By way of motivating example, someone came up with a completely evil user interface where the user was required to select their ten digit phone number by moving a sliding window over the digits of $\pi$:
Given that $\pi$ is infinite and non-repeating, my gut says that all possible combinations of ten digits must exist somewhere in the digits of $\pi$, but I don't know how to prove that.
Help appreciated.
Best,
Glenn
On
It may well be true, but you cannot prove it simply using that fact that it is infinite and non-repeating. So is the number $0.1001000010\ldots$ (it has a $1$ at the $n$th digit after the dot if $n$ is a perfect square and $0$ otherwise), but it clearly doesn't contain all phone numbers.
On
Infinite and non-repeating does not necessarily imply to contain all ten digits combinations.
Consider the infinite nonrepeating sequence $$101001000100001....$$ or $$12233344445555566666....$$
On
Independently of guts, this is a property called normality. It is unknown whether $\pi$ is normal (in any base, including base 10). We can't even show (currently) that there is no place in the decimal expansion $\pi$ after which only the digits $0$ and $1$ appear.
As DecimalTurn pointed out in the comments, this OEIS page says that the first 241641121048 digits of $\pi$ contains all ten-digit strings.