Sorry if this has been asked before, but I have a query about the notion that the decimal expansion of $\pi$ contains every possible string of numbers (please note that I am only a "casual" maths enthusiast). If it does, then would the expansion not contain $\pi$ itself? (I.e. 3.1415926…31415926…) That would make $\pi$ a repeating decimal which could, in theory, be represented as an exact fraction. If we constrict the argument to say that $\pi$ only contains every finite sequence of numbers, then wouldn't that be contradictory (we would see 3, 31, 314, 3141, 31415… so why not all the way? We could always add another digit to create a longer finite string ad infinitum)?
Thank you in advance.
Best Answer
The claim is only about finite strings (and apart from this, it is only conjectured, has not been proven). In fact your what-if argument is sound and would show that $\pi$ is rational. The fact that it is not rational (in fact, transcendental) shows that it cannot contain itself in a nontrivial manner.
Regarding the second question: No, all finite strings does not imply a given infinite string. In fact, the number $$0.123456789101112131415161718192021222324\ldots $$ obtained by concatenating all natural numbers provably contains every finite string, and among these $3$, $31$, $314$, $3141$ and so on, but certainly (though perhaps not obviously) not the full expansion of $\pi$.