How about ANY FINITE SEQUENCE AT ALL?
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?
it’s actually unknown. It looks like it, but it is not proven
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.
I’m going to say yes to both versions of your question. Infinity is still infinitely bigger than any expressible finite number. Plenty of room for local anomalies like long repeats and other apparent patterns.
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 ...fun fact, “most” real numbers have this property. If you were to mark each one on a number line, you’d fill the whole line out. Numbers that don’t have this property are vanishingly rare.
The term for what you’re describing is a “normal number”. As @[email protected] correctly pointed out it is still an open question whether pi is normal. This is a fun, simple-language exploration of the question in iambic pentameter, and is only 3 minutes and 45 seconds long.
Merry Christmas!
It’s remarkable how there are uncountably many non-normal numbers, yet they take up no space at all in the real numbers (form a null set), since almost all numbers are normal. And despite this, we can only prove normality for some specific classes of examples.
It helps me to think, how there are many “totally random” or non computable numbers, that are not normal because they don’t contain the digit 1.
My birthday in American MMDDYYYY format shows up in the first few ten-million digits, but in standard DDMMYYYY format, it’s not in any of the digits that site is able to check.
Self-doxxing! Makes me wonder how many of the dates would fit.
standard DDMMYYYY
🙄
A number for which that is true is called a normal number. It’s proven that almost all real numbers are normal, but it’s very difficult to prove that any particular number is normal. It hasn’t yet been proved that π is normal, though it’s 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.
Technically to meet OPs criteria it needs only be a rich number in base 10, not necessarily a normal one. Although being normal would certainly be sufficient
Yes.
And if you’re thinking of a compression algorithm, nope, pigeonhole principle.
It’s almost sure to be the case, but nobody has managed to prove it yet.
But simply being infinite and non-repeating doesn’t guarantee that all finite sequences should appear. For example, you could have an infinite non-repeating number that doesn’t have any 9s in it.
Very rare in the sense that they have a probability of 0.
There are lot that fit that pattern. However, most/all naturally used irrational numbers seem to be normal. Maths has, however had enough things that seemed ‘obvious’ which turned out to be false later. Just because it’s obvious doesn’t mean it’s mathematically true.
Exceptions are infinite. Is that rare?
Yes. The exceptions are a smaller cardinality of infinity than the set of all real numbers.
Rare in this context is a question of density. There are infinitely many integers within the real numbers, for example, but there are far more non-integers than integers. So integers are more rare within the real.
There is not density in infinity
I think you mean “I don’t understand density in infinity”.
They should look up the classic example of rationals in the real numbers. Their statement could hardly be more wrong.
we were talking about probability
I most assuredly am talking about your false statement regarding density.
https://github.com/philipl/pifs
πfs is a revolutionary new file system that, instead of wasting space storing your data on your hard drive, stores your data in π! You’ll never run out of space again - π holds every file that could possibly exist! They said 100% compression was impossible? You’re looking at it!
I enjoyed this linked text:
If you compute it, you will be guilty of:
- Copyright infringement (of all books, all short stories, all newspapers, all magazines, all web sites, all music, all movies, and all software, including the complete Windows source code)
- Trademark infringement
- Possession of child pornography
- Espionage (unauthorized possession of top secret information)
- Possession of DVD-cracking software
- Possession of threats to the President
- Possession of everyone’s SSN, everyone’s credit card numbers, everyone’s PIN numbers, everyone’s unlisted phone numbers, and everyone’s passwords
- Defaming Islam. Not technically illegal, but you’ll have to go into hiding along with Salman Rushdie.
- Defaming Scientology. Which IS illegal–just ask Keith Henson.
Yes, this is implied. It’s also why many people use digits of pi as passwords and make the password hint “easy as pi”.
I have a slightly unique version of this.
When I was in high school, one of the maths teachers had printed out pi to 100+ digits on tractor feed paper (FYI I am old) and run it around the top of the classroom as a nerdy bit of cornice or whatever.
Because I was so insanely clever(…), I decided to memorise pi to 20 digits to use as my school login password, being about the maximum length password you could have.
Unbeknownst to me, whoever printed it had left one of the pieces of the tractor feed folded over on itself when they hung it up, leaving out a section of the first 20 digits.
I used that password all through school, thinking i was so clever. Until i tried to unrelatedly show off my knowledge of pi and found I’d learned the wrong digits.
I still remember that password / pi to 20 wrong digits. On the one hand, what a waste of brain space. On the other hand, pretty secure password I guess?
Not sure if this is sarcasm, but I sure hope so…
I just use the last 10 digits of pi for all my passwords.
0123456789
Right?
It’s a Criminal Minds reference, though people do use this method, including me.
I use encryption and… modern… 2024 standards.
Pi, tho. I mean, you do you.
0.101001000100001000001 . . .
I’m infinite and non-repeating. Can you find a 2 in me?You can’t prove that there isn’t one somewhere
You can, it’s literally the way the number is defined.
Defined where ?
It’s implicitly defined here by its decimal form:
0.101001000100001000001 . . .
The definition of this number is that the number of 0s after each 1 is given by the total previous number of 1s in the sequence. That’s why it can’t contain 2 despite being infinite and non-repeating.
Pi is often defined as 3.141 592 653… Does that mean Pi does not contain any 7s or 8s?
0.101001000100001000001 . . .
Might very well be :
0.101001000100001000001202002000200002000002 …
Real life, is different from gamified questions asked in student exams.
Implicitly defining a number via it’s decimal form typically relies on their being a pattern to follow after the ellipsis. You can define a different number with twos in it, but if you put an ellipsis at the end you’re implying there’s a different pattern to follow for the rest of the decimal expansion, hence your number is not the same number as the one without twos in it.
assumption ≠ definition
Why couldn’t you?
Because you’d need to search through an infinite number of digits (unless you have access to the original formula)
Are you trying to say the answer to their question is no? Because if so, you’re wrong, and if not I’m not sure what you’re trying to say.
The conclusion does not follow from the premises, as evidenced by my counterexample. It could be the case that every finite string of digits appears in the decimal expansion of pi, but if that’s the case, a proof would have to involve more properties than an infinite non-repeating decimal expansion. I would like to see your proof that every finite string of digits appears in the decimal expansion of pi.
Well that’s just being pointlessly pedantic, obviously they fucking know that a repeating number of all zeros and ones doesn’t have a two in it. This is pure reddit pedantry you’re doing
You might want to stay away from higher maths and all discussions around it, like this one.
You might want to play in traffic
It kind of does come across as pedantic – the real question is just that “Does pi contain all sequences”
But because of the way that it is phrased, in mathematics you do a lot of problems/phrasing proofs where you would be expected to follow along exactly in this pedantic manner
It’s a math proof. Chill.
Are you Pi?
Shit, opsec fail.
Yeah. This is a plot point used in a few stories, eg Carl Sagan’s “Contact”
Replace numbers with letters, and you have Jorge Luis Borges’ The Library of Babel.
https://libraryofbabel.info/ kinda blows my mind.
Not accurate. Pi needs to be a normal number for that to happen, something yet to prove/disprove.
Yes
Can you prove this? Or link a proof?
I don’t know of one but the proof is simple. Let me try (badly) to make one up:
If it doesn’t go into a loop of some kind, then it necessarily must include all finite strings (that’s a theoretical compsci term).
Basically, take a string of any finite length, and then view pi in inrements of this length. Calculate it out to double the amount of substrings of length of your target string’s interval you have [or intervals]. Check if your string one of those intervals. If not, do it again until it is, doubling how long you calculate each time.
Because pi is non-repeating, each doubling in intervals must necessarily include at least one new interval from all other previous ones. And because your target string length is finite, you have a finite upper limit to how many of these doublings you have to search. I think it’s n in the length of your target string.
Someone please check my work I’m bad at these things, but that’s the general idea. It’s also wildly inefficient This doesn’t work with Infinite strings because of diagnonalization.
Your first sentence asserts the claim to be proved. Actually it asserts something much stronger which is also false, as e.g. 0.101001000100001… is a non-repeating decimal which doesn’t include “2”. While pi is known to be irrational and transcendental, there is no known proof that it is normal or even disjunctive, and generally such proofs are hard to come by except for pathological numbers constructed specifically to be normal/disjunctive or not.
Let me give another counterexample. Let x be the binary expansion of pi i.e. the infinite string representing pi in base 2.
Now you will not find 2 in this sequence by definition but it’s still a non-repeating number.
Now one can validly say that we restricted our alphabet and we should look only for finite strings with digits that actually occure in the number. The answer is the string “23456789” concatenated with x.
That’s like saying your car is busted because it can’t drive on a road made of broken glass.
That’s mathematics. It do be like that sometimes. Counterexamples can be stupid but still valid.
It’s on you to prove your claims.
No this does not work. Counter example can be found in the comments here of a non-repeating number that definitely does not contain all finite strings.
Edit: I think the confusion is about the word non-repeating. Non repeating does not mean a subsequence cannot repeat but that you cannot write the number as a rational.
Are you talking about a different base/character set? I think every single person understands that.
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
