it's actually unknown. It looks like it, but it is not proven
this post was submitted on 26 Dec 2024
231 points (98.7% liked)
Asklemmy
44186 readers
1167 users here now
A loosely moderated place to ask open-ended questions
Search asklemmy π
If your post meets the following criteria, it's welcome here!
- Open-ended question
- 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.
- Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
- Not ad nauseam inducing: please make sure it is a question that would be new to most members
- An actual topic of discussion
Looking for support?
Looking for a community?
- Lemmyverse: community search
- sub.rehab: maps old subreddits to fediverse options, marks official as such
- !lemmy411@lemmy.ca: a community for finding communities
~Icon~ ~by~ ~@Double_A@discuss.tchncs.de~
founded 5 years ago
MODERATORS
Also is it even possible to prove it at all? My completely math inept brain thinks that it might be similar to the countable vs uncountable infinities thing, where even if you mapped every element of a countable infinity to one in the uncountable infinity, you could still generate more elements from the uncountable infinity. Would the same kind of logic apply to sequences in pi?
Man, you're giving me flashbacks to real analysis. Shit is weird. Like the set of all integers is the same size as the set of all positive integers. The set of all fractions, including whole numbers, aka integers, is the same size as the set of all integers. The set of all real numbers (all numbers including factions and irrational numbers like pi) is the same size as the set of all real numbers between 0 and 1. The proofs make perfect sense, but the conclusions are maddening.
load more comments
(1 replies)
My guess would be that - depending on the number of digits you are looking for in the sequence - you could calculate the probability of finding any given group of those digits.
For example, there is a 100% probability of finding any group of two, three or four digits, but that probability decreases as you approach one hundred thousand digits.
Of course, the difficulty in proving this hypothesis rests on the computing power needed to prove it empirically and the number of digits of Pi available. That is, a million digits of Pi is a small number if you are looking for a ten thousand digit sequence
But surely given infinity, there is no problem finding a number of ANY length. It's there, somewhere, eventually, given that nothing repeats, the number is NORMAL, as people have said, and infinite.
The probability is 100% for any number, no matter how large, isn't it?
Smart people?
In theory, sure. In practice, are we really going to find a series of ten thousand ones? I would also like to hear more opinions from smart people
This is what allows pifs to work!
Thats very cool. It brings to mind some sort of espionage where spies are exchanging massive messages contained in 2 numbers. The index and the Metadata length. All the other spy has to do is pass it though pifs to decode. Maybe adding some salt as well to prevent someone figuring it out.
Thanks. I love these kind of fun OpenSource community projects/ideas/jokes whatever. The readme reminds me of ed
load more comments
(5 replies)
No, the fact that a number is infinite and non-repeating doesn't mean that and since in order to disprove something you need only one example here it is: 0.1101001000100001000001... this is a number that goes 1 and then x times 0 with x incrementing. It is infinite and non-repeating, yet doesn't contain a single 2.
This proves that an infinite, non-repeating number needn't contain any given finite numeric sequence, but it doesn't prove that an infinite, non-repeating number can't. This is not to say that Pi does contain all finite numeric sequences, just that this statement isn't sufficient to prove it can't.
you are absolutely right.
it just proves that even if Pi contains all finite sequences it's not "since it oa infinite and non-repeating"
load more comments
(16 replies)
It's almost sure to be the case, but nobody has managed to prove it yet.
Simply being infinite and non-repeating doesn't guarantee that all finite sequences will appear. For example, you could have an infinite non-repeating number that doesn't have any 9s in it. But, as far as numbers go, exceptions like that are very rare, and in almost all (infinite, non-repeating) numbers you'll have all finite sequences appearing.
load more comments
(15 replies)
It has not been proven either way but if pi is proven to be normal then yes. https://en.m.wikipedia.org/wiki/Normal_number
A number for which that is true is called a normal number. Itβs proven that almost all real numbers are normal, but itβs very difficult to prove that any particular number is normal. It hasnβt yet been proved that Ο is normal, though itβs generally assumed to be.
I love the idea (and it's definitely true) that there are irrational numbers which, when written in a suitable base, contain the sequence of characters, "This number is provably normal" and are simultaneously not normal.
load more comments
(1 replies)
The jury is out on whether every finite sequence of digits is contained in pi.
However, there are a multitude of real numbers that contain every finite sequence of digits when written in base 10. Here's one, which is defined by concatenating the digits of every non-negative integer in increasing order. It looks like this:
0 . 0 1 2 3 4 5 6 7 8 9 10 11 12 ...
load more comments
(1 replies)
0.101001000100001000001 . . .
I'm infinite and non-repeating. Can you find a 2 in me?
load more comments
(26 replies)
view more: next βΊ