$(-3)^3 = -27$. $f(x) = x^3$ is a bijective function, $g(x) = \sqrt[3]x$ is defined for all values of $x$. But my maths teacher said, that $\sqrt[n]a = b \iff b^n = a, a\geq0$. She said that allowing the notation e.g. $-3 =\sqrt[3]{-27}$ leads to contradictions, and the she went and did this:
$$-3 = \sqrt[3]{-27} = (-27)^\frac {1}{3} = (-27)^\frac {2}{6} = \sqrt[6]{(-27)^2} = \sqrt[6]{27^2} = 27^\frac{2}{6} = 27^\frac{1}{3}=\sqrt[3]{27}=3$$
Is there a mistake in the “proof”, or is an $\sqrt[n]a$ really only defined for $a\geq 0$?
I do know that $(-3)^3 = -27$ and all that. I am aware of real and complex roots, and of the attributes of even and odd orders of roots and powers. I searched online and found no mention of $\sqrt[n]a$ only being defined for $a\geq0$.
I am asking whether the DEFINITION of the $n$-th root allows us to take the root of a negative number (I presume it does), and if so, then how can I prove the wrongness of the "proof" above?
You do have to be careful manipulating rational exponents and negative bases. You can't have $a^{bc}=(a^b)^c$ true in general; there must be some restrictions, either to the base or to the exponents, or to whether you can apply the rule. Your teacher thinks the restriction should be in the base (the base should be non-negative). As I read you, you think it should be a combination (as long as you don't have both negative base and exponent with even denominator, it's fine). Both approaches solve the problem.