this post was submitted on 26 Dec 2024
223 points (98.7% liked)

Asklemmy

44240 readers
622 users here now

A loosely moderated place to ask open-ended questions

Search asklemmy πŸ”

If your post meets the following criteria, it's welcome here!

  1. Open-ended question
  2. Not offensive: at this point, we do not have the bandwidth to moderate overtly political discussions. Assume best intent and be excellent to each other.
  3. Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
  4. Not ad nauseam inducing: please make sure it is a question that would be new to most members
  5. An actual topic of discussion

Looking for support?

Looking for a community?

~Icon~ ~by~ ~@Double_[email protected]~

founded 5 years ago
MODERATORS
 

How about ANY FINITE SEQUENCE AT ALL?

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 7 points 1 week 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.

[–] [email protected] 2 points 1 week 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.

[–] [email protected] 3 points 1 week ago (1 children)

Please read it all again. They didn't rely on the conversion. It's just a convenient way to create a counterexample.

Anyway, here's a simple equivalent. Let's consider a number like pi except that wherever pi has a 9, this new number has a 1. This new number is infinite and doesn't repeat. So it also answers the original question.

[–] [email protected] 2 points 1 week ago (1 children)

"please consider a number that isnt pi" so not relevant, gotcha. it does not answer the original question, this new number is not normal, sure, but that has no bearing on if pi is normal.

[–] [email protected] 2 points 1 week ago* (last edited 1 week ago) (1 children)

OK, fine. Imagine that in pi after the quadrillionth digit, all 1s are replaced with 9. It still holds

[–] [email protected] 1 points 1 week ago (1 children)

"ok fine consider a number that still isn't pi, it still holds." ??

[–] [email protected] 1 points 1 week ago (1 children)

Prove that said number isn't pi.

[–] [email protected] 1 points 6 days ago (1 children)
[–] [email protected] 1 points 5 days ago (1 children)

Hmm, ok. Let me retry.

The digits of pi are not proven to be uniform or randomly distributed according to any pattern.

Pi could have a point where it stops having 9's at all.

If that's the case, it would not contain all sequences that contain the digit 9, and could not contain all sequences.

While we can't look at all the digits of Pi, we could consider that the uniform behavior of the digits in pi ends at some point, and wherever there would usually be a 9, the digit is instead a 1. This new number candidate for pi is infinite, doesn’t repeat and contains all the known properties of pi.

Therefore, it is possible that not any finite sequence of non-repeating numbers would appear somewhere in Pi.

[–] [email protected] 1 points 5 days ago

i am aware nobody has proven pi is normal.

[–] [email protected] 1 points 1 week ago (1 children)

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

[–] [email protected] 1 points 1 week ago (1 children)

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

[–] [email protected] 3 points 1 week ago* (last edited 1 week 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