Talk:Integer to Rational Power is Irrational iff not Integer or Reciprocal

Thanks for the edits.

After signing up I had a really hard time finding the house style page again (in fact, I failed to), but now that I have the link I'll get this into the house style by 5/1/16. For the 4-tuple (p, q, r, s) I was using the wrong notation just because without parenthesis is the style I'm used to. I've changed it and added explanation for why Z_{>0} is used, the terms are all positive and it is important to avoid the possibility of 0^0. This also facilitates use of the fundamental theorem of arithmetic since every element of the set Z_{>0} is a natural number.

As far as the title goes, I'm completely open to suggestion. Originally, I had it expressed as iff as in the theorem, but making a precise and meaningful statement in that way produced and even longer and more awkward title. Perhaps simply "Rational Powers of Positive Integers"? But that ignores the important fact that many of these numbers are irrational.


 * How about: Integer to Rational Power is Irrational iff not Integer or Reciprocal? Again, let me know if there is a subtlety I'm missing. --prime mover (talk) 15:27, 30 April 2016 (UTC)

Great idea. I tried to make the change but I don't seem to be able to rename the page and I couldn't figure out how to in the help pages, would you mind making the change or letting me know how to on my own? --SethBorgo (talk) 16:11, 30 April 2016 (UTC)


 * Under "More" (top right) you should have a "Move" option. Do you have that? --prime mover (talk) 16:22, 30 April 2016 (UTC)

Yes, thanks! Moved the page. Sorry I couldn't find that on my own, I saw the move option when I was searching but thought it meant something else and kept searching. --SethBorgo (talk) 16:34, 30 April 2016 (UTC)


 * You're all right. Don't get too hung up about the house style for the moment. I will go through and "stylise" it in due course, then you will see how it works. Page is a good one. --prime mover (talk) 18:26, 30 April 2016 (UTC)


 * The main point about house style you might want to take on board is that the form: "$A$, $B$ therefore $C$" is preferred to "$C$ because $A$, $B$" or "$A$, therefore $C$ since $B$" -- it forces the train of thought forward without mental back-tracking. --prime mover (talk) 18:48, 30 April 2016 (UTC)


 * Excellent, thanks for the explanation that makes a lot of sense, and thanks for putting the article in house style, that would have been much harder for me to do myself but I learned a lot from reading your restatement. I will try to find a proof that implies n^m\in\R and reference it. --SethBorgo (talk) 20:15, 30 April 2016 (UTC)

Been doing my research regarding $n^m$. There is always real number that is equal to $n^m$, but complex numbers are also valid "roots" for some m values. When I wrote the proof I implicitly assumed a universe of real numbers, and I thought $n^m\in\R$ was true without this, but it doesn't appear to be. For rigor, I think $n^m\in\R$ must be included in the hypothesis. I'll update the page in that way, but let me know if you feel there's another alternative.--SethBorgo (talk) 02:52, 1 May 2016 (UTC)


 * If $n$ is a positive real number, then it is definitely the case that there exists a positive real number $r$ such that $n^m = r$. (This is not the case when $n$ is negative.) See, if $n > 0$ we have $n^m = e^{m \ln n}$, and then $\ln n \in \R_{>0}$ and $e^{m \ln n} \in \R_{>0}$. And I'm fairly sure that result exists somewhere in, I just haven't looked for it. --prime mover (talk) 08:44, 1 May 2016 (UTC)


 * I believe I found it, https://proofwiki.org/wiki/Existence_of_Positive_Root [note that they prove cases for positive and negative exponents and the result for the zero case is irrelevant]. This extends to rational powers because (n^p/q)=(n^p)^1/q and n^p is a natural number, so its qth root is in R. I'll add this in, but if I fail on style please do edit. Thanks for your help on this, again it's been a great learning experience.