this post was submitted on 17 Jul 2024
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[–] driving_crooner@lemmy.eco.br 83 points 2 months ago (30 children)

That's not how it's works. Being "infinite" is not enough, the number 1.110100100010000... is "infinite", without repeating patterns and dosen't have other digits that 1 or 0.

[–] HatchetHaro@lemmy.blahaj.zone 38 points 2 months ago* (last edited 2 months ago) (6 children)

to be fair, though, 1 and 0 are just binary representations of values, same as decimal and hexadecimal. within your example, we'd absolutely find the entire works of shakespeare encoded in ascii, unicode, and lcd pixel format with each letter arranged in 3x5 grids.

[–] driving_crooner@lemmy.eco.br 31 points 2 months ago (4 children)

Doesn't, the binary pattern 10101010 dosen't exists on that number, for example.

[–] TdotMatrix@lemmy.ml 5 points 2 months ago (1 children)
[–] Turun@feddit.de 11 points 2 months ago

No, because you can't mathematically guarantee that pi contains long strings of predetermined patterns.

The 1.101001000100001... example by the other user was just that - an example. Their number is infinite, but never contains a 2. Pi is also infinite, but does it contain the number e to 100 digits of precision? Maybe. Maybe not. The point is, we don't know and we can't prove it either way (except finding it by accident).

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