How about ANY FINITE SEQUENCE AT ALL?

  • Umbrias@beehaw.org
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    1 day ago

    it’s not a good example because you’ve only changed the symbolic representation and not the numerical value. the op’s question is identical when you convert to binary. thir is not a counterexample and does not prove anything.

    • orcrist@lemm.ee
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      20 hours 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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        18 hours 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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          15 hours 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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          15 hours ago

          The question is

          Since pi is infinite and non-repeating, would it mean…

          Then the answer is mathematically, no. If X is infinite and non-repeating it doesn’t.

          If a number is normal, infinite, and non-repeating, then yes.

          To answer the real question “Does any finite sequence of non-repeating numbers appear somewhere in Pi?”

          The answer depends on if Pi is normal or not, but not necessarily