How about ANY FINITE SEQUENCE AT ALL?
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.
Exceptions are infinite. Is that rare?
Yes, compared to the infinitely more non exceptions. For each infinite number that doesn’t contain the digit 9 you have an infinite amount of numbers that can be mapped to that by removing all the 9s. For example 3.99345 and 3.34999995 both map to 3.345. In the other direction it doesn’t work that way.
Very rare in the sense that they have a probability of 0.
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.
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.
I’m a layman here and not a mathematician but how does it store the complete value of pi and not rounded up to a certain amount? Or do one of the libraries generate that?
You generate it when needed, using one of the known sequences that converges to π. As a simple example, the
pi()recipe here shows how to compute π to arbitrary precision. For an application like pifs you can do even better and use the BBP formula which lets you directly calculate a specific hexadecimal digit of π.
I want that project continues so hard. Sounds amazing
Thanks. I love these kind of fun OpenSource community projects/ideas/jokes whatever. The readme reminds me of ed
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.
its been proven for some other numbers, but not yet for pi.
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
Yes.
And if you’re thinking of a compression algorithm, nope, pigeonhole principle.
Not just any all finite number sequence appear in pi
Can you prove this? Or link a proof?
no. it merely being infinitely non-repeating is insufficient to say that it contains any particular finite string.
for instance, write out pi in base 2, and reinterpret as base 10.
11.0010010000111111011010101000100010000101...it is infinitely non-repeating, but nowhere will you find a 2.
i’ve often heard it said that pi, in particular, does contain any finite sequence of digits, but i haven’t seen a proof of that myself, and if it did exist, it would have to depend on more than its irrationality.
Isnt this a stupid example though, because obviously if you remove all penguins from the zoo, you’re not going to see any penguins
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.
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In terms of formal logic, this…
Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating digits from 0-9 should appear somewhere in Pi in base 10?
…and this…
Does any possible string of infinite non-repeating digits contain every possible finite sequence of non repeating digits?
are equivalent statements.
The phrase “since X, would that mean Y” is the same as asking “is X a sufficient condition for Y”. Providing ANY example of X WITHOUT Y is a counter-example which proves X is NOT a sufficient condition.
The 1.010010001… example is literally one that is taught in classes to disprove OPs exact hypothesis. This isn’t a discussion where we’re both offering different perspectives and working towards a truth we don’t both see, thus is a discussion where you’re factually wrong and I’m trying to help you learn why lol.
Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating digits from 0-9 should appear somewhere in Pi in base 10?
Does any possible string of infinite non-repeating digits contain every possible finite sequence of non repeating digits?
Let’s abstract this.
S = an arbitrary string of numbers
X = is infinite
Y = is non-repeating
Z = contains every possible sequence of finite digits
Now your statements become:
Since S is X and Y, does that mean that it’s also Z?
Does any S that is X and Y, also Z?"
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.
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.
“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.
OK, fine. Imagine that in pi after the quadrillionth digit, all 1s are replaced with 9. It still holds
“ok fine consider a number that still isn’t pi, it still holds.” ??
They didn’t convert anything to anything, and the 1.010010001… number isn’t binary
then it’s not relevant to the question as it is not pi.
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
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It does contain a 2 though? Binary ‘10’ is 2, which this sequence contains?
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
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.
this is correct but i think op is asking the wrong question.
at least from a mathematical perspective, the claim that pi contains any finite string is only a half-baked version of the conjecture with that implication. the property tied to this is the normality of pi which is actually about whether the digits present in pi are uniformly distributed or not.
from this angle, the given example only shows that a base 2 string contains no digits greater than 1 but the question of whether the 1s and 0s present are uniformly distributed remains unanswered. if they are uniformly distributed (which is unknown) the implication does follow that every possible finite string containing only 1s and 0s is contained within, even if interpreted as a base 10 string while still base 2. base 3 pi would similarly contain every possible finite string containing only the digits 0-2, even when interpreted in base 10 etc. if it is true in any one base it is true in all bases for their corresponding digits
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
Yeah. This is a plot point used in a few stories, eg Carl Sagan’s “Contact”
Not accurate. Pi needs to be a normal number for that to happen, something yet to prove/disprove.
Replace numbers with letters, and you have Jorge Luis Borges’ The Library of Babel.
https://libraryofbabel.info/ kinda blows my mind.
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 or with a finite decimal representation. I.e. it’s not 3.ba repeating. Where a is a finite sequence that repeats infinitely and b is a finite sequence.
Edit edit: another assumption you make is that pi does not go into a loop of some kind. You would need to prove that.
Are you talking about a different base/character set? I think every single person understands that.
See my other comment
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?
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.
