How about ANY FINITE SEQUENCE AT ALL?
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
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.
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.
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.
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
🙄
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.
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.
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.
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.
Yes, this is implied. It’s also why many people use digits of pi as passwords and make the password hint “easy as pi”.
Not sure if this is sarcasm, but I sure hope so…
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.
I just use the last 10 digits of pi for all my passwords.
0123456789
Right?
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?
Yes.
And if you’re thinking of a compression algorithm, nope, pigeonhole principle.
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.
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
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.
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.
What about in the context of Pi?
1/3 is infinite in decimal form, as a more common example. 0.333333333….
it repeats
Doesn’t the sequence “01” repeat? Or am I misunderstanding the term.
A nonrepeating number does not mean that a sequence within that number never happens again, it means that the there is no point in the number where you can predict the numbers to follow by playing back a subset of the numbers before that point on repeat. So for 01 to be the “repeating pattern”, the rest of the number at some point would have to be 010101010101010101… You can find the sequence “14” at digits 2 and 3, 104 and 105, 251 and 252, and 296 and 297 (I’m sure more places as well).
yeah, but non-repeating in terms of decimal numbers usually mean: you cannot write it as 0.(abc), which would mean 0.abcabcabcabc…
Wouldn’t binary ‘10’ be 2, which it does contain? I feel like that’s cheating, since binary is just a mode of interpreting information …all numbers, regardless of base, can be represented in binary.
They’re not writing in binary. They’re defining a base 10 number that is 0.11, followed by a single 0, then 1, then two 0s, then 1, then three 0s, then 1, and so on. The definition ensures that it never repeats, but because it only contains 1 and 0, it would never contain any sequence with the numbers 2 through 9.
And you can strongman this by first using the string 23456789 at the start. It does contain all base 10 digits but not 22.
Thanks for the consideration for my pronouns XD
he/him if it ever matters
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”
That was quite an elegant proof
But didn’t you just give a counterexample with an infinite number? OP only said something about finite numbers.
They were showing that another Infinite repeating sequence 0.1010010001… is infinite and repeating (like pi) but doesn’t contain all finite numbers
You mean infinite and non- repeating?
This! Fixed it. Thanks!
“2” is a finite sequence that doesn’t exist in the example number
Not just any all finite number sequence appear in pi
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.
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
The explanation is misdirecting because yes they’re removing the penguins from the zoo. But they also interpreted the question as to if the zoo had infinite non-repeating exhibits whether it would NECESSARILY contain penguins. So all they had to show was that the penguins weren’t necessary.
By tying the example to pi they seemed to be trying to show something about pi. I don’t think that was the intention.
i just figured using pi was an easy way to acquire a known irrational number, not trying to make any special point about it.
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.
It’s misunderstanding the question even if unintentionally.
Clearer: 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?
They somehow interpreted it as Does any possible string of infinite non-repeating digits contain every possible finite sequence of non repeating digits?
It’s like if I ask “since the dictionary contains every word that means it contains every letter right?” And someone answers, actually you’ll find if you translate it to Japanese and only use kanji it actually doesn’t contain these letters. It fundamentally isn’t what I’m asking, and yes, you can argue I didn’t say IN ENGLISH, but just like the pi question, I feel like it’s pretty intuitive that I wasn’t referring to non English letters in the question.
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.
Is the 1.0010101 just another sequence with similar properties? And this sequence with similar properties just behaves differently than pi.
Others mentioned a zoo and a penguin. If you say that a zoo will contain a penguin, and then take one that doesn’t, then obviously it will not contain a penguin. If you take a sequence that only consists of 0 and 1 and it doesn’t contain a 2, then it obviously won’t.
But I find the example confusing to take pi, transform it and then say “yeah, this transformed pi doesn’t have it anymore, so obviously pi doesn’t” If I take all the 2s out of pi, then it will obviously not contain any 2 anymore, but it will also not be really be pi anymore, but just another sequence of infinite length and non repeating.
So, while it is true that the two properties do not necessarily lead to this behavior. The example of transforming pi to something is more confusing than helping.
The original question was not exactly about pi in base ten. It was about infinite non-repeating numbers. The comment answered the question by providing a counterexample to the proffered claim. It was perfectly good math.
You have switched focus to a different question. And that is fine, but please recognize that you have done so. See other comment threads for more information about pi itself.
We need to start teaching formal logic in grade schools I’m going insane.
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.
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
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.” ??
It does contain a 2 though? Binary ‘10’ is 2, which this sequence contains?
Like the other commenter said its meant to be interpreted in base10.
You could also just take 0.01001100011100001111… as an example
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.
I don’t think the example given above is an apples-to-apples comparison though. This new example of “an infinite non-repeating string” is actually “an infinite non-repeating string of only 0s and 1s”. Of course it’s not going to contain a “2”, just like pi doesn’t contain a “Y”. Wouldn’t a more appropriate reframing of the original question to go with this new example be “would any finite string consisting of only 0s and 1s be present in it?”
They just proved that “X is irrational and non-repeating digits -> can find any sequence in X”, as the original question implied, is false. Maybe pi does in fact contain any sequence, but that wouldn’t be because of its irrationality or the fact that it’s non-repeating, it would be some other property
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.
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.
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.
