# A maze about $\pi$.

I know that $\pi$ is a transcendental number, it will never repeat, every combination of numbers may occur. Since the digits appear randomly, is it possible that at the $n$th digit of $\pi$, the digits start repeating (coincidentally) to infinity? Although the chance for this to occur is infinitely small, $\pi$ has infinitely many digits.

Does the fact that "$\pi$ has infinitely many digits" provide a chance large enough for "$\pi$ has recurring decimals" to happen?

Thank you.

p.s. Can you support your view without using the fact "$\pi$ is not a rational number"?

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"Since numbers appear randomly"? What? – Chris Eagle Oct 23 '12 at 14:29
I believe the question is whether the digits or pi repeat after the $n$-th digit. Silly example: $0.23872349872373737373737...$. – axblount Oct 23 '12 at 14:31
No irrational number has a chance to have a repeating $d$-ary expansion for any integer $d \geq 2$. This is easily proved by observing the fact that any real number having repeating $d$-ary expansion for some $d \geq 2$ must reduce to a rational number. – Sangchul Lee Oct 23 '12 at 14:33
The decimal expansion is a representation of real numbers which is far from natural in mathematical point of view. So when working with the property of 'having recurring decimals' mathematically, one has to translate this artifact into a natural one, which exactly coincides with the notion of rationality. This is what thwarts you from achieving your goal. – Sangchul Lee Oct 23 '12 at 14:59
A transcendental number does not necessarily have random digits. – user02138 Oct 23 '12 at 18:18

No, it does not. It's possible for some string to repeat an arbitrarily long but still finite number of times, but the digits of $\pi$ cannot perfectly repeat after a certain point. If this were the case, then $\pi$ would be a fraction, which we know it's not. It will probably have strings that repeat any arbitrarily high finite number of times somewhere in it, but it will never repeat exactly past some point.

Say I'm trying to figure out the decimal expansion of some fraction, like $\frac56$. Let's call the decimal expansion $0.b_1b_2b_3b_4....$ where the $b_n$ are digits. Multiplying by $10$, we find that $\frac{50}6=8\frac26=b_1.b_2b_3b_4...$. Matching integer parts we find that $b_1=8$. Matching fractional parts we find that $0.b_2b_3b_4...=\frac26$. Multiplying by $10$ again gives us $b_2.b_3b_4b_5...=\frac{20}6=3\frac26$, so by similar reasoning $b_2$=3 and $0.b_3b_4b_5...=\frac26$. But look! Our new set of decimals is equal to $\frac26$ just like our last one! And we just computed that the first digit of $\frac 26$ (here $b_2$) was $3$, meaning that this new digit has to be three as well. But we already know that "subtracting three and multiplying by 10" will give us another $3\frac26$, meaning that these threes must continue indefinitely. So we've figured out that

$$\frac56=0.8333333 . . .$$

This always has to happen. Because the numerators can only take on values from $0-5$, and if we ever get the same numerator twice we know that the expansion will continue exactly like it did the last time we got that numerator. It had to repeat. You can reverse this process as well, so that any repeating decimal can be converted back into its fractional form. So $\pi$ not being a fraction is exactly the same as saying that $\pi$ never has a recurring decimal sequence.

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@ Robert Mastragostino: Can you support your view without using the fact "$\pi$ is not a fraction"? – ᴊ ᴀ s ᴏ ɴ Oct 23 '12 at 14:34
He can't, because it's precisely that fact that makes this impossible. – user123123 Oct 23 '12 at 14:44
@jasoncube No I can't, because what you're asking is equivalent to $\pi$ being a fraction. Edited with an explanation. – Robert Mastragostino Oct 23 '12 at 14:56

Suppose that this happened with $\pi$. Then $$\pi=3.a_1a_2\ldots a_n(b_1\ldots b_k)(b_1\ldots b_k)\ldots$$ Here the $b_1\ldots b_k$ are the repeated decimals. Then $$10^n\pi=3a_1a_2\ldots a_n.(b_1\ldots b_k)(b_1\ldots b_k)\ldots$$ Also $$10^{n+k}\pi=3a_1a_2\ldots a_n(b_1\ldots b_k).(b_1\ldots b_k)\ldots$$ Subtracting, we get that $$10^{n+k}\pi - 10^n\pi=3a_1a_2\ldots a_n-3a_1a_2\ldots a_n(b_1\ldots b_k)$$ But, the number on the right hand side is an integer. Thus $$\pi=\frac{3a_1a_2\ldots a_n-3a_1a_2\ldots a_n(b_1\ldots b_k)}{10^{n+k}-10^n}$$ because $10^{n+k}\pi-10^n\pi=\pi(10^{n+k}-10^n)$. This says that $\pi$ is rational. The proof that $\pi$ is not rational is a bit involved.

The issue with your demand is that a number having no repeated decimal is effectively the definition of irrational.

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No.

As Robert pointed out, the easiest way to see that the digits will not start repeating after a certain point is to argue that $\pi$ is irrational.

You ask for an argument that does not use the fact that $\pi$ is irrational. This is pretty hard, since being rational is the same thing as having repeating digits past a certain point, or having a finite decimal expansion. So yes, maybe there is an awkward way to say that $\pi$ does not have repeating digits past a certain point, but it would only be a different formulation of the claim that $\pi$ is irrational.

Remark. Statements such as "numbers appear randomly" are very misleading and somewhat confusing, because the digits that appear are not influenced by chance. Pi is pi, and there is a very precise mathematical definition for it, completely fixing all the digits. To me, this is very similar to statements such as "the chance for the number $n$ to be prime is approximately $1/\log(n)$". If you pick any number, whether or not is prime is not a matter of chances. It is either prime or it is not. A better statement would probably be "taking a random set of integers around $n$, about $1/\log(n)$ of those will be prime" (though this is still not a very clear statement, of course). My point is: this question has nothing to do with probability, and one should try to avoid words such as 'chance' in discussing it.

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