Closure of exponentiation of real algebraic numbers.
Given two real, nonzero algebraic numbers a and b, with a > 0 (so that it excludes complex numbers), is there any named subset of the reals S such that (a^b) belongs to S forall a,b? I know it's not all the reals since there should be countably many a^b's, since a,b are also countable.
8 replies
Do your own homework, buddy.
Who says a and b are countable
https://en.m.wikipedia.org/wiki/Algebraic_number
They're given as algebraic, which are countably infinite since they can be mapped 1-to-1 with integers.
I see. I missed that word in the question, and I didn't remember that definition anyway.
ℕ
Fun question! I don't know the answer other than to say it's not just the algebraics because of the Gelfond-Schneider constant
Are you sure this is well-defined? You say that a and b are algebraic but "closure" implies that they could also be any members of S. This might mess up your proof that it's not all the reals if you do mean the closure.
My mistake, in that case it's not the closure what I mean. But then how are those kinds of sets called?
You could say something like "the image of exponentiation over...." to mean the set of values created by applying the function once, but it sounds slightly clunky.
Looks like there aren't really very many sets of mostly transcendental numbers that have names. Computational numbers and periods are two of them, I'd guess that both probably contain your set, so you could compare with those to see where it gets you.