Context: I teach at a small liberal arts school in the US. The students of interest in this question are generally in their first year.

In courses like precalculus and calculus I like to use functions involving fractional powers like $x^\frac{2}{3}$ and $x^\frac{7}{5}$. Occasionally that leads to the evaluation of expressions like $(-8)^\frac{2}{3}$. Of course, this is equal to $4$, but not all calculators are able to resolve this expression, unfortunately.

Graphing calculators like the TI-83 and TI-84 have no issues, and the TI-89 does not complain, provided that you have its complex format set to "real" instead of "rectangular".

However, I have found that basic calculators like the TI-30XIIS and TI-36 respond to this expression with domain errors.

Most curiously, these basic calculators can evaluate expressions with composite powers like $((-8)^\frac{1}{3})^2$ and $((-8)^2)^\frac{1}{3}$. In fact, they seem to be able to handle negative numbers to the power of $\frac{1}{n}$ (of course, provided that $n$ is odd).

This leaves me with a few questions...

  1. Is there any workaround for this other than telling me students to (a) understand rational exponents more thoroughly, (b) buy a better calculator, or (c) break every fractional exponent into repeated powers?

  2. What is it about the TI-30XIIS that allows it to evaluate the decomposed expressions but not the original? I've heard it said that it uses logarithms to handle rational exponents, so that might explain why the first expression ran into issues--but that does not explain why the decomposed expressions were ok! (Maybe they're coded differently?)

Any insight that anyone can provide would be much appreciated! (Also, I'm not sure if this question is appropriate here, or if it should be migrated to math.stackexchange.)

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    $\begingroup$ Hmm, which definition of fractional powers do you use which makes $(-8)^{2/3}$ "of course [...] equal to $4$"? $\endgroup$ Commented Nov 12, 2021 at 20:30
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    $\begingroup$ @SueVanHattum: Yes, but an order of interpretation (for instance, $(-8)^{2/3} = \big((-8)^2\big)^{1/3}$) is not a definition of fractional powers; it is a mathematical result which has to be derived from the definition - and for this, one needs a definition to start with. Of course, one could solve this issue ad hoc by simply defining $x^{p/q}$ for negative $x$, integer $p$ and odd $q$ by means of such a formula, but there are various disadvantages of using this ad hoc definition (for instance, it is inconsistent with more common approaches from complex analysis). [...] $\endgroup$ Commented Nov 12, 2021 at 21:36
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    $\begingroup$ [...] And indeed, in my experience, it is quite uncommon to define fractional powers of negative numbers at all (maybe with the exeception of the exponent $1/q$ for $q$ odd). So my point is that there are good reasons why some calculators refuse to compute $(-8)^{2/3}$, and I was thus surprised why OP takes it as granted that there should be an obvious result. $\endgroup$ Commented Nov 12, 2021 at 21:36
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    $\begingroup$ [Hmm, now that I think about it, I should probably write this down in more detail in an answer. I'll do so as soon as I have a bit of time left.] $\endgroup$ Commented Nov 12, 2021 at 21:41
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    $\begingroup$ @AdamRubinson, (-8)^1/3 does not need complex numbers. -2 is a fine real number answer for that, since (-2)^3 = -8. $\endgroup$
    – Sue VanHattum
    Commented Nov 13, 2021 at 0:56

1 Answer 1


Answer: The calculators are fine, it is the question's premise "$(-8)^{2/3} = 4$" that needs to be fixed.

Now this was a bit polemic, of course. Here's a more detailed description of what I mean:

About definitions

Since results of computations are not dictated to us be a higher power, but are consequences of the definitions of the involved operations, the question is: What definition of fractional powers of negative numbers yields the equality $(-8)^{2/3} = 4$? One can certainly give a definition which yields this result (see below), but in my experience such a definition is hardly ever used, for the following reason:

Whenever we define a mathematical concept, there's a number of issues that we should typically consider to evaluate whether the definition - let's call it $D$ - that we are putting forth is useful (or "good"):

  1. $D$ should be well motivated.

  2. $D$ should be useful in some sense.

  3. $D$ should have (to a certain extent) nice mathematical properties. (At least if we intend to build rich theory about it.)

  4. $D$ should be consistent with previous notions that are generalized by $D$.

  5. In case that even more general concepts than $D$ are available (or even common) in the literature, it is desirable that $D$ be consistent with these more general concepts.

It's not always possible to satisfy all these points, and $D$ might well turn out to be fruitful if only a subset of them are satisfied. But the smaller this subset is, the less likely it become that it is a good choice to use $D$.

Fractional powers $x^{\alpha}$ for positive numbers $x$.

(Here, $\alpha$ is a rational number.)

I think it is useful to first point out why things work so well for $x > 0$. There are, of course, various different approaches to define fractional powers of positive numbers - but they all turned out to be equivalent (in the sense that they all yield the same result). Here's what happens to the points 1)-5) above:

  1. Depending on the approach one uses to define $x^\alpha$ for $x > 0$, there are various ways to motivate the definition. (I guess there's no point in listing some of them here.)

  2. The (equivalent) definitions of $x^\alpha$ are, for instance, useful because they tell use something about solutions of equations: For a non-zero integer $q$ and a real number $x > 0$, the number $y = x^{1/q}$ is the only positive real number that satisfies $y^q = x$. This shows that positive numbers to the power $1/q$ are a useful concepts. This idea that "fractional powers solve equations" is consistent with more general fractional exponents: For non-zero rational $\alpha$ and for positive $x$, the number $y = x^{1/\alpha}$ is the only positive solution of $y^\alpha = x$.

  3. For fixed $x > 0$, there are various nice properties of the mapping $\alpha \mapsto x^\alpha$. For instance, it is a group homomorphism from the additive group of rational numbers into the multiplicitive group of positive real numbers. Moreoever, it is continuous, and it's continuous extension to $\alpha \in \mathbb{R}$ is even differentiable, and in fact even real analytic.

  4. The definition is consistent with the definition of $x^n$ for integer $n$.

  5. The definition is also consistent with the definition of $x^\alpha$ for real $\alpha$ and $x > 0$ (that one obtains either by continuous continuation of fractional powers, or by using the real exponential function and logarithm). It is even consistent with the definition of $x^\alpha$ that one obtains for $x \in \mathbb{C} \setminus (-\infty,0]$ and $\alpha \in \mathbb{C}$ by using the principal branch of the complex logarithm.

Fractional powers $x^\alpha$ for negative $x$.

Now, let $x < 0$ and $\alpha = p/q \in \mathbb{Q}$ for integers $q \not= 0$ and $p$. There are serious problems to give a general ad hoc definition of $x^\alpha$ since, for even $q$, there is no real number $y$ which satisfies $y^q = x$. So an ad hoc definition of $x^{p/q}$ can apparently only work for odd $q$ - in which case we can define $x^{p/q}$ as $(x^{1/q})^p$, where we define $x^{1/q}$ to be the only real number whose $q$'s power is $x$. This ad hoc definition does indeed yield $(-8)^{2/3} = 4$. So let's check what happens to points 1-5 with this definition:

  1. One can certainly find a reasonable motivation for such a definition (for instance, that one whises to have a similar concept as for positive $x$). But even here there's a first problem: Since we have, for instance, also $\alpha = \frac{2p}{2q}$, this definition only works if we first cancel all common factors of $p$ and $q$ - so the definition of $x^{\alpha}$ now relies on a certain representation of the number $\alpha$ (instead of giving the definition for every representation and just showing that this is consistent). Strictly speaking, that's not a problem, but it makes the definition of $x^\alpha$ at least a bit less "natural" than for positive $x$.

  2. For $\alpha = 1/q$, where $q$ is odd, one can say as in the case of positive $x$ that it makes sense to define $x^{1/q}$ to be the unique real number whose $q$'s power is $x$. However, in contrast to the positive case, this does not generalise to fractional exponents whose numerator is even: Indeed, if we use the definition described above - and thus obtain $(-8)^{2/3} = 4$ -, then $y = (-8)^{2/3} = 4$ is not a solution of the equation $y^{3/2} = -8$. So the idea that $y = x^{1/\alpha}$ solves $y^\alpha = x$ breaks down with this definition - and thus one of the major reasons why fractional powers are useful is lost.

  3. This is where it becomes really bad. From an algebraic point of view we can at least still say that $\alpha \mapsto x^\alpha$ is a group homomorphism from the additive group $G := \{p/q: \, p,q \in \mathbb{Z}, \; q \text{ odd}\}$ into the multiplicative group $(0,\infty)$ (although the domain of the group homomorphism is, well, a bit special). But from an analytic point of view, everything breaks down: When we endow the set $G \subseteq \mathbb{R}$ with the topology induced by $\mathbb{R}$, the mapping $\alpha \mapsto x^\alpha$ is far from being continuous: If $p,q$ are both odd, then $x^{p/q}$ will be negative. But if $p$ is even and $q$ is odd, $x^{p/q}$ will be positive. Since both sets $\{p/q: \; p,q \text{ odd}\}$ and $\{p/q: \; p \text{ even}, q \text{ odd}\}$ are dense in $G$, the mapping $\alpha \mapsto x^\alpha$ is nowhere continuous.

  4. This point seems to be ok. The definition is still consistent with the definition of $x^n$ for integer $n$.

  5. This doesn't go well, either. Since $x^\alpha$ does not depend continuously on $\alpha$, we can't even extend the definition to real $\alpha$ by a continuity argument. Our definition of $x^\alpha$ is also completely at odds with complex analysis, for two reasons: (i) The principal branch of the complex logarithms gives us a definition of $x^\alpha$ for all complex numbers $x$ except for those in $(-\infty,0]$. So we chose to define $x^{\alpha}$ for precisely those numbers $x$ for which the principal branch of the logarithm doesn't yield a result. (ii) If we choose another branch of the logarithm, we will get a definition of $x^\alpha$ for negative $x$ - but with this definition, $x^\alpha$ will depend continuously on $\alpha$ and hence, it cannot be consistent with the definition that we gave.


For the reasons pointed out above, it is in my experience very uncommon to define $x^{p/q}$ for negative $x$, odd $q$ and general integers $p$.


  • Please let the basic calculators live :-), they are apparently doing a good job (at least with respect to this question).

  • Please don't tell your students that $(-8)^{2/3}$ were "of course" equal to $4$ - this does them a disservice because the underlying definition is not useful with respect to solving equations, very badly behaved from an analytic point of view (continuity!), and incosistent with common concepts from complex analysis.

Additional remarks:

  • As I mentioned above, some people certainly define $x^{1/q}$ for negative $x$ and odd $q$ to be the only real number whose $q$'s power is $x$. This has some of the disadvantages mentioned above (in particular, the inconistency with the complex logarithm), but it also has the advantage that it scores highly on point 2: Solving equations is one of the most fundamental things in maths, so a notation which helps us to discuss solutions to certain equations can often be useful. Moreover, this definition is not so badly behaved with respect to 3: Since the set $\{1/q: \, q \text{ odd}\}$ has no accumulation points in $(0,\infty)$, there are no continuity problems.

  • It is quite illuminating to check how the common definition $0^0 := 1$ scores on points 1-5. It might be somewhat ambivalent with regard to 3 (in particular, with respect to continuity issues) but it is absolutely unbeatable with regard to 2: All typical formulas from basic algebra or analysis that include sums over powers (for instance: the binomial expansion, the geometric sum, Taylor expansions) would immediately break down - or at least require extremly bizzare exception handling - if we didn't define $0^0$ as $1$. So this is an nice example where certainly not all of the points 1-5 are satisfied, but where some of the points outrank the others so stronlgy that the definition is completely common.

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    $\begingroup$ [...] The reason why I claim that this is not a good idea (and why, in my experience, almost all books on real analysis define $0^0$ as $1$), is as follows: Consider, for instance the exponential series $e^x = \sum_{k=0}^\infty x^k / k!$ for real $x$. The series contains the expression $x^0/0!$. So if we did not define $0^0$, we were no allowed to substitute $x=0$ in the series. And if we defined $0^0$, but as the number $0$, then we were allowed to subsitute $x=0$ in the series, but the result would be equal to $0$, while $e^0$ is $1$ - so the equality wouldn't be true for $x=0$. [...] $\endgroup$ Commented Nov 13, 2021 at 2:24
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    $\begingroup$ [...] The same phenomenon occurs for many further formulas. For instance, for real numbers $x,y$ and a positive integer $n$, one has the binomial formula $(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$. The $0$th expression under the sum is $\binom{n}{0} x^0 y^n$, and the last expression is $\binom{n}{n}x^ny^0$. So again, if $0^0$ wasn't defined, we weren't allowed to substitute $x=0$ or $y=0$. And if we chose to define $0^0$ as $0$, we were allowed to substitute $0$ for $x$ or $y$, but the equality would then be wrong for $x=0$ or $y=0$. [...] $\endgroup$ Commented Nov 13, 2021 at 2:25
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    $\begingroup$ [...] So if we want these formulas to be true in the form they are usually stated, we have to go with the definition $0^0 = 1$. That's what I meant by "The definition $0^0 = 1$ is absolutely unbeatable with regard to point 2": The definition is very useful since it makes many important formulas true in the way we would like them to be true. $\endgroup$ Commented Nov 13, 2021 at 2:25
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    $\begingroup$ And so it makes sense to wait on defining it until you will be using one of those expressions. $\endgroup$
    – Sue VanHattum
    Commented Nov 13, 2021 at 2:30
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Commented Nov 13, 2021 at 2:32

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