Is this proof that $\pi$ does not contain all possible combination correct?

Question

I came across this meme today:

The counterproof is very trivial, but I see no one disproves it. Some even say that the meme might be true. Well, $\pi$ cannot contain itself.

Well, everything means $\pi$ might contain $\pi$ somewhere in it. Say it starts going $\pi=3.1415...31415...$ again on the $p$ digit. Then it will have to do the same at the $2p$ digit, since the "nested $\pi$" also contains another $\pi$ in it. $\pi$ then will be rational, which is wrong. Thus $\pi$ does not contain all possible combination.

Is this proof correct? I'm not a mathematician so I'm afraid I make silly mistakes.

Answer 1

I am sure it has $31415$ again in the decimal expansion, but why should it continue $926535$ after that? Sometimes it will, but it will eventually diverge from the decimals at the start. You have not made any argument that when you see $31415$ it should repeat from there and in fact it will not.

You are correct that $\pi$ cannot contain itself. The claim, not known to be true, is that $\pi$ contains all finite sequences of digits.

Answer 2

An infinite non-repeating decimal does not imply that every possible number combination exists somewhere. Consider the number: $0.101001000100001\ldots$, the pattern is easy to spot, but this is an irrational number because...you guessed it...it's an infinite, non-repeating decimal.

It is conjectured that $\pi$ is a normal number, but this has not been proven. Here is an old question from MathOverflow with some more details.

Answer 3

Let $a_{n} = tr_{n}(\pi)$

$a_{1} = 3.1$

$a_{2} = 3.14$

$a_{3} = 3.141$

$a_{4} = 3.1415$

$lim_{n \to \infty} a_{n} = \pi$

Then, suppose that $\pi$ contains all finite sequences of digits.

If $\pi$ not contains himself, then there exist $m = max\{ n \in \mathbb{N} : a_{n} \in$ digits of $\pi \}$

But then $a_{m+1}$ is a finite sequence, so is in digits of $\pi$

Then $m$ is not the max, that is a contradiction.

Then $\pi$ contain $\pi$

Well this is in the hypothetical case in which it happens that $\pi$ contains all finite sequences of digits