# Repetition in pi

If there are infinite digits in $\pi$ and any group of digits occurs in $\pi$. Then does all the digits of pi occur in itself infinite times over? Therefore $\pi$ repeats. What is wrong with my reasoning.

• Search for repeating patterns: angio.net/pi. Also, read the bottom snippet: mathforum.org/dr.math/faq/faq.pi.html – Amzoti Mar 21 '15 at 1:31
• Side note: as far as I know it's unproven (but strongly suspected) that every group of digits occurs in $\pi$. – user7530 Mar 21 '15 at 2:59
• It does not make sense to say "π repeats". For example, "011000111100000111111..." has both "0" and "1" repeating infinitely many times, but would you say that the entire sequence repeats, and what exactly would you mean if you did? – user21820 Mar 21 '15 at 4:15

An irrational number that contains all finite strings of digits is called normal. It is not known if $\pi$ is normal. Even if it was normal, your argument would not apply, since it only applies to finite strings.
First of all, we don't know that "any group of digits occurs in $\pi$", although we suspect that this is true, and moreover that it occurs infinitely often. So for example $12345$ might occur at positions $53256$ and $814324$ and $2534246$ and ... (these are not the actual numbers: I just made them up). But that's a lot different from saying the whole thing repeats.
Consider the Champernowne constant in base 10, defined by concatenating representations of successive integers $C_{10}=0.12345678910111213141516\ldots$. You can see that it is not a repeating decimal. Yet it is normal, that is, its digits follow a uniform distribution: all digits being equally likely, all pairs of digits equally likely, all triplets of digits equally likely, etc. As mentioned in the other answers, it is not known if $\pi$ is normal, but even if it was, it is not necessarily the case that it would have a repeating decimal.