How about ANY FINITE SEQUENCE AT ALL?

  • Umbrias@beehaw.org
    link
    fedilink
    arrow-up
    3
    arrow-down
    1
    ·
    2 days 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
      link
      fedilink
      arrow-up
      3
      ·
      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.

        • spireghost@lemmy.zip
          link
          fedilink
          English
          arrow-up
          3
          ·
          edit-2
          2 days 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