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?

  • $\begingroup$ 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. $\endgroup$ – crf Mar 1 '13 at 4:13
  • 1
    $\begingroup$ @crf Yes I do, I just do not know how the negative sign will behave when simplifying. $\endgroup$ – Kot Mar 1 '13 at 4:14
  • 3
    $\begingroup$ Good question! You have put your finger on a subtle issue. $\endgroup$ – MJD Mar 1 '13 at 4:15
  • $\begingroup$ 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!) $\endgroup$ – Namaste Mar 1 '13 at 5:45
  • $\begingroup$ possible duplicate? math.stackexchange.com/q/608023/75231 $\endgroup$ – CODE Mar 11 '16 at 9:17

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.

  • 1
    $\begingroup$ Why do you say $\omega\sqrt[3]{25}$ is the principal value and not $\sqrt[3]{25}$? $\endgroup$ – MJD Mar 1 '13 at 4:47
  • 1
    $\begingroup$ Well done. Probably TMI. Maybe it will prompt further investigation. $\endgroup$ – Ross Millikan Mar 1 '13 at 4:47
  • 1
    $\begingroup$ @MJD: The standard choice of complex angle for a negative number is $\arg z = \pi$, and the principal $n$-th root has angle $n \pi$. $\endgroup$ – Hurkyl Mar 1 '13 at 4:57
  • 3
    $\begingroup$ @Hurkyl: sorry for resurrecting this thread, but just for posterity, did you mean that the principal $n^\text{th}$ root has angle $\pi/n$? $\endgroup$ – robjohn Feb 18 '14 at 2:51
  • 1
    $\begingroup$ @Brian: Nope; the expression I defined is only for odd denominators. (and if a fraction can't be written with an odd denominator, the exponentiation isn't defined) $\endgroup$ – Hurkyl May 17 '14 at 15:35

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.

  • 3
    $\begingroup$ Thanks for answering 'at the level of the question', your answer is the only one that I understood. $\endgroup$ – mattst Sep 29 '18 at 19:38

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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Kot Mar 1 '13 at 4:54
  • $\begingroup$ @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 $-1$ $\endgroup$ – Ross Millikan Mar 1 '13 at 5:00

The answer to your question depends on the assumed definition of exponentiation, but I think the most reasonable and consistent way to look at it is in the light of complex numbers. Thinking in polar coordinates, we have to ask what happens the magnitude and what happens to the angle. To answer that, we split the problem in the following way:


The term $(-1)^{2/3}$ specifies the angle and is equal to $e^{(2/3)i\pi}$, while $5^{2/3}$ specifies the real-valued magnitude. As such, the result according to how I interpret the operation is $$(-5)^{2/3} = 5^{2/3}*e^{(2/3)i\pi} = 5^{2/3}*(-\frac12+\frac{\sqrt3}{2}i)$$ which is roughly equal to $-1.462+2.532*i$.

Transformations interpreting the rational-valued exponent as a composition of exponentiation and extracting the root are problematic because of the failure of some logarithm and power identities on complex numbers.

On that note, it should be mentioned that even the transformation I used in the first equation is not generally allowed, it just works with splitting a negative real base into a positive real and a negative real, in the way I used it here.

EDIT: Pondering over the other answers, I think the reason for confusion is the assumed identity $\sqrt[a]{x}=x^{1/a}$, which isn't entirely correct. Taking the a-th root of a number is the inverse relation to taking a number to the power of a, it's supposed to return all numbers which, when taken to the power of a, become $x$. Raising $x$ to the power of the multiplicative inverse of a, on the other hand, is still just exponentiation with any real-valued exponent, where you raise the magnitude to the specified power and multiply the complex angle with the exponent, yielding one unique result.

This mostly comes down to interpretation, I propose it's just convenient to do the distinction to remove ambiguity.

  • $\begingroup$ It is interesting to view as a movie the complex graphs of y=x^r for r=-infinity to r=infinity (with frame rate varying to prevent an infinite duration of the movie). $\endgroup$ – amI Oct 14 '16 at 18:26

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?


protected by Zev Chonoles Aug 23 '16 at 19:41

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.