r/askmath • u/Loose-Balance3225 • May 06 '26
Discrete Math Can infinity contain infinity
If pi has no end it has to have every combination of numbers but could it hold an infinite combination? Like 1 2 3 4... To infinity
Please help this is missing my brain up
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u/2ndcountable May 06 '26
If pi is a normal number, as we suspect, it 'must' contain any finite sequence of digits. That is, given any finite sequence of digits, like '3333' or '123456789', it must exist within pi. However, even if pi is normal, it does not have to contain a given infinite sequence of digits. For example, we know for sure that pi does not contain the infinite sequence '333333333...', since that would mean its decimal expansion is eventually periodic and hence that pi is rational, which we know to be false.
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u/mousicle May 06 '26
As far as normal numbers, right now we have no test as to whether a number is normal or not. The only numbers we know for sure are normal are ones specifically constructed to be normal like 0.123456789101112 ... We do know that almost all numbers are normal though.
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u/Bills_afterMATH May 06 '26
What do you mean by test? For example, it’s trivial to show that x is normal in base b iff the sequence (x*b^n) is uniformly distributed mod 1. The set of numbers normal in base b was proven to be Pi_3^0 complete by Ki and Linton in 1994. So any condition that can characterize normality has to be a Pi_3^0 condition. Are you talking about some condition that checks (but can sometimes not determine) if a number is normal based on a finite amount of information?
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u/KattyTheEnby May 06 '26
For example, we know for sure that pi does not contain the infinite sequence '333333333...', since that would mean its decimal expansion is eventually periodic
How come pi containing an infinite series ov 3s, or any number, at some point implicate that it is periodic?
Can an entropic number, like pi, not contain a series ov random numbers and then have, sandwiched within it, an infinite series ov numbers?*
(* I am necessarily implying that there is some order ov magnitude to the infinities, since there would be an infinitely long sequence ov random numbers on both sides ov the infinitely long sequence ov 3s.)
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u/Head_Evening_5697 May 06 '26
How can you sandwich an infinite sequence inbetween two other sequences?
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u/2ndcountable May 06 '26
The fractional part of the decimal expansion of a number is indexed by N. That is, to the natural number 1 corresponds the first digit of the fractional part, to 2 corresponds the second digit, and so on. Let ai be the i-th digit of the fractional part, i.e. the digit that the natural number i corresponds to. Suppose pi contains the infinite sequence 3333..., starting at the index t. Then, we must have that a_t = 3, a(t+1) = 3, a(t+2) = 3, and so on. Then we have a_i = 3 for all i >= t; Indeed, when i >= t, a_i = a(t+d) = 3, where d = i-t >= 0. It is, however, possible to consider "an infinite sequence with another infinite sequence sandwiched within it", just not within a decimal expansion; For this, you can consider the concept of ordinal numbers. For example, a sequence indexed by w+w could indeed have an infinite sequence of 3s, followed by a different infinite sequence.
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u/KattyTheEnby May 07 '26
For example, a sequence indexed by w+w
w+w?
I'm guessing you mean in a similar to how, for example, complex numbers work (i.e.
2i + 5), rather than literally multiplying by two. Is that right?•
u/jsundqui May 06 '26
Decimal expansion can only have one infinite sequence, there cannot be different infinite sequences within one decimal expansion. So if, starting at some point, the decimal expansion contains infinite 3's, that's it, there is not room for anything else.
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u/trutheality May 06 '26
(* I am necessarily implying that there is some order ov magnitude to the infinities, since there would be an infinitely long sequence ov random numbers on both sides ov the infinitely long sequence ov 3s.)
That's not going to work here. A contiguous infinite subsequence of an infinite sequence is necessarily one of the latter's tail sequences.
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u/Fickle_Engineering91 May 06 '26
Just because pits decimal string is infinitely long doesn't mean that it contains everything. Consider 1/3 = 0.33333... or 1/7 = 0 142857142857142857... Both are infinitely long but neither contains 6.
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u/Iksfen May 06 '26
Since pi infinite digit expantion it contains every finite string of digits.
That is not true in principle. For example a number 0.101001000100001000001... (each time the string of 0s is one longer) has infinite expansion, but it does not contain finite string 2. Pi is also not proven to contain each finite string of digits. It is only strongly suspected.
Let's define mi = 0.123356789101112131415... (each finite string of digits written one after the other). Most infinite strings of digits will not appear in it. For example "11111111..." will not appear since no matter how far you go there will be a finite string at some point that will contain another digit
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u/u8589869056 May 06 '26
No, it doesn’t have to have every possible subsequence in it.
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u/Loose-Balance3225 May 09 '26
Not infinite one's
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u/u8589869056 May 09 '26
It doesn’t even have to contain every finite subsequence. It might, it might not.
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u/ButterSlicerSeven May 06 '26
Yes. Here's a 7th grade maths proof.
Take the natural numbers: 1, 2, 3 ... and so on to infinity. The set of natural numbers is infinite. Now take integers: ... -3, -2, -1, 0, 1, 2 ... and so on to infinity. It's clear that the set of integers includes the set of natural numbers. Thus, infinity can contain infinity.
Alternatively, we can take a geometric approach: take a straight line. Straight line goes to infinity in two directions. Now divide it in 2, the two resulting halves will go to infinity in opposite directions.
As per Pi, this is not really a related concept. What you want to know is whether Pi is a normal number or not, which we do not know for certain. It might be, or it might be not, we have no clue.
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u/Mishtle May 06 '26
If pi has no end it has to have every combination of numbers
This isn't true. Such a number is called "normal". Most non-repeating numbers are normal and pi is believed to be one of them. It's not been proven to be one though.
The number 0.1010010001... never repeats but only contains 0s and 1s.
but could it hold an infinite combination? Like 1 2 3 4... To infinity
Any sequence has what are called subsequences. Ignoring any number of items from the original sequence gives a new sequence.
Using the non-repeating number above, both the numbers 0.111... and 0.000..., along with many others, could be made from a subsequence of its digits.
If a number is normal, then it would contain every infinite sequence as a subsequence in this way. You might need to wait arbitrarily long for each next element to appear though.
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u/Bills_afterMATH May 06 '26
Not correct. A number for which every block of digits appear is called either rich or disjunctive. Not normal. Of course all normal numbers are rich. Normality is a significantly stronger and more complicated condition than richness.
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u/Exciting_Audience601 May 06 '26
it has to have every combination of numbers
this does not follow.
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u/Loose-Balance3225 May 09 '26
Do you even know what infinity is?
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u/Exciting_Audience601 May 19 '26 edited May 19 '26
ok give an argument why an infinite non repeating decimal has to have every possible combination of numbers and why that argument is not refuted by 0.10110111011110... and so on. or give a proof that pi is normal in base 10.
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u/TheTurtleCub May 06 '26
If pi has no end it has to have every combination of number
No it doesn't have to.
How about 0.101001000100001....
But yeah, an infinite set can contain many infinite sets inside. The reals contain the naturals, and the rationals, and the multiples of Pi, etc
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u/Infobomb May 06 '26
1/7 has no end (when you spell out its decimal expansion) but it does not include every combination of digits. Just because a string of digits is infinite does not mean it has to contain a specific pattern.
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u/Suitable-Elk-540 May 06 '26
First, Pi doesn't have to have "every combination of numbers".
Whatever "infinite combination" of numbers it has would be part of the definition of Pi. I'm not sure what you're asking, but you can't just come up with some infinite "combination of numbers" and expect to find that in Pi. That combination would have to just be equal to Pi after some finite number of leading digits. Otherwise, every infinite sequence would be part of every other infinite sequence somehow.
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u/jajwhite May 06 '26
As far as I know that’s a definition of infinity sometimes used, that you can remove an infinite set without making it smaller. So there are the whole numbers and you can remove the infinite set of even numbers leaving the size intact.
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u/Earthhorn90 May 06 '26
We should build an infinitely big hotel to hold an infinite number of guests... and infinitely more.
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u/HotPepperAssociation May 06 '26
There are different kinds of infinities. Countable and uncountable. https://youtu.be/Efj1ZSsHVcw
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u/fallen_one_fs May 06 '26
No.
It can contain any finite combination of numbers, no matter how large, but not an infinite combination.
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May 06 '26
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u/jsundqui May 06 '26
Pi can't contain Pi because then Pi would be a rational number.
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May 06 '26
[removed] — view removed comment
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u/jsundqui May 06 '26
I meant that "can you also find pi somewhere within the digits of pi" the answer is no because it would force pi to be a rational number.
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u/tbdabbholm Engineering/Physics with Math Minor May 06 '26
Pi could be 3.1415...123456789101112... yes. It would be very strange for that to be the case but there's nothing that would make it impossible
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u/2ndcountable May 06 '26
Surprisingly, we can prove that it is impossible. The base-10 champernowne constant 0.123456789101112... has irrationality measure(https://en.wikipedia.org/wiki/Irrationality_measure) equal to 10. By the definition of the irrationality measure, any number with base-10 decimal expansion terminating in 1234567... has the same irrationality measure of 10. Hence, if pi were to be 3.1415...123456..., it too would have irrationality measure 10, but it is known that this is not the case(https://arxiv.org/abs/1912.06345).
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u/mpaw976 May 06 '26
As stated, no, that's not guaranteed because there's a difference between having every finite string, and every infinite string.
As a simple example,
0.101101110111101111101...
Will contain every finite string of 1s, but it won't contain 111... repeating forever.
In fact, any given decimal number only "has room" to end in a single infinite string. E.g. it can't end in both all 0s and all 1s.
Isn't that neat?
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u/mattynmax May 06 '26
Well infinity isn’t a number, so no. It’s like saying “does infinity contain a potato inside it?”
Your question of “does pi contain all the numbers 1,2,3,4….”, the answer to that is “we don’t know”
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u/RecognitionSweet8294 Philosophy ∧ Math May 06 '26
No it doesn’t have to have any combination of numbers. That’s a theorem that hasn’t been proven yet.
It depends on what you mean by „holding“. Does it have to be in consecutive order…?
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u/badalya1 May 07 '26
just because it has no end doesn’t mean it must have every combination of numbers
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u/ikonoqlast May 09 '26
Yes. The infinite set of integers contains the infinite set of even integers as a subset for instance.
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u/idksomerandomcrap May 06 '26
Yes, you don't need complicated math to prove it. You have a set of numbers, 1 to infinity. This set contains all numbers. You have another set that is all even numbers. The set of all even numbers would still be infinite. The set of even numbers will be contained within the first set. There are different levels of infinity.
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u/idksomerandomcrap May 06 '26
Oh I didn't read the post fully, idk about pi specifically. I dont think we can prove anything about that set.
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u/shwilliams4 May 06 '26
The set of even numbers is not larger than the set of numbers so it is not contained. They are the same cardinality.
A correct example would be the reals and the integers.
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u/Bills_afterMATH May 06 '26
Whether it’s “larger” depends on the notion. They have the same cardinality, but the set of even numbers has density 1/2 compared to the set of natural numbers which has density 1.
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u/shwilliams4 May 06 '26
My math degree says that’s wrong but I’m old. Can you send a paper on this?
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u/Bills_afterMATH May 06 '26
My math degree says there's a lot more than cardinality that's frequently used to compare infinite sets (measure, Hausdorff dimension, Hausdorff measure (when Hausdorff dimensions are equal), many variants of asymptotic densities (when comparing subsets of the natural numbers), Packing dimension, winning for Schmidt's game, meagre vs comeagre, etc.). All of these notions are fairly old and I'm surprised that you only considered cardinality.
Specifically, for the vanilla version of density, you consider any A \subset \mathbb{N}$. You can define the density of $A$ to be
$d(A) := \lim_{n \to \infty} \frac{A \cap [1,n]} {n},$
assuming it exists (you can easily construct examples of subsets of the natural numbers that don't have a density). For the example being considered, you can check that $d(2\mathbb{N})=1/2$. You can also show that the density of the set of square free numbers is $6/\pi^2$. There are also variants of density taking a liminf or limsup (to make sure every subset has a density) and there are also uniform versions (for example, upper Banach density shows up a lot in ergodic Ramsey theory).
Edit: forgot to mention that density is NOT $\sigma$-additive. For example, every singleton has density 0 and the set of natural numbers has density 1. So, you don't want to use this as a measure.
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u/idksomerandomcrap May 07 '26
I cant bother to check if that is actually notation for anything or not, but this looks like we broke the bot lmao
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u/tottasanorotta May 06 '26
Cardinality without this idea of density seems very misleading for describing sizes of infinite sets. It seems extremely unintuitive when someone says to me that the even natural numbers have the same size as the natural numbers. You can keep pairing them up to as far as you like successfully, but you can't tell me that there isn't half as many evens for every finite n.
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u/idksomerandomcrap May 07 '26
Youre so close. Thats kind of exactly the point. Yes there are half as many, but it is still an infinite set. Even though there is half the numbers in the set, half of infinity is infinity. Thus there are different sizes of infinity. You call it density, sure. I put it in laymans terms because reddit users are not rocket scientists.
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u/Bills_afterMATH May 07 '26
It captures this intuition of being “half as big” for some nice situations like the set of even numbers. Just like measure does. For example, the Lebesgue measure of (0,1/2) is 1/2 and of (0,1) is 1. Just need to be really careful since density is often not well behaved. Probably one of the reasons that it’s not usually introduced in something like an intro to proof course.
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u/GatePorters May 06 '26
Infinity isn’t a number, but a boundary of a system. It has no fixed value. So yes, an infinity can contain another. But not every infinity can contain every infinity.
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u/Idksonameiguess May 06 '26
Nope. Take 0.3333... for example, has no end and yet doesn't have every combination of numbers. We have no proof that pi contains every combination of numbers, but it is hypothesized.
We do, however, know the following:
Let a be some infinite combination of digits. If pi contains a, then
pi = n/10^k + a, with n and k being whole numbers (this is just by definition). Therefore, it can't contain 2 different infinite combinations which do not contain each other. This is true in general for any number, not just pi.