There have been multiple accounts created with the sole purpose of posting advertisement posts or replies containing unsolicited advertising.

Accounts which solely post advertisements, or persistently post them may be terminated.

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.

deegeese ,

Do your own homework, buddy.

ns1 ,

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.

Ad4mWayn3 OP ,

My mistake, in that case it’s not the closure what I mean. But then how are those kinds of sets called?

NeatNit ,

Who says a and b are countable

Deestan ,

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.

Septimaeus ,

ℕ

NeatNit ,

I see. I missed that word in the question, and I didn’t remember that definition anyway.

  • All
  • Subscribed
  • Moderated
  • Favorites
  • •
  • [email protected]
  • random
  • lifeLocal
  • goranko
  • All magazines
    •