Changing digits of an irrational allowed? Suppose you change every instance of a specific digit of π, e.g., suppose you make every "4" a "6" instead. I realize that this too would be irrational, but what I want to know is (1) on what basis is this allowed, and (2) what kind of irrational would this be?
So for (1), what I mean is, clearly this is not an arithmetical operation (changing a number), but what then is it? What do we call a rule like that? Is it a recursive definition?
(2) This might be answered in the answer to (1), but what kind of irrational is this redefined π? Is it a computable number (that would be the case if the answer to (1) is that it is a recursive definition)? Is it a definable number? ( http://en.wikipedia.org/wiki/Definable_real_number ) Or is it (something mentioned in that wikipedia page) an "unambiguously defined" number?
If you just want to point me to sources instead of answering, that's fine too. I'm basically just a little unclear on what qualifies as a recursive definition.
 A: Well, strictly speaking, it is unknown whether $\pi$ with every $4$ changed to $6$ would be irrational (since we don't know even that every digit of $\pi$ occurs infinitely often), but if it's a normal number, then it would still be irrational. Notice that if we took a number like
$$\alpha = \sum_{i=1}^{\infty}10^{-i!}$$
which is irrational, but only uses digits 1 and 0, we could replace every 1 by 0, and it would equal $0$, which is clearly rational. So, in general, making a substitution like this might not preserve irrationality (and unless we can prove that $\pi$ definitely doesn't end in a sequence of $4$'s and $6$'s, we can't prove that your number is irrational). The thing to notice is that, if you let $x$ be a number and $x'$ be a number with digits replaced, it's clear that, if $x$ is rational, so is $x'$ (although it may not be well-defined since, $0.\bar{9}=1$), and further, that the difference $x-x'$ only uses one sort of digit (i.e. the difference could be $.10110$, but not $.123$).
However, if $x$ is computable, so is $x'$. This is obvious, since "computable" essentially means that there is an algorithm which, takes any natural number $n$ as input, and outputs the first $n$ digits of $x$. So, if you can do that for $x$, then an algorithm for $x'$ must exist, where we use the same algorithm, except apply the replacement as we output things.
