot involved in pi. //

    Yes I know it seems like very ignorant, I think the known misconception is defined in different way

     – Micheal Brain Hurts Jun 12 '18 at 20:46
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    "We do not know whether π consists every real number in its decimals, or not." Yes, we do. We know it does not. What we don't know is if $\pi$ contains ever integer in its decimals or not (although it probably does). – fleablood Jun 12 '18 at 20:47
  • Alright, why every integer combination doesnot mean the real numbers because every real numbers can be written as sum of integers (infinitely many of course). – Micheal Brain Hurts Jun 12 '18 at 20:49
  • It is hypothetically possible that $\pi$ does start with $n$ digits and then repeats those exact same digits and then continues on. But those $n$ digits are not $\pi$ but pit to the first n digits. But the main reason $\pi$ can't contain $\pi$ within it, is that... that just doesnt make any sense. $\pi$'s digits are infinite and to be "in" something means there is something after it. Or it means it repeats itself at the end infinitely. – fleablood Jun 12 '18 at 20:56
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    Actually that argument would be fine if $\pi$ were rational. $\frac 17 = 0.14285714285714285714285714285714...$ "contains itself". Hypothetically it's possible that if $\pi$ ends with the infinite digits of $e$ that $\pi$ contains $e$ (or vice versa) but that is almost certainly not the case. The only irrational numbers $\pi$ can contain are the irrational numbers that are an infinite tail of $\pi$. Which is to say "almost none of them". Anyway, good job figuring out that something is hinky +1. – fleablood Jun 12 '18 at 21:03
  • Can we exactly say that $\pi$ has no the $e$ tail? In addition, actually, we can observe as many as infinity number of irrational at the tail of $\pi$ why not there is a $e$ ? – Micheal Brain Hurts Jun 12 '18 at 21:15

2 Answers2

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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.

Jaroslaw Matlak
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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.