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Alright, recently there was a question on 9gag whether the digits of $\pi$ may contain $\pi$ itself here's the original. One user had - in my opinion - a really plausible answer:

Here's his answer.

Nevertheless, I would still argue that $\pi$ can "contain" itself. Since $\pi$ is a non-ending sequence of decimal places, consider that the starting index of the repetition can be an infinitely large natural number. We therefore have an infinite non-repeating sequence prior to that index, thus it would still fulfill the requirement of non-repetitiveness in the sense of $ \pi \not\in \{a/b\} \; \forall \; a,b \in \mathbb{N}$.

I wanted to know if this argument is reasonable and what you think about it.

@5xum As for your first remark:

Let that starting index be $\aleph_0$, the smallest infinite cardinal number.

As for your second remark:

Assume the $(n+1)$-th digit is $\aleph_0$.

David Seres
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    By definition, every sequence is a subsequence of itself. So yes, the set of digits of $\pi$ contains itself. – orion Jan 11 '16 at 13:41
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    Such a sequence is called a fractal sequence. https://en.wikipedia.org/wiki/Fractal_sequence – CommonerG Jan 11 '16 at 13:44
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    See also the existing question Decimal Expansion of Pi. – epimorphic Jan 11 '16 at 13:47
  • what does make sense, because for every irrationals there is a sequence of rationals converging to it, is that the decimal representation of those rationals converge in some sense to an "infinite length repeating sequence". – reuns Jan 11 '16 at 13:49
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    The original question starts with "If $\pi$ contains every single number combination possible". I'm not sure if this has been proven (surely it has not been established whether $\pi$ is normal or not). In any case, in order to have some mathematical meaning, it refers to finite strings of digits, and the full expansion of $\pi$ is definitely not a finite string. – Paolo Franchi Jan 11 '16 at 13:52
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    "the starting index of the repetition can be an infinitely large natural number" Well... no. – Did Jan 11 '16 at 14:08
  • The indices of a sequence are natural numbers, it makes no sense to write $n = \aleph_0$. – Paolo Franchi Jan 12 '16 at 00:54
  • Hey, who said $\pi$ contained every (finite) sequence of digits? That's an open problem. After the quadrillionth digit, it might be all twos and nines for all we know. – Milo Brandt Jan 12 '16 at 01:01
  • @MiloBrandt If it would end in twos and nines, then it would be solvable by a polynomial, and since $\pi$ is a transcendential number this can't be the case. Thus due to the law of large numbers the decimals in $\pi$ do contain every possible sequences almost surely, even if after the quadrillionth digit it "ends" in 2s and 9s. – David Seres Jan 12 '16 at 01:30
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    @DavidSchlesingerSeres Liouville's constant has only zeros and ones and is transcendental. Multiply that by $7$ and add $\frac{2}9$ and you get a transcendental number with only twos and nines. Also, if you chose a random real in $[0,10]$ it contains every possible sequences almost surely (though this isn't the law of large numbers). It's the "almost" part that gives us trouble, since we don't know if $\pi$ is among the exceptions. (And if, after the quadrillionth digit, it has only 2s and 9s, it doesn't contains a sequence of a quadrillion 7s) – Milo Brandt Jan 12 '16 at 01:39
  • please tell us what is the aleph-0-th digit of $\pi$. – mercio Jan 12 '16 at 01:44
  • Lol, infinity just does not work like that... – Sean English Jan 12 '16 at 01:46
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    @MiloBrandt Thanks for pointing that out, I understand. Alright, assume it ends in 2s and 9s. Then just switch to a different base. Though then I guess you could make any irrational number contain any possible sequence, except itself. – David Seres Jan 12 '16 at 02:00

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Your argument requires further elaboration before it can be considered a "mathematical argument":

  • In particular, you need to explain what a "infinitely large natural number" is.
  • You then need to explain, in your new mathematical system, what the technical definition of "containing itself" means. In the standard sense, it means that there exists some $n\in\mathbb N$ such that the digits $n+1,n+2,\dots$ are equal to digits $1,2,3,\dots$, respectively. If you allow infinite numbers, then you need to define what the "$n+1$-th digit" is for $\pi$, where $n$ is an infinitely large natural number.

Oh, and another thing. The answer you link to is not only plausible, it is correct. There is no strict subsequence of digits of $\pi$ which would be equal to the entire sequence of digits of $\pi$, and this is a mathematically provable fact. Calling this fact "plausible" is basically insulting to it.

5xum
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    You are right, $\infty \not\in \mathbb{N}$. – David Seres Jan 11 '16 at 13:44
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    Calling an answer "plausible" before you have verified it completely strikes me as a perfectly reasonable thing to do. – Ben Millwood Jan 11 '16 at 14:36
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    @BenMillwood Thank you – David Seres Jan 11 '16 at 18:33
  • @BenMillwood I just think it's overkill saying "i think this answer is plausible". You can say "this answer is plausible" or "I think this answer is correct". I think the way the comment is worded now implies that the most the answer can be is "plausible". – 5xum Jan 11 '16 at 19:33
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    That's an assumption. As @BenMillwood pointed out, I have not verified the answer yet, yet you insinuate that I have already known it beforehand and that I insult the the author of the comment. – David Seres Jan 12 '16 at 00:12
  • @DavidSchlesingerSeres Actually, I am saying you are insulting the fact itself. I was a bit overdramatic, but what I wanted to say is that calling a true fact plausible is like saying that mount Everest is quite big. It's an insult to the biggest mountain to just call it "quite big". I didn't mean you are insulting the author, and I see now how I was not clear in my words. Sorry. – 5xum Jan 12 '16 at 06:54
  • Since I am only a hobby mathematician, I wasn't able to do a full research on that topic. Nevertheless, I do understand your argument, but then again, Mount Everest might be the highest mountain on earth, but surely not the highest in our solar system. – David Seres Jan 12 '16 at 13:42
  • @DavidSchlesingerSeres Sure, but I hope you understand what I wanted to say and that I was not trying to imply you are insulting to the author, even though it may seem like I was. – 5xum Jan 12 '16 at 13:44
  • As I said, I do understand your argument :) – David Seres Jan 12 '16 at 13:45