Exponentiation with rationals - why algebraic?

Question

Why is exponentiation with rational numbers considered an algebraic operation? I get why exponentiation with integers is since that's just a finite number of applications of multiplication, but this doesn't extend to roots.

Answer 1

I believe this is due to the fact they are required for the solution of polynomial equations with rational (or even integer) coefficients, which was the main concern of algebra for several centuries.

Answer 2

This is for historical reasons, the operation $x \mapsto \sqrt[n]{x}$, extracting a ($n$:th) root, is considered algebraic. (And consequently, raising something to a rational number is, too.)

Answer 3

Look at the statement of Abel's Impossibility Theorem that you linked to in a comment. It says something about a finite number of certain operations which include multiplication and root extraction (which in this context means $n$th root, where $n$ is an integer).

Given a number $r$, with a finite number of multiplications you can produce $r^m$. Then take the $n$th root: $$\sqrt[n]{r^m} = r^{m/n}.$$