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This question already has an answer here:

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 $\aleph_0$.

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marked as duplicate by Eric Wofsey, Milo Brandt, Shailesh, user228113, user296602 Jan 12 '16 at 2:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ By definition, every sequence is a subsequence of itself. So yes, the set of digits of $\pi$ contains itself. $\endgroup$ – orion Jan 11 '16 at 13:41
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    $\begingroup$ Such a sequence is called a fractal sequence. en.wikipedia.org/wiki/Fractal_sequence $\endgroup$ – CommonerG Jan 11 '16 at 13:44
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    $\begingroup$ See also the existing question Decimal Expansion of Pi. $\endgroup$ – epimorphic Jan 11 '16 at 13:47
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    $\begingroup$ 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. $\endgroup$ – Paolo Franchi Jan 11 '16 at 13:52
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    $\begingroup$ "the starting index of the repetition can be an infinitely large natural number" Well... no. $\endgroup$ – Did Jan 11 '16 at 14:08
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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.

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    $\begingroup$ You are right, $\infty \not\in \mathbb{N}$. $\endgroup$ – David Seres Jan 11 '16 at 13:44
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    $\begingroup$ Calling an answer "plausible" before you have verified it completely strikes me as a perfectly reasonable thing to do. $\endgroup$ – Ben Millwood Jan 11 '16 at 14:36
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    $\begingroup$ @BenMillwood Thank you $\endgroup$ – David Seres Jan 11 '16 at 18:33
  • $\begingroup$ @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". $\endgroup$ – 5xum Jan 11 '16 at 19:33
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    $\begingroup$ 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. $\endgroup$ – David Seres Jan 12 '16 at 0:12

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