Seems rational

Considering the fact that we are taking into consideration non-bijective relations in this problem, one must get accustomed in dealing with them.

A function $f(x) = x^{\frac{1}{m}}$ is a one-to-many relation in $\mathbb{C}$ . That is, if $x = 1$ , then there are $m$ values to $f(x)$ , (essentially, the $m^{th}$ roots of unity). On the other hand, a function $g(x) = x^{n}$ , is a many-to-one relation in $\mathbb{C}$ .

When we resolve fractional exponents, note that we do not confine ourselves to the "obvious" solutions. Instead, we get a set of solutions.

For instance, why don't we try evaluating

$(-4)^{\frac{2}{4}}$

Acknowledging that $\frac{2}{4} = \frac{1}{2}$ ,

$(-4)^{\frac{1}{2}} = \{ 2i, -2i \}$

Well then, we can also do it this way:

$(-4)^{\frac{1}{4}} = \{ 1+i, 1-i, -1-i, -1+i \}$

then we can square them and get

$( (-4)^{ \frac{1}{4}})^2 = \{ (1+i)^2, (1-i)^2, (-1-i)^2, (-1+i)^2 \}$

$= \{ 2i, -2i, 2i, -2i \}$

$= \{ 2i, -2i \}$

But, note that we are not bound to solve it this way

$(-4)^2 = 16$

$(16)^{\frac{1}{4}} = \{ 2, -2, 2i, -2i \}$

as we will be getting extraneous values.

Extraneous solutions arise when we carry out a many-to-one, then one-to-many operations. Thus, the convention is to carry out the one-to-many operation first, then the many-to-one. Note that this will only apply when the fractional exponents have a common factor, i.e., it is not reduced to its lowest terms. In such case, both ways will yield the same answer.