I came across this meme today:

enter image description here

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.

Kim Dong
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    The claim is that it contains all finite substrings, which while conjectured, has not yet been proven. – Cheerful Parsnip Jul 28 '19 at 01:06
  • Yeah, but I don't claim to prove that. I guess I only attempt to prove (if correct) that it doesn't contain one infinite substring, which mean it does not contain "everything". – Kim Dong Jul 28 '19 at 01:10
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    well, it does contain the ASCII-encoded text for the definition of $\pi$... – Cheerful Parsnip Jul 28 '19 at 01:12
  • So "all possible combination" only applies to finite sequence of numbers? I'm loose on the mathematical terms (obviously lol) – Kim Dong Jul 28 '19 at 01:18

3 Answers3


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.

Ross Millikan
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    @D.Tran It's not you, it's the meme. It's romanticised to the point of ridiculousness, and is not interested in actually stating the result it's claiming, or in acknowledging that the result is merely an open problem. What it doesn't say is that *most* numbers actually do have this property (which is known)! It's just unknown whether $\pi$ is one of these numbers (called "normal numbers"). – Theo Bendit Jul 28 '19 at 01:26

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.

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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

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