this post was submitted on 26 Dec 2024
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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
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?”
Then put 23456789 at the start. Doesn't contain 22 then but all digits in base 10.
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
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"
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.
op's question was focued very clearly on pi, but sure.