this post was submitted on 26 Dec 2024
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How about ANY FINITE SEQUENCE AT ALL?

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[–] juliebean@lemm.ee 7 points 6 months ago (35 children)

no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.

for instance, write out pi in base 2, and reinterpret as base 10.

11.0010010000111111011010101000100010000101...

it is infinitely non-repeating, but nowhere will you find a 2.

i've often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven't seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.

[–] tetris11@lemmy.ml 8 points 6 months ago (24 children)

Isnt this a stupid example though, because obviously if you remove all penguins from the zoo, you're not going to see any penguins

[–] Lanthanae@lemmy.blahaj.zone 7 points 6 months ago (3 children)

Its not stupid. To disprove a claim that states "All X have Y" then you only need ONE example. So, as pick a really obvious example.

[–] Umbrias@beehaw.org 2 points 6 months ago (2 children)

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.

[–] Lanthanae@lemmy.blahaj.zone 1 points 6 months ago (1 children)

They didn't convert anything to anything, and the 1.010010001... number isn't binary

[–] Umbrias@beehaw.org 1 points 6 months ago (1 children)

then it's not relevant to the question as it is not pi.

[–] spireghost@lemmy.zip 3 points 6 months ago* (last edited 6 months 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

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