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Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi? (lemmy.dbzer0.com)

submitted 1 day ago* (last edited 1 day ago) by Melatonin@lemmy.dbzer0.com to c/asklemmy@lemmy.ml

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How about ANY FINITE SEQUENCE AT ALL?

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[–] gerryflap@feddit.nl 10 points 22 hours ago (2 children)

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

[–] Umbrias@beehaw.org 1 points 11 hours ago* (last edited 11 hours 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"

[–] CaptSneeze@lemmy.world 4 points 22 hours ago (2 children)

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?”

[–] weker01@sh.itjust.works 1 points 2 hours ago* (last edited 2 hours ago)

Then put 23456789 at the start. Doesn't contain 22 then but all digits in base 10.

[–] Phlimy@jlai.lu 4 points 19 hours 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