# How do you compute negative numbers to fractional powers?

My teachers have gone over rules for dealing with fractional exponents. I was just wondering how someone would compute say: $$(-5)^{2/3}$$ I have tried a couple ways to simplify this and I am not sure if the number stays negative or turns into a positive. I know that if a negative number is raised to an odd power it is negative, but fractional powers are neither odd or even. Is there a general rule for dealing with these types of problems?

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Do you know what $5^\frac{2}{3}$ is? Also, do you know that a negative number does not have, for instance, a square root, and in general a negative number doesn't have an n-th root where n is even (in the real numbers)? Combining those ideas should help you work out the answer. – crf Mar 1 '13 at 4:13
@crf Yes I do, I just do not know how the negative sign will behave when simplifying. – Kot Mar 1 '13 at 4:14
Good question! You have put your finger on a subtle issue. – MJD Mar 1 '13 at 4:15
Steven - feel free to accept answers that you find to be helpful. You can accept one answer per question (and upvote as many as you'd like). To accept an answer, click on the $\checkmark$ to the left of the answer you want to accept. (You get two reputation points for every answer accepted, too!) – amWhy Mar 1 '13 at 5:45

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.

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Why do you say $\omega\sqrt[3]{25}$ is the principal value and not $\sqrt[3]{25}$? – MJD Mar 1 '13 at 4:47
Well done. Probably TMI. Maybe it will prompt further investigation. – Ross Millikan Mar 1 '13 at 4:47
@MJD: The standard choice of complex angle for a negative number is $\arg z = \pi$, and the principal $n$-th root has angle $n \pi$. – Hurkyl Mar 1 '13 at 4:57
@Hurkyl: sorry for resurrecting this thread, but just for posterity, did you mean that the principal $n^\text{th}$ root has angle $\pi/n$? – robjohn Feb 18 '14 at 2:51
@robjohn: Ah yes. Or maybe I was thinking that the $m$-th power has angle $m \pi$ and had $m = 1/n$ in mind, but either way the same thing. – Hurkyl Feb 18 '14 at 8:01

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.

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This is a bit confusing to me. Would I be able to factor out a negative 1 then simplify? All of these answers are a little advanced to me. This problem came up when I was working with limits to negative infinity. – Kot Mar 1 '13 at 4:54
@StevenN: as long as the exponent is rational and the denominator is odd, yes. The issue is that $x^n$ is monotonic if $n$ is odd, but not if $n$ is even. This implies that $x^{\frac 1n}$ is a well defined function as long as $n$ is odd, but not if $n$ is even. This is what MJD was alluding to when s/he said this is a subtle issue. If you ask for $(-1)^{\frac 26}$ you might see something different from $(-1)^{\frac 13}$ because we don't know how (in the reals) to take a sixth root of $-1$ – Ross Millikan Mar 1 '13 at 5:00

Hint: we know that $a^\frac{b}{c}=(a^b)^\frac{1}{c}$. What conditions do you need on $a^b$ so that $(a^b)^\frac{1}{c}$ is defined when $c$ is even? What about odd? Then, what conditions can you put on $a$ and $b$ that will satisfy those last conditions?

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I'm pretty rusty, so I could be wrong, but I think the correct answer can be found using the formula

i^x = cos(xpi) = isin(x*pi)

So in this case, you could rewrite the equation as 5^(2/3) * i^(2/3)

You know 5^(2/3), and by the above equation i^(2/3) is cos(2/3*pi) + i*sin(2/3*pi)

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