How do you compute negative numbers to fractional powers?
A negative base is a point of conflict between the three commonly used meanings of exponentiation.
- For the continuous real exponentiation operator, you're not allowed to have a negative base.
- For the discrete real exponentiation operator, we allow fractional exponents with odd denominators, and $$(-a)^{b/c} = \sqrt[c]{(-a)^b}= \left( \sqrt[c]{-a} \right)^b = (-1)^b a^{b/c} $$ (and this is allowed because every real number has a unique $c$-th root)
- For the complex exponentiation operator, exponentiation is multivalued. An exponentiation with denominator $n$ generally takes on $n$ distinct values, although one is generally chosen as the "principal" value.
For $(-5)^{2/3}$, these three exponentiation operators give
- Undefined
- $\sqrt[3]{25}$
- $\omega \sqrt[3]{25}$ is the principal value. The other two are $\sqrt[3]{25}$ and $\omega^2 \sqrt[3]{25}$, where $\omega = -\frac{1}{2} + \mathbf{i} \frac{\sqrt{3}}{2}$ is a cube root of $1$.
Unfortunately, which meaning of exponentiation is meant is rarely ever stated explicitly, and has to be guessed from context.
I'm guessing that the second one is meant.
In case you're curious, here is part of the rationale for the first and third conventions.
In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other negative. And it's not even clear that $(-5)^{\pi/5}$ should be meaningful!
For the third convention, the whole thing is like the idea of $\pm 2$ being the 'square root of 4', but for the fact the complexes cannot be cleanly separated into "negative" and "positive" to let us choose a specific one nicely.
A method is chosen for the principal value, based trying to get positive bases 'right' and trying to keep continuity as much as possible, but alas this convention gets the negative bases 'wrong'.
In some sense, this can be viewed as the principal value of $(-5)^{2/3}$ chosen to be "two-thirds of the way" from positive to negative.
This question is several years old. Nevertheless, I'm bothered that none of the answers provided were at an introductory level. Let me then contribute an answer, which is not rigorous, but will serve as a general heuristic for students who have only recently encountered fractional exponents with real numbers.
We can interpret a base raised to a simplified$^\dagger$ fractional exponent with this heuristic: $$x^{\frac ab} = x^{\frac{\text{'power'}}{\text{'root'}}}$$ That's saying that $a$ acts like a standard integer power and $b$ acts like a standard integer root.$^{\dagger\dagger}$
Your example, $(-5)^{2/3}$, can be interpreted as squaring $-5$ and then taking the third root. Or, in the opposite order, taking the cube root of $-5$ and then squaring that result.
\begin{align} (-5)^{2/3} &= ((-5)^2)^{1/3} = \sqrt[3]{25} \approx 2.92 \\\\ \text{or}& \\\\ (-5)^{2/3} &= (\sqrt[3]{-5})^{2} \approx (-1.71)^2 \approx 2.92 \end{align}
Notice that in this particular example our base was negative. Since the denominator of the fraction was odd, we were able to solve for a real number. If the denominator were even, though, we would have no real solution, since the even root of a negative number is undefined for real numbers. Instead, we would have to turn to complex numbers for a more adequate interpretation (see the accepted answer by Hurkyl).
$^\dagger$ The fractional exponent must be simplified for the upcoming process to make sense. To see why, consider the example $(-8)^\frac 24$. What happens if you don't simplify? If you do?
$^{\dagger\dagger}$ We're assuming $a$ and $b$ are integers such that $a/b$ is a rational number. That's likely what a student has seen when first encountering fractional exponents. If $a,b$ are not integers, then the meaning is less obvious.
Odd roots of negative numbers are well-defined. $(-5)^{\frac 13}=-(\sqrt[3]5)$ is well defined as you can check: it is the only real number that satisfies $x^3=-5$. You can then square it to get $(-5)^{\frac 23}$ and if you square first and ask for $\sqrt[3]{25}$ you get the same result. If the denominator isn't odd you have a problem.