The infinite sequence of the digits of pi

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Delta2
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TL;DR Summary
How many zeros are there in the infinite decimal sequence of pi
Is there a way to prove if the 0s (or 1s 2s or 3s or ... or 9s) that are present in the infinite sequence of digits of pi in the decimal system are finite or infinite?
If they are infinite they are countable infinite or uncountable infinite?
My intuition tells me that they are countable infinite but cant find a way to prove it.

They must be countable, because the number of digits in a decimal expansion is countable, since we can number (count) each one based on how many places after the decimal point it comes.
I intuitively expect each of the ten numerals to occur infinitely often in the expansion. But I can think of no way to go about proving that must be the case.

It's easy to prove that at least two must occur infinitely often, because otherwise we'd end up with a recurring decimal, which would mean pi was rational, and the irrationality of pi has been proven.
But there are uncountably many decimal expansions with just two available numerals, so there's no easy (for me) way to prove that more than two of the ten numerals must occur infinitely often.
EDIT: It occurs to me that an easier way to start on this would be to investigate whether in the ternary (base-3) expansion, each of the numerals 0,1,2 must occur infinitely often, and what would be the consequences if one did not.

We might be able to pick up something from work on the Cantor Set, which is the set of all numbers in [0,1] that have ternary expansions that do not contain the numeral 1.
The wiki article on the Cantor Set lists a number of interesting properties, some of which might be useful. In particular, I wonder whether the fact of its being nowhere dense night help.

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FactChecker, DeBangis21, Bosko and 2 others
andrewkirk said:
It occurs to me that an easier way to start on this would be to investigate whether in the ternary (base-3) expansion, each of the numerals 0,1,2 must occur infinitely often, and what would be the consequences if one did not.
I thought of something based on what you said, can we say that in the binary representation there must be infinite zeros (and 1s) otherwise it would be a rational number?

whit3rd
Delta2
I'd say each digit countably infinite with probability one. I don't see how this could be proved because the digits are artifacts of base ten, which has nothing to do with pi.
Certainly not uncountable.

Delta2 and fresh_42
Hornbein said:
I'd say each digit countably infinite with probability one. I don't see how this could be proved because the digits are artifacts of base ten, which has nothing to do with pi.
Certainly not uncountable.
There can be some surprising relationships between the decimal digits of ##\pi## and physical phenomena. I was very surprised by this:

Why do colliding blocks compute pi?​

Hornbein said:
I don't see how this could be proved because the digits are artifacts of base ten, which has nothing to do with pi.
Hmmm, it could still be that pi has finite number of zeros in the decimal system. Or at least the "cousin of pi" could have.

What is the cousin of pi?

Lets suppose that pi has infinite number of zeros. We give birth to the cousin of pi by leaving the first 100 zeros as they are and replace the rest infinite zeros with 1. This new number will still be irrational and would have finite number of zeros. And i think it would differ from pi less than ##10^{-100}##.

FactChecker and andrewkirk
FactChecker said:
There can be some surprising relationships between the decimal digits of ##\pi## and physical phenomena. I was very surprised by this:

Why do colliding blocks compute pi?​

I would expect that to work with any large mass.

I think that ##\pi## couldn't be pseudo-random with only finitely many zeroes.

Delta2
fresh_42 said:
I think that ##\pi## couldn't be pseudo-random with only finitely many zeroes.
What's the definition of pseudo-random in mathematics?

Delta2 said:
This new number will still be irrational
How do we know this?

FactChecker
Hill said:
How do we know this?
Hmmmm because it will still have infinite non periodic structure of digits, I don't think that the modification I do creates periodic structure of digits.

Yeah I think it is easy to prove that if we had periodic structure with the inserted 1s we would have periodic structure in pi too that has 0s there instead of 1, but we know pi is irrational.

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Delta2 said:
I don't think that the modification I do creates periodic structure of digits.
It does not seem so, but proof is not immediately obvious to me. What is obvious is that if we replace nine digits by the same one, we get periodic structure.

Delta2 said:
What's the definition of pseudo-random in mathematics?
https://en.wikipedia.org/wiki/Pseudorandomness

It is only a conjecture that the decimal digits of ##\pi## are pseudo-random, but it is a bit of a common view.

Hill said:
It does not seem so, but proof is not immediately obvious to me.
Sounds a bit like Liouville's theorem about transcendent numbers. I haven't looked closer at it, but it would be my starting point. However, the irregular occurrence of zeroes could be a problem.

https://de.wikibooks.org/wiki/Beweisarchiv:_Algebra:_Körper:_Approximationssatz_von_Liouville

It is possible to have a transcendental nuber that does not have the digits equally populated: 0.101001000100001000001000000100000001000000001....

A number is normal base 10 if the digits are equally populated. the pairs equally populated, the trplets, and so on. A number is normal if it is normal ion all bases. There are many more normal numbers than non-normal numbers (a set of measure zero) but AFAIK, there is no single example.

It is possible to have a transcendental nuber that does not have the digits equally populated: 0.101001000100001000001000000100000001000000001....
I know that this example is irrational, but can it be proven to be transcendental? It's my understanding that, although there are uncountable transcendental numbers, there are very few examples that are proven to be transcendental. I don't know what those proofs look like.

FactChecker said:
I know that this example is irrational, but can it be proven to be transcendental? It's my understanding that, although there are uncountable transcendental numbers, there are very few examples that are proven to be transcendental. I don't know what those proofs look like.
There are known examples on the Wikipedia page. I don't see that one there, although there is something similar.

https://en.m.wikipedia.org/wiki/Transcendental_number

FactChecker
FactChecker said:
I know that this example is irrational, but can it be proven to be transcendental? It's my understanding that, although there are uncountable transcendental numbers, there are very few examples that are proven to be transcendental. I don't know what those proofs look like.
Not sure whether this number is a Liouville number like
$$\sum_{k=1}^\infty \dfrac{1}{10^{k!}}$$
https://oeis.org/A012245
but all Liouville numbers
$$\sum_{k=1}^\infty \dfrac{a_k}{b^{k!}}\, , \,a_k\in \{0,1,\ldots,b-1\}\, , \,b\ge 2$$
are transcendental according to Wikipedia.

FactChecker and PeroK
fresh_42 said:
Not sure whether this number is a Liouville number like
$$\sum_{k=1}^\infty \dfrac{1}{10^{k!}}$$
https://oeis.org/A012245
but all Liouville numbers
$$\sum_{k=1}^\infty \dfrac{a_k}{b^{k!}}\, , \,a_k\in \{0,1,\ldots,b-1\}\, , \,b\ge 2$$
are transcendental according to Wikipedia.
Thanks! @Vanadium 50 's example doesn't look like a Liouville number to me, but very similar ones (just larger strings of zeros) would be. Those are good examples!
PS. I don't want to hijack the thread.

Delta2 said:
TL;DR Summary: How many zeros are there in the infinite decimal sequence of pi

If they are infinite they are countable infinite or uncountable infinite?
Not only infinite, but there is also no bound on the number of consecutive zeros in the decimal expansion of pi.

No way to prove that of course, but any length of finite run has a probability of 1.
What is interesting about this is if they are countable. I mean, it's a subset of the total digits of pi, but there is no 'the longest run of zeroes', so where do you begin the count?

FactChecker said:
I know that this example is irrational, but can it be proven to be transcendental?
It sure doesn't look algebraic.

But try this one (no proof either, but can see how one might set out to prove it): 3.141582653588... (replace all the 9's in the decimal expansion of pi with 8's)

bhobba and FactChecker
Delta2 said:
Lets suppose that pi has infinite number of zeros. We give birth to the cousin of pi by leaving the first 100 zeros as they are and replace the rest infinite zeros with 1. This new number will still be irrational and would have finite number of zeros. And i think it would differ from pi less than 10−100.
If we tried this in binary, the result would certainly not be irrational.

andrewkirk said:
the number of digits in a decimal expansion is countable
What would happen if you explicitly constructed a number with an uncountably infinite number of digits?

Algr said:
What would happen if you explicitly constructed a number with an uncountably infinite number of digits?
It would be something like an infinite series of infinite series.

Algr said:
What would happen if you explicitly constructed a number with an uncountably infinite number of digits?
How would you define or construct that?

Delta2 said:
Hmmm, it could still be that pi has finite number of zeros in the decimal system. Or at least the "cousin of pi" could have.

What is the cousin of pi?

Lets suppose that pi has infinite number of zeros. We give birth to the cousin of pi by leaving the first 100 zeros as they are and replace the rest infinite zeros with 1. This new number will still be irrational
IMO, not necessarily. If the irrational-ness depends on the pattern of zeros, making them finite could make the modified number rational. For example, take the pattern 0.101001000100001000001.... Clearly it is irrational, but changing the zeros to 1's would make it a rational number. Something similar could happen to pi unless the other digits also make it irrational, but I don't know if that is a "given".

jbriggs444
PeroK said:
How would you define or construct that?
- Put all the digits of Pi into a grid, like Cantor's diagonal.
- Copy the diagonal digits and advance each one to the front. (In base 10, you would advance the 9's to zeros but not carry a one.)
- Place the resulting digits in front of the pi.
- Repeat infinitely.

Algr said:
If we tried this in binary, the result would certainly not be irrational.What would happen if you explicitly constructed a number with an uncountably infinite number of digits?
An uncountable sum won't converge unless only countably many are nonzero.

That's curious. I understood that once you had a countable number of digits after the decimal point, nothing else mattered.

Algr said:
- Put all the digits of Pi into a grid, like Cantor's diagonal.
This part is straightforward.
Algr said:
- Copy the diagonal digits and advance each one to the front.
So now we are shuffling the digits?

Algr said:
(In base 10, you would advance the 9's to zeros but not carry a one.)
Huh? So if we have a 9 on the diagonal, we shuffle up a 0 in its place? You are trying to avoid the possibility of creating an all 9's string?

Algr said:
- Place the resulting digits in front of the pi.
Huh? We already shuffled the digits to the front. Why are we doing it a second time? Surely we are not putting them to the left of the leading "3". Such a construction would yield a numeral with infinitely many digits to the left of the decimal point. That is not allowed.

So it seems that I have failed to grasp the proposal.
Algr said:
- Repeat infinitely.
Unless this converges to some result, the proposed process does not produce a decimal string.

Note that a countable sequence of countable sequences of decimal digits has only a countably many decimal digits.

Note that there is no standard way to interpret an uncountable array of decimal digits as a real number.

jbriggs444 said:
Huh? So if we have a 9 on the diagonal, we shuffle up a 0 in its place? You are trying to avoid the possibility of creating an all 9's string?
Yes. In the binary or true/false version of Cantor's diagonal, there is only one choice as to what to change each digit to. This makes it similarly unambiguous in decimal, while preventing any digit from remaining the same.

jbriggs444 said:
So now we are shuffling the digits?
Not yet, just preparing them to be added to the pi string.

jbriggs444 said:
Huh? We already shuffled the digits to the front. Why are we doing it a second time? Surely we are not putting them to the left of the leading "3". Such a construction would yield a numeral with infinitely many digits to the left of the decimal point. That is not allowed.
Oops, I meant to say that they were inserted just to the right of the decimal point, leaving the 3 in the one's place. I wanted to avoid the ambiguity of placing them "after infinity" if I appended the digits to the far left. (Although it occurs to me that the original Cantor's Diagonal does exactly this, placing the newly created string at the end of an infinite list. So perhaps I should have stuck to that.)

But the point is this: As you go from countable to uncountable, the value of the string changes from exactly equal to Pi, to "won't converge". This can't just suddenly happen, so it seems like their ought to be some intermediate value of digits between countable and uncountable that would cause the string to equal to any arbitrary finite number larger than Pi. "4" for example.
To get around that, you might have to say that the digits of Pi are uncountable, and any countable string would fall short by an infinitesimal value.

I tried to Google the question of how we know that the digits of Pi are countable, and this thread was the top result.

Algr said:
But the point is this: As you go from countable to uncountable, the value of the string changes from exactly equal to Pi, to "won't converge".
Even if the construction that you describe were a construction, it does not go from countable to uncountable. A countable collection of countable collections is countable.

FactChecker and PeroK
But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.

Algr said:
But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.
It shows that any countable list of countable decimal strings must be incomplete. Every such list must miss at least one decimal string.

It does not show anything about the cardinality of the set of digit positions in a decimal string. It does serve to show that regardless of the cardinality of the set of digit positions, the cardinality of the set of all strings with decimal digits in those positions must exceed that.

If you decide to consider decimal "strings" with ##\aleph_1## many digit positions then you will have a set of decimal "strings" with a cardinality in excess of ##\aleph_1##.

If you decide to consider decimal strings with 2 digit positions then you will have a set of decimal strings with a cardinality in excess of 2. Indeed, 100 is greater than 2.

It is not clear what "putting ##\pi## in that format" even means.

Edit: Some people glance over Cantor's diagonal argument and think that it is an algorithm for creating an uncountable list by iteratively appending more and more numbers at the end of a countable list. That is not how the argument works. It works as @FactChecker outlines below.

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Algr said:
But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.
The diagonal proof constructs one number that is not on the proposed countable list. That number is still represented by a countable number of digits. It is not the countable digits that keeps the list countable. The thing that keeps the list countable is that you can count the members on the list.

PS. If you look at a Cantor diagonal proof that uses the decimal system to represent numbers, you can see that there are enormously more ways to construct numbers not on the list than there are numbers on the list.

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jbriggs444
The sequence of decimal digits of any number written in decimal digits is countable per construction. Just count them, one by one. This will be short in the case you count ##0.5## and will take a bit longer in the case of ##\pi.##

However, it could be interesting to discuss the term pseudo-randomness or whether ##\pi## is or is not pseudo-random, but this would be another subject. The digits themselves are only one - and insufficient - way to write ##\pi.## We still do not know enough members of A000796 and its properties. Countability, at least, is certain.

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