How about ANY FINITE SEQUENCE AT ALL?

  • Umbrias@beehaw.org
    link
    fedilink
    arrow-up
    0
    ·
    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.

        • spireghost@lemmy.zip
          link
          fedilink
          English
          arrow-up
          0
          ·
          edit-2
          22 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

    • orcrist@lemm.ee
      link
      fedilink
      arrow-up
      0
      ·
      1 day 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
        link
        fedilink
        arrow-up
        0
        ·
        1 day 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.