How about ANY FINITE SEQUENCE AT ALL?

    • gerryflap@feddit.nl
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      2 days ago

      They also say “and reinterpret in base 10”. I.e. interpret the base 2 number as a base 10 number (which could theoretically contain 2,3,4,etc). So 10 in that number represents decimal 10 and not binary 10

      • CaptSneeze@lemmy.world
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        2 days ago

        I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”

        • Phlimy@jlai.lu
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          2 days ago

          They just proved that “X is irrational and non-repeating digits -> can find any sequence in X”, as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn’t be because of its irrationality or the fact that it’s non-repeating, it would be some other property

      • Umbrias@beehaw.org
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        1 day ago

        that number is no longer pi… this is like answering the question “does the number “3548” contain 35?” by answering “no, 6925 doesnthave 35. qed”

        • PatheticGroundThing@beehaw.org
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          19 hours ago

          It was just an example of an infinite, non-repeating number that still does not contain every other finite number

          Another example could be 0.10100100010000100000… with the number of 0’s increasing by one every time. It never repeats, but it still doesn’t contain every other finite number.

    • tomi000@lemmy.world
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      2 days ago

      Like the other commenter said its meant to be interpreted in base10.

      You could also just take 0.01001100011100001111… as an example