How about ANY FINITE SEQUENCE AT ALL?

  • orcrist@lemm.ee
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    2 days ago

    Please read it all again. They didn’t rely on the conversion. It’s just a convenient way to create a counterexample.

    Anyway, here’s a simple equivalent. Let’s consider a number like pi except that wherever pi has a 9, this new number has a 1. This new number is infinite and doesn’t repeat. So it also answers the original question.

    • Umbrias@beehaw.org
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      2 days ago

      “please consider a number that isnt pi” so not relevant, gotcha. it does not answer the original question, this new number is not normal, sure, but that has no bearing on if pi is normal.

      • spireghost@lemmy.zip
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        2 days ago

        OK, fine. Imagine that in pi after the quadrillionth digit, all 1s are replaced with 9. It still holds

              • spireghost@lemmy.zip
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                34 minutes ago

                Hmm, ok. Let me retry.

                The digits of pi are not proven to be uniform or randomly distributed according to any pattern.

                Pi could have a point where it stops having 9’s at all.

                If that’s the case, it would not contain all sequences that contain the digit 9, and could not contain all sequences.

                While we can’t look at all the digits of Pi, we could consider that the uniform behavior of the digits in pi ends at some point, and wherever there would usually be a 9, the digit is instead a 1. This new number candidate for pi is infinite, doesn’t repeat and contains all the known properties of pi.

                Therefore, it is possible that not any finite sequence of non-repeating numbers would appear somewhere in Pi.