# Pi's Recursiveness [duplicate]

This question already has an answer here:

I don't know if this will make sense, but:

If $\pi$ is infinite and contains all strings of numbers including those of infinite length, then it must contain $\sqrt2$, and if $\sqrt2$ is infinite and contains all strings of numbers including those of infinite length then $\sqrt2$ contains $\pi$. This means that there is $\pi$ inside of $\pi$, and therefore it's recursive.

I probably went wrong somewhere (I am a secondary school student) and I don't know everything about these numbers. I am assuming that $\pi$ is infinite and contains all strings although I know this has been called into question. So can anyone help me out, because this just seems wrong yet the sum of all natural numbers is $-1/12$ so, anything is possible.

## marked as duplicate by Semiclassical, John B, Stefan Mesken, Mike Pierce, Daniel W. FarlowMar 4 '16 at 1:32

• $\pi$ doesn't contain all strings of infinite length. No number does. – 5xum Mar 3 '16 at 21:13
• BTW, we don't actually know if pi contains all finite strings or not. We suspect it does (and if it doesn't there are uncountably others that do) but we don't actually know this. – fleablood Mar 3 '16 at 21:17
• Same question math.stackexchange.com/questions/216343/… – MCMastery Mar 3 '16 at 21:18
• Why do you think pi contains all strings ? – Uri Goren Mar 3 '16 at 21:19
• Another side note... the sum of all numbers is not equal to $-\frac{1}{12}$. It just isn't. There are some ways to assign a value to infinite expressions of the type $a_1+a_2+\cdots$, and these ways align with sums if the expressions are infinite, but to call them sums is a stretch. – 5xum Mar 3 '16 at 21:20

$\pi$ is widely believed to contain all strings of finite length, although so far nobody has been able to prove this.

But obviously $\pi$ can't contain all strings of infinite length. For instance, if $\pi$ contains the infinite string $11111\ldots$, then all its digits after a certain point must be $1$; but if $\pi$ contains the infinite string $22222\ldots$, then all its digits after a certain point must be $2$. So $\pi$ can't possibly contain them both.

It's a good question and it shows a inquisitive and logical mind that you thought of it.

So bear that in mind when I point out what is wrong with this.

1) When people informally say "\pi's decimal expansion is infinite, and given infinite options all possibilities must occur so every possible possible string of digits must occur in pi" they mean all finite possible strings occur. To fit an infinite string in, such infinite strings never end and thus you can only put in the strings that happen to be pi with some front end cut off. It just doesn't make sense for the reasons you think it wouldn't. There's no fancy mind numbing strange counterintuitive cardinality thinky explanation to make it work.

2) We strongly suspect pi is "normal" meaning that its digits (in any base) will each occur infinitely often. That is that the digits occur with "normal" frequency and distribution. We don't actually know if pi is normal.

If pi is normal, then, statistically, all finite strings of digits must occur eventually. If pi isn't normal than that needn't be the case.

3) You are probably aware of two cardinalities of infinity. There is "countably" infinite which means an infinite set can be indexed and "counted" or, in other words there is a one-to-one corespondence between the infinite set and the Natural numbers. The digits of pi are, because the are in order of place value countably infinite.

Then there is "bigger" uncountable infinity. The real numbers for instance are uncountable.

The set of all infinite countable strings of digits is uncountable. But set of all finite strings is countable. The set of strings within pi (even the infinite ones) is countable. So there are "more" infinite strings then can possibly be in pi.

Maybe that was more than you needed.

4) The sum of 1+2+3 +... = -1/12 is kind of a misstatement. 1+2+3+ ... diverges and has no sum as is intuitively obvious. Some infinite sums converge such as 1 + 1/2 + 1/4 + 1/8 + ..... = 2, and do have sums but those that diverge to not. There is a function called the "zeta function" (you can google it) that evaluates characteristics of infinite sums. If the infinite sum converges then the zeta function will result in the sum. This is a coresponding overlap. The zeta function is not the sum but something that coincides when there is a sum. If the infinite sum does not converge the zeta function still returns a result but it isn't the sum. The zeta function evaluating the sum 1+2+3+... results in -1/12. But that isn't the same thing as the sum being -1/12 which is obviously absurd. (Clearly each partial sum is positive... and bigger than the previous one...)

• +1 for your comment about the zeta function. – Lawrence Jun 3 '18 at 17:38