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.
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2Where have you heard that it is? – Chappers Sep 15 '15 at 12:23
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As I understood it, it is referred to as such in the context of the Abel-Ruffini theorem. – G. Bach Sep 15 '15 at 12:28
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2Probably because the function $x\mapsto x^{1/n}$ defined on $[0,\infty)$ is the reciprocal of the function $x\mapsto x^n$ which is algebraic. – Tom-Tom Sep 15 '15 at 12:40
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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.
Shahar Even-Dar Mandel
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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.)
mrf
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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: $$\sqrt[n]{r^m} = r^{m/n}.$$
David K
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