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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.

Andrés E. Caicedo
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user1801325
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    In mathematics, we don’t “allow”. Rather, we insist that any statement you make should be true. Notice that if you change all the $4$’s in $\pi$ to sixes, you haven’t told a lie, because you haven’t made a statement. That is, there has been no equals sign or any other verb. So anybody may make that change in $\pi$, but then it’s up to that person to try to say something true about the resulting number. Like: “it’s no longer normal”. That’s a true statement about your new number. – Lubin Oct 13 '14 at 03:53

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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.

Milo Brandt
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  • Actually, if $4$ occurs finitely often then the result would definitely be irrational since it is not eventually periodic. It is the other case which is more difficult. – Yuval Filmus Oct 13 '14 at 03:09
  • I wrote that the difficult case would be if only 4 and 6 occurred infinitely often (since $\pi$ is irrational, at least two digits must occur infinitely often); you're right that $x'$ is still irrational if the replaced digit occurs only finitely often (since $|x-x'|$ would be have only finitely many non-zero digits and be rational). – Milo Brandt Oct 13 '14 at 03:14
  • So if π' IS irrational, then would we call it a strictly computable irrational? I don't know if that makes sense, but what I'm trying to ask is whether the actual digit change itself can ONLY be stated as a recursive algorithm definition. Or perhaps even more simply, is this kind of digit change necessarily a "recursive" algorithm. – user1801325 Oct 13 '14 at 04:02
  • @Meelo Forgot to tag you in that^ – user1801325 Oct 13 '14 at 04:08
  • @user Well, given how little we know about the decimal expansions of any irrational, it would be a stretch to say that the algorithm of "read off digits. Make the relevant replacements" is the only way to do this (though, in the general case, it is) - but, since we could take any other computable number and make this change, unless we can know something deep about any algorithm spitting out digits, we're sort of stuck to a single, algorithmic definition. (For instance, no algebraic property could exist, since $x$ and $|x-x'|$ could be forced to be algebraically independent) – Milo Brandt Oct 14 '14 at 02:41