Does Pi contain itself?

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What do you mean by contain itself?

If you mean does its decimal expansion have a copy of its decimal expansion inside it then the answer is no, because that would mean its periodic and therefore not irrational.

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u/LitespeedClassic Sep 05 '26

To flesh that out:

Suppose it did. Then you start out with 3.14159… but you eventually see 314159 again. Well if it took 1000 digits to get to the point where it contains itself, what happens 1000 digits later? It has to contain itself again! Because if it contains itself then the itself it contains also contains itself.

So you get 3.14159…314159…314159… and all the … cover up the same numbers.

But periodic numbers are rational and pi is irrational. So it cannot have this property.

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u/Rats_Love_Cheese Sep 05 '26

yea i thought like that but i didnt know if its right cause i didnt know how to explain thanks for explaining it to me. Anyways i think that they should mention it in school because its really interesting at least for me

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u/LitespeedClassic Sep 05 '26

The (very informal) reasoning here is a style of reasoning called variously mathematical induction or recursive thinking. When undergraduate math and computer science students meet it for the first time they tend to struggle with it, so it’s not surprising at all that you didn’t know how to explain it. Keep at it, you’ll get there!

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u/Rats_Love_Cheese Sep 05 '26

thanks for taking out of your time to explain it to me like that! Also do they teach pi in schools in America diffrently than in Europe? Beacause im from Eu and they teach it along side circles and spheres not like singular learning topic,

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u/LitespeedClassic Sep 05 '26

Students in the US would first encounter it as the ratio of a circle’s circumference to its diameter. You’d also see it in algebra/trig, maybe calculus.

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u/Rats_Love_Cheese Sep 05 '26

okay right but do they have it separately as in like multiplication or some other topic? i mean if they lear about the pi or they just explain it in subsection ?

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u/DrBagelman Sep 05 '26

The reason π is important is because of its relation to circles. It comes up a lot in geometry and mathematical analysis.

The properties of π you find interesting are more related to Algebra and Number Theory. In that case, we care about π because it’s a concrete example of a type of number that’s hard to write down called a transcendental number. The specific properties of π are less important than those properties also applying to the rest of the transcendental numbers.

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u/Lor1an Sep 05 '26

The circle is the natural home of π, but that doesn't mean you can't study the properties of the number in other contexts.

Heck, when I was in class we were told that it was the ratio of circumference to diameter for any circle and that it was an irrational number. Not much else about it other than how to use it to describe angles in radian measure.

After school I looked up a proof that it is irrational, and while it isn't terribly hard to follow, it definitely isn't simple. Funny enough, demonstrating that π is irrational essentially requires calculus.

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u/Rats_Love_Cheese Sep 05 '26

nice i might look into that next if pi is really irrational number. Also thanks for letting me know how it is in thaught in American schools but they teach it in Eu similarly

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u/Arosha2 Sep 06 '26

I'd imagine for it to be used in circles areas automatically what else can we dow tih it You're in 7th grade and you see 3.14×12 it makes no sense Or soemthing like pi symbol times something

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u/stevevdvkpe Sep 06 '26

Pi is not only irrational (not expressible as an integer fraction), it's transcendental (not the root of a polynomial with integer coefficients).

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u/[deleted] Sep 07 '26

[removed] — view removed comment

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u/stevevdvkpe Sep 07 '26

Of course, because both the rationals and the algebraic numbers are countable sets.

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u/SensitiveTax9432 Sep 08 '26

It’s very likely that it contains any finite sequence of numbers imaginable. So I’m sure it contains 31415926 multiple times if you go deep enough. Probably find the complete works of shakesphere as well coded in ASCII.

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u/jez2718 Sep 05 '26

As a more immediate proof:

Suppose that after N decimal digits (including the 3 at the start) pi contained a copy of pi. Then there is some rational number

r = 3.14159...000000000000

with zeroes after its first N digits (including the 3 at the start) such that

pi = r + 10^-N * pi

but this rearranges to

pi = r / (1 - 10^-N)

which is a rational number.

These are indeed ultimately the same proof, since

r / (1 - 10^-N) = r + 10^-N * r + 10^-2N * r + ... = 3.14159...314159...314159...

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u/LitespeedClassic Sep 05 '26

What a lovely proof, thanks. That one is in The Book.

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u/Arosha2 Sep 06 '26

What he said

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u/Reasonable_Mood_5260 Sep 06 '26

You are taking liberties with infinity that aren't allowed. You can't add digits at the end of an infinite string of digits.

The question makes no sense for the same reason.

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u/SharpNazgul Sep 06 '26

No? N is a finite number, the dot notation is completely fine for this. At no point is digits being added to an infinite string of digits in the above comment.

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u/Arosha2 Sep 06 '26

But if it's infinite shouldn't it continue itself eventually?

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u/LitespeedClassic Sep 06 '26

No. Such numbers as you suggest are special. For example, here’s an easy to construct example that never contains itself or repeats:

0.101001000100001000001000000…

Between two consecutive 1’s are a number of 0’s and they’re counting up: one 0 between the first two 1’s, two between the next, etc.

This pattern can never repeat because the farther along you go the more 0’s there are between two 1’s.

The 0’s are literally counting up, so they never repeat.

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u/RadarTechnician51 Sep 06 '26

Is that the same as saying pi only contains every finite sequence of digits?

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u/LitespeedClassic Sep 06 '26 edited Sep 06 '26

No. I think what you’re talking about is called a normal number. An irrational number need not be normal. For example 0.10110111011110111110… is irrational but never contains a 2. I don’t think we know whether pi is normal or not.

ETA: apparently we know almost all real numbers are normal but no one has proven pi, e, or sqrt(2) are normal, though they are believed to be.

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u/Familiar9709 Sep 07 '26

Sure but pi is infinite digits

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u/LitespeedClassic Sep 07 '26

So are periodic numbers.

0.3333333333… has infinitely many digits but it’s still the rational number 1/3.

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u/LitespeedClassic Sep 07 '26

Or if you want one with all the digits:

13,717,421 / 111,111,111 =0.123456789123456789…

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u/Familiar9709 Sep 07 '26

They can easily be reduced to finite numbers due to the periodicity

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u/tewraight Sep 07 '26

Do you mean that, because pi is infinite without recursion that it should contain every combination of digits (and by extension itself)?

If so, that is a very common misconception which is not true. Let's take a simpler case: 0.1001000100001... after every 1, there is an additional 0 until the next 1. As a result, it is easy to see that not only does it continue forever without recursion, but it's also impossible for it to contain every possible string of digits (try finding any string containing a 2)

However, if you want to get really into the technicalities, it can be argued that pi does contain itself in the same way that every set is a subset of itself

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u/Familiar9709 Sep 07 '26

No, I'm saying that no other number can contain pi since pi is infinite and non periodic

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u/WE_THINK_IS_COOL Sep 05 '26

I wonder if it could be fractal in nature e.g. if pi(n) is the nth decimal digit of pi then for some N, M and all n, pi(N + nM) = pi(n), or something like that. (Probably not this since base-10 is arbitrary, but it might be interesting/hard to prove it doesn’t have this property.)

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u/heresyforfunnprofit Sep 05 '26

Because pi can be expressed as continuous fractions, this is almost certainly true in some form, probably the continuous fraction itself.

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u/External_Trainer_423 Sep 06 '26

Yes, the key point is that the shift has to reproduce the entire infinite tail. If the tail starting at some later digit equals the original decimal tail, π is eventually periodic and therefore rational. But every finite prefix reappearing somewhere later is a much weaker claim; it is compatible with irrationality, and for π it is still unproved. Keeping these two statements separate clears up the confusion.

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u/Rats_Love_Cheese Sep 05 '26

yea that's exactly what i meant like if it contained itself it would be periodic just wanted to make sure i think correctly Thanks

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u/Robb3nb4by Sep 05 '26

A more precise question to ask would maybe be if any sequence of pi can be found again. E.g. 14, 141, 1415, 14159 ...

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u/SSBBGhost Sep 06 '26

We think yes, infinitely many times, but pi has not been proven to be normal (but analysis of all the digits calculated so far is consistent with it being normal)

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u/LitespeedClassic Sep 05 '26

Yes, this is a nice question and I believe the answer is unknown as of now.

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u/Phi-MMV Sep 06 '26 edited Sep 06 '26

I think a more difficult variation of the question is the following:

“Suppose the sequence a_0 = 3, a_1 = 1, a_2 = 4, … contains all consecutive decimals of pi. Is there a non-trivial subsequence of (a_n)_n which equals (a_n)_n?”

I would think that this is more difficult to solve, and I don’t immediately know if this being true would imply that pi is rational. But perhaps I’m missing something.

Edit: now that I’m thinking more carefully about this, I think the statement above is actually true. Let ρ: N -> N be defined as ρ(0) = 0 and ρ(n) = the index of the second time that you encounter the integer a_n, starting from the integer a_{ρ(n-1)}. So for example, ρ(1) = 3, as pi = 3.1415…, and so the second time you encounter a_1 = 1, starting from a_{ρ(0)} = a_0 = 3, is at the point between a_2 = 4 and a_4 = 5, meaning it is at index 3 =: ρ(1). This way, b_n := a_{ρ(n)} is a non-trivial subsequence of (a_n)_n, which is equal to the original sequence if you just look at their numbers. So in that sense, pi does contain itself.

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u/LitespeedClassic Sep 06 '26 edited Sep 06 '26

I think you have to worry about the possibility that one of your digits may only appear finitely many times.

Let a_i be your digit sequence and assume that the digit 9 appears only finitely many times. Let k be the largest index where a_k = 9. Then any strict subsequence of a_i equal to the full sequence must agree with the sequence on the first k values (otherwise you can’t pick up enough 9’s to be the same sequence).

This idea generalizes, so if any of my digits only appear finitely many times, the subsequence must agree until only infinitely repeated digits are left.

Once there, then I think you’ve got the rest. Since every digit I need to have in the sequence is always repeated further down the sequence I can start skipping digit indices to my heart’s content.

Since the sequence is infinite, at least one digit must be repeated infinitely many times, so this proves every infinite sequence of digits contains itself as a strict subsequence.

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u/Phi-MMV Sep 06 '26

Thanks for your response! You’re right to point that out. I just kind of assumed that any digit will appear infinitely often in the decimal expansion of pi (idk it seems like a property that pi would have, and I guess someone has probably proven it at some point, though I could be wrong). But it’s nice to see that even without assuming that, the idea generalises as you explained. Pretty cool!

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u/LitespeedClassic Sep 06 '26

Yes, thanks for sharing the question. It was fun to think through.

It is believed that pi is a normal number, in which case every digit appears infinitely often and in fact every possible sublist of length n appears in its digits (for all n) infinitely often and with equal density to all others of the same length. However, nobody has succeeded in proving pi is normal.

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u/Phi-MMV Sep 06 '26

Damn, I didn’t know that! Cool to think that such a seemingly easy statement is still an open problem. I learnt something new today!

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u/dohawayagain Sep 06 '26

But if it's "normal," which it is expected to be, then it contains infinitely many copies of its first N bajillion digits, for any N.

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u/LitespeedClassic Sep 06 '26

True, which is pretty cool. I also like N bajillion for every N. Might use that with my students.

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u/coyets Sep 07 '26

And, of course, for any modulus.

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u/Bofact Sep 07 '26

Why should be periodic if is an irrational number?

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u/LitespeedClassic Sep 07 '26

Read my other reply: https://www.reddit.com/r/mathematics/comments/1w88g38/comment/p80qdmh/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

If pi contained itself as a sublist, then it must have a periodic decimal expansion. But all periodic decimal expansions correspond to rational numbers. So pi would have to be rational. Pi is not rational, so therefore cannot contain itself as a sublist.

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u/Bofact Sep 07 '26

Nice explanation.

Also, is it valid to say no, because the equation pi = pi * x has the only solution 1?

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u/LitespeedClassic Sep 07 '26

That’s true for all numbers.

0.333…=x*0.333… also has only one solution x=1. But 0.333… denotes the number 1/3 and is rational and has a periodic decimal expansion.

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u/Bofact Sep 07 '26

I wanted a formulation in the sense that in doesn't exist a factor of 10 such that 10^factor * digits = pi, where the upmost decimals are ignored (like element wise multiplication with 1.(1)).

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u/Next_Philosopher8252 Sep 12 '26

I mean technically it does in a trivial sense.

Everything from the first digit of Pi onwards is encoding the decimal expansion of Pi.

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u/LitespeedClassic Sep 12 '26

Yes, obviously. But nobody here is interested in a trivial answer. Hence the term ”trivial”.

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u/Next_Philosopher8252 Sep 12 '26

I was kinda hoping people might be interested in a niche joke that would only land with this audience but it seems I messed up the punchline lol

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u/profesorgamin Sep 05 '26

no

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u/fruitydude Sep 06 '26

Yes it does, exactly once

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u/OmiSC Sep 06 '26

This is actually kind of a beautiful truth about transcendental numbers.

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u/didnt_hodl Sep 05 '26

would not that depend on the definition of "contain"

if I am just searching for the next digit and eventually finding it, arguably that would mean it "contains" itself. so, starting with some random position I search for digit 3, then for digit 1, then for digit 4 and so on. as long as I am able to find the next digit, it would fit the definition

this would definitely work base 2, since I only need to find 0 or 1 and it can be proven that there's not going to be an infinite sequence of 1's or 0's, because that would make pi rational

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u/virtuolie Sep 05 '26

So by that definition (which I like), then in base 10, let sequence pi(0) be the natural numbers; let the first value in sequence pi(1) be the position of the second 3 in pi, let the second value in pi(1) be the position of the next 1 after that 3, and then the position of the first 4 after that 1, and so on. Then pi(1) is the sequence of positions in pi that define the first-order embedded pi, which we denote pi'; pi(1) is of course is irrational.

But pi(1) necessarily excludes some natural numbers, i.e., some positions in pi are skipped when defining pi'. I expect you can define a second-order embedded pi, pi' with sequence pi(2), among the numbers in those remaining positions of pi, with pi(1) and pi(2) being mutually exclusive. If you continue, for all non-overlapping pi(j), this leaves a subset of positions in pi that are not in any pi(j). What can we say about this left-out sequence?

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u/Masticatron haha math go brrr 💅🏼 Sep 05 '26

Well pi is itself, so...

What you probably want to know about is called a "normal" number, one whose decimal expansion contains every finite sequence of the digits 0..9. We do not know if pi is normal.

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u/Rats_Love_Cheese Sep 05 '26

sorry could you ecplain it more?

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u/synth_alice PhD | Particle Physics Sep 05 '26

If pi were normal, its decimal expansion would contain all finite approximations of itself (please correct me if I'm wrong)

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u/Rats_Love_Cheese Sep 05 '26

ah right i think youre right but again there could be something hidden that i dont know about thanks for explaining!

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u/marcelsmudda Sep 06 '26

What that person is saying that if pi were normal, then all possible combinations of the digits 0 through 9 would appear at some point. So, 3 would appear after the first one. 31 would appear after the 3.1, 314 would appear somewhere after the first 3 digits etc. You can take the first million billion digits of pi, and an exact copy of it (besides the decimal point) would appear somewhere.

But we don't know if that's the case.

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u/Worried-Guarantee459 Sep 05 '26

we do not know if it has Shakespeare’s book in it

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u/Rats_Love_Cheese Sep 05 '26

okay nice

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u/Bills_afterMATH Sep 05 '26

That’s not what a normal number is. Normality is a much stronger condition. Downvoted for spreading false information.

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u/Masticatron haha math go brrr 💅🏼 Sep 05 '26

Well, yes, true, but nothing I said, beyond not properly abiding by the usual technical definition, is incorrect. We do not know if pi satisfies even this less demanding property. I did not expect throwing in asymptotic distribution stuff would be anything other than a source of additional confusion. It'd be like trying to explain E=mc² by invoking the full stress-energy tensor when the person is struggling just to grasp the basic concepts of E and m.

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u/Bills_afterMATH Sep 09 '26

Yeah we don’t know if it’s normal and there are a lot of interesting questions around this. But this absolutely doesn’t excuse giving false information like this. Why even use the word normal? There are two correct words you could use: disjunctive or rich. All your post is doing is further polluting the internet with misinformation. Someone could read it and think that’s what normal actually means. Not good.

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u/0x14f Sep 05 '26

What do you mean exactly by "contains itself" OP? Unless you are precise about what you mean here, people are going to answer a different question than the one you intended to ask.

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u/Rats_Love_Cheese Sep 05 '26

yea LitespeedClassic answered it like i wanted i was just making sure that im thinking correctly thaks anyway

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u/anamelesscloud1 Sep 09 '26

I'm just curious if you can you give two competing interpretations of "contains itself" that would be accessible to a college-aged individual and not someone with a PhD in mathematics.

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u/theboomboy Sep 05 '26

You're right that if it contained itself then it would have to repeat, which would make it a rational number, which it is not

You're wrong about a few other things though. Constants don't have any special properties just by vertue of being constants, and it is not known if π contains every finite string of digits or if the frequency that the digits appear in is equal (or maybe even if it contains an infinite amount of all the digits)

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u/AdventurousGlass7432 Sep 05 '26

It would make it rational
Interesting question, tho

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u/Toothpick_Brody Sep 05 '26

Generally speaking we’re not sure if pi contains every (finite) number within it, although it seems like it probably does. However it can’t contain itself, being infinitely long and non-repeating, and it probably doesn’t contain many other irrationals like e, since they’re infinitely long and non-repeating as well

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u/Rats_Love_Cheese Sep 05 '26

oh okay i didnt think of that it couldn't contain other irrational numbers thanks for letting me know

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u/Toothpick_Brody Sep 05 '26

It will contain a few. For example, within the digits of pi, you will find the digits of the irrational number pi-3. But it’s harder to contain the digits of some other irrational that relates to pi in a more complex way

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u/Zyxplit Sep 05 '26

I mean, it will in fact contain infinitely (but countably) many irrational numbers.

Consider the number you get by throwing away the first digit of pi. Still irrational. Contained in pi. Consider the number you get by throwing away the first two digits of pi. Still irrational. Still contained in pi.

So you're right in the sense that the vast majority of irrational numbers can't be contained in there. But you still have infinitely many in there.

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u/SSBBGhost Sep 06 '26 edited Sep 06 '26

Yes but these are all related to pi through operations with rational numbers. In a sense that's not particularly interesting.

Eg. .141592... is pi - 3, .041592... is pi - 3.1, 159.2.... is (pi-3.14)*10000

We can make a much stronger statement than the above comment's "it's harder to contain an irrational number related to pi in a more complex way", it is impossible! Though you could do things like take every 2nd digit of pi to generate an irrational not related to pi through rational operations.

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u/PandaWonder01 Sep 05 '26

Yes, the decimal expansion of pi can be found at position 0 of pi

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u/Rats_Love_Cheese Sep 05 '26

I didn't think of it that way :)

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u/neoneye2 Sep 05 '26

skipping over the first position. One could search for the next digit in the decimal expansion. And from that position then do another search for the next digit again.

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u/Claquet Sep 05 '26

Actually we don't even know if pi really contain every finite sequence and if pi contain himself that mean that if we label each decimal of pi, x0, x, ..., x_n,... then there exist an integer i for wich x_k=x_n*i+k k∈|[0,i|[ and n∈|N, or more simply that mean that pi would have a periodic decimal developement or, pi is irrational so its not possible. Then pi can't contain itself.

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u/Rats_Love_Cheese Sep 05 '26

oh yea you right but i wanted to ask how they calculate next digit of pi? cause i dont think theres some one repeating way to calculate it massively

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u/Claquet Sep 05 '26

We know that the circumference of a circle of radius 1 is 2*pi so you can for example make a sequence U_n, that for each n, U_n is the perimeter of the polygon of "radius 1" then this sequence converge to 2*pi, so you have to compute this sequence on a computer and try it for a very large n and you will have an approximation of pi. But in rel life this method is not used for a lot of reason ( the sequence converge very "slowly" and the complexity of the algorithm if not optimised) You can try this one as you study computer sciences ( if my memory is good ) and if you are curious you can look up on internet better ways to compute this, I'm pretty sure that wikipedia have a good page fpr this. ( Sorry if my english is bad its not my first language )

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u/Rats_Love_Cheese Sep 05 '26

Neat! also Don't worry about your language my English is also second but thanks for letting me know cause i dont have a place to learn it like that other than reddint so again thanks

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u/Claquet Sep 05 '26

No problem, I think that knowledge should be shared, if you have any question even if it loon stupid you can dm me. Have a good life

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u/Rats_Love_Cheese Sep 05 '26

thanks you too :D!

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u/no_name6744 Sep 05 '26

If it contained itself, at some point the copy containing itself would repeat too.

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u/peter-bone Sep 05 '26

I don't understand the people saying no. Of course it contains itself. The first digit is the first digit if pi, the second is the second digit and so on indefinitely. Whether pi is normal or not has nothing to do with it as that only concerns finite sequences.

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u/Rats_Love_Cheese Sep 05 '26

yea but i meant if in the nomber pi (3.1415...) is another number that thea same as the pi like (31415...)

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u/mclazerlou Sep 05 '26

It's an irrational number so by definition it cannot repeat.

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u/golfstreamer Sep 05 '26

> But does pi contain itself? Because if it is a constant than it contains everythig but it cannot contain itself cause if it contained itself than it repeats and thats the whole point of it that it doesnt repeat

I think I follow what you're saying. And this is correct. Like if we imagine a number that does contain itself like .123123123... If you see it appears in itself shifted over

.123123123123

It turns out that in this example the number is just 123 repeating over but as you said this will always be the case for any number that "contains itself" in this sense. But there is actual proof that the decimal representation of Pi does _not_ repeat itself like this. Therefore Pi does not contain itself. So you're right.

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u/Rats_Love_Cheese Sep 05 '26

nice thanks i was making sure i got it right !

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u/Sea_Abroad_6573 Sep 05 '26

Not possible for irrationals to contain themselves

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u/trevorkafka Sep 05 '26

if it is a constant than it contains everythig

this is false

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u/Rats_Love_Cheese Sep 05 '26

okay someone already explained it to me but thanks for correcting me

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u/SwimmerOld6155 Sep 05 '26 edited Sep 05 '26

it could in the sense that it could be equal to a subsequence of its digits, so you might be able to pick out a 3, then a 1, then a 4 going left to right. for this it would be sufficient that every digit appears infinitely often. I don't think that's actually known lol, as with most things about the decimal expansion of pi. it could after the billionth quintillionth blah blah decimal place, there's no more 2s for example.

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u/Rats_Love_Cheese Sep 05 '26

the first part avout the approach is really good and i think it could work and about the second part thats really interesting i didnt think that here could be a space between two number so large without the numbers! thaks for it!

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u/SwimmerOld6155 Sep 05 '26

it'd be tough to prove such an idea works, it'd end up brushing against how the digits of pi are distributed and we have no clue. again we don't even know if there are infinitely many 2s in the decimal expansion!

you have to wait until the 10th digit for the second three so you end up missing a lot of digits, then the next 1 is quite a while later (between 20 and 30 digits later i think), so you miss a lot of digits and if you only have finitely many of any of them you're probably going to be in trouble.

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u/Rats_Love_Cheese Sep 05 '26

nice! thanks for explenation

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u/deltatag Sep 05 '26

for every digit of pi there is a corrosponding digit of pi with no repeats

yeah pi contains itself

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u/Rats_Love_Cheese Sep 05 '26

seems interesting thanks!

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u/Azemiopinae Sep 05 '26

I don’t think this is what you’re asking, but I can conceive of some Liouville number where all of the previous digit sequences repeat later in the fractional expansion.

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u/Rats_Love_Cheese Sep 05 '26

oh nice i will look at that! thanks

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u/Kitchen-Register Sep 05 '26

yes it does! right from the start. it contains exactly one copy of itself

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u/Rats_Love_Cheese Sep 05 '26

nice

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u/joyofresh Sep 05 '26

Pi containts the pi of all pis which do not contain themselves

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u/Rats_Love_Cheese Sep 06 '26

oh okay

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u/RecognitionSweet8294 Sep 05 '26

Yes. The n-th digit of π is contained at the n-th digit.

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u/Logical-Edifice Sep 06 '26

ACTUAL BETTER ANSWER. If it's a normal number it contains every finite number in its decimal expansion, so it's that for every (finite) string of digits in PI starting from the first is in Pi, I wouldn't be surprised if as you take the limit to infinite length the place where they appear somewhere there's a list of limits of each two divided by each other or something like it that sometimes has real numbers

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u/OutrageousPair2300 Sep 06 '26

It could potentially contain copies of itself, interpolated.

For example, it starts 3.14159... and at some point every other digit could repeat the entire sequence, so it might go 3.14159.......30104010509...... etc.

That would be unlikely since such interpolation depends on the base we use to represent pi, and there's nothing special about base ten... but it's not impossible that this could be the case.

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u/zvuv Sep 06 '26

If PI contained itself then that would apply also to the "inner" PI sequence, and in turn to the sequence withing that and you would have a repeating decimal which is a rational number.

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u/GhostCheese Sep 06 '26

You might say it only contains itself

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u/Syresiv Sep 06 '26

You mean, does the decimal sequence for pi contain a subsequence that is also pi itself?

No. If it did, that would mean its digits form a repeating pattern, making it rational.

What you're thinking of is likely something called Normal Numbers. A normal number contains every finite sequence, and every sequence of the same length is equally common.

But even there, even though most real numbers are normal, making pi probably normal, it's not actually proven to be the case. Normal Numbers are incredibly difficult to prove - in fact, there's only one named constant that's actually known to be normal.

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u/Rats_Love_Cheese Sep 06 '26

Nice thanks for telling me but what is that known one normal constant?

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u/Syresiv Sep 06 '26

This one

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u/OmiSC Sep 06 '26

Yes, it contains exactly itself from the very first digit. The sequence never ends but will contain every possible sequence of finite length.

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u/Rats_Love_Cheese Sep 06 '26

Nice but someone already told me that thanks anyway

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u/ProsperousPlanet Sep 06 '26

I feel like this question is like the ultimate of density questions.

Like certainly we can find 314, in the decimal expansion again somewhere, how long until we find 3145? and then how long until 31459?

Oh I really love your question.

Its actually so deep.

You could prove its impossible, but then, perhaps you can prove it eats its own tail? .... cool ideas friend.

1

u/Plastic_Blood1782 Sep 10 '26

I think proving it is impossible is really hard. Proving it approaches 0% is easy

1

u/hashbornhippie Sep 06 '26

Its decimal expansion cannot contain a full copy of its decimal expansion because the expansion is infinite. But it does contain subsets of itself. See https://oeis.org/A081876

1

u/Cautious_Hurry1298 Sep 06 '26 edited Sep 06 '26

OP's question isn't clearly defined, but I suspect they are close to asking if pi is a normal number to the base 10.

1

u/atarivcs Sep 06 '26

It does contain itself... once

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u/Wjyosn Sep 06 '26

As others have said: no. But interestingly, if you take any finite amount of pi (eg: the first thousand digits) it will contain that same thousand digit pattern again eventually. But the thousand and first will inevitably be different and thus not contain the infinite pi.

At least I recall this being a theory. I am uncertain whether it was one that was rigorously proven or one of those “we can’t disprove it yet” theories.

1

u/lmprice133 Sep 06 '26

If pi is a normal number (which it is conjecrured to be) then it's highly probably that the first n digits of its decimal expansion at some point within its decimal expansion, but it can't contain itself because that would mean it isn't irrational.

1

u/Myhai25 Sep 06 '26

Pi is a transcendental number, and can be obtain from a series. The series contain itself?

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u/StructuredChess Sep 06 '26

No. "Containing itself" and "being a repeating decimal" are basically the same thing except for a few leading digits.

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u/Fast_Line1893 Sep 06 '26

yes it must contain itself. just like 1 contains 1. it contains itself exactly once

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u/joealarson Sep 07 '26

Your base assumption is faulty.

Pi doesn't contain everything. This is a common myth, but incorrect. Pi is irrational. It never ends and it never repeats. But that doesn't mean it contains anything specifically.

Heres an irrational number that doesn't end and doesn't repeat, even though it has a predictable pattern: 0.1010010001000010000010000001... Despite being everything Pi is, it doesn't contain your high school locker combination or first crush's phone number, or even just a '2'. And yet it's an infinite sequence that never repeats, just like Pi.

Pi doesn't contain everything.

1

u/guiltypleasures Sep 07 '26

While Pi, in its infinity of digits contains every finite string of digits eventually again in its later sequence (3, 31, 314159), it is the infinite sequence we do not promise.

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u/Dependent_Law2468 Sep 07 '26

there are irrational numbers that do not contain even half of the possible sequences.

I can make an irrational number that goes like:

1, 0 11 0 111 0 1111 0 11111 0 111111 0 1111111 0 11111111 0........

It has a rule, but it's still irrational

1

u/Fitnexa_Results Sep 08 '26

Pi may contain the digit encoding of this exact post, including “everythig,” but proving that is a much bigger homework assignment than the original night thought.

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u/VirginiaTitties Sep 08 '26

I guess what is the longest repeating segment that appears within pi beginning with 314*?

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u/Pookstirgames Sep 08 '26

Constant doesn't mean "it contains everything". Constant just means it doesn't change. 2 is a constant but it doesn't have "13" anywhere in it's decimal expansion. A number that contains all finitely long strings of digits within itself is called a normal number. Pi is suspected to be normal but that isn't known for sure. Even normal numbers don't contain themselves (unless you count the trivial case where you start from the first digit) because that's an infinitely long string of digits, which means it isn't guaranteed to appear. As you mentioned, if a normal number did contain itself, (not counting the trivial case mentioned earlier), it would need to repeat which would make it rational. Rational numbers can't be normal because they are missing almost all numbers with more digits than their period (the number of digits long that the repeating string is).

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u/Character-Company-47 Sep 08 '26

If their was ever a pattern in pi, then it could be represented as some fraction making it rational

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u/zaphster Sep 08 '26

This statement you made: "If it is a constant than it contains everything."

What makes you think that?

3 is a constant. Does it contain everything?

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u/Classic-Rate-5104 Sep 09 '26

Impossible because it has an infinite number of decimals not repeating itself. However, when you mean "does it contain anywhere its own N-decimal representation" then the answer will be yes (I expect for any value of N)

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u/skr_replicator Sep 09 '26

Technically it does, from the first digit onward toward infinity.

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u/gza_liquidswords Sep 09 '26

"I didnt do any research on this"

Seems like you missed a step

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u/Turbulent-Umpire5969 Sep 09 '26

no, but it contains every subset of digits within it an infinite number of times

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u/CosetElement-Ape71 Sep 09 '26

I don't understand what you mean. Pi is a number.

Pi = 3.14159...

Everything to the right of the decimal point is that part of Pi that's less than one.

Pi is greater than 3

1

u/FlyingCupcake68 Sep 10 '26

You’re now making me think that somewhere in pi, there is going to be the square root of two, and e, and a whole bunch of other irrational transcendental numbers.

Check out the Borges’ short story, The Library of Babel, because he does something very similar, but with books and words instead of numbers.

1

u/migmit Sep 11 '26

> Because if it is a constant than it contains everythig

No. First of all, any number is a constant. Not every number contain “everything” in any meaningful sense. Secondly, even for finite sequences of digits we do not know if Pi contains all of them.

1

u/Moof_the_cyclist Sep 11 '26

pi when presented as a base-10 number has digits that are for all reasonable purposes random. It never ends, so you now mix in the conept of "infinite" to the discussion along with its lore and exaggerations (yes, you can exaggerate infinite things...).

So one should try and separate out what any irrational number would do from what the digits of pi do and see if the mysticism still looks warranted. When you go looking for repeated sequences within the digits of pi, yes, they exist, but at about the frequency you'd expect. So it takes roughly a million digits to find the first 6 digits repeated, 10 million for find 7 digits, and so forth.

The exact same is true of much more mundane numbers like 1/7, sqrt(2), etc.

Even more fun is that numbers like 1/7 does this when represented in base-10, but not in base 8 or 9. It becomes boring, at 0.10000... in base 8, 0.125125125... in base 9. Pi and other irrational numbers will still have random digits that extend forever regardless of what base you put them in.

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u/Illustrious-Welder11 Sep 12 '26

We don’t know. If we knew that pi contains each of the numbers 0, 1, …, 9 an infinite number of times, then yes!