One of my high school students who has ZERO knowledge on complex numbers and the modulus function has showed me the following algebra:


Hence $$16=-16 \qquad\text{and so}\qquad 1=-1.$$

Should we impose that $$(a^m)^n=a^{mn}$$ only when $a \gt 0$?

  • 5
    $\begingroup$ You may be interested in this question from MESE's sister site, MSE. $\endgroup$
    – Nick C
    Feb 24, 2020 at 16:01
  • 1
    $\begingroup$ Since you mention high school, I have added the (secondary-education) tag. If you feel this is unnecessarily restrictive, please remove it. $\endgroup$
    – J W
    Feb 25, 2020 at 10:58
  • $\begingroup$ I have withdrawn my answer. $\endgroup$ Mar 6, 2020 at 17:06
  • $\begingroup$ If the student doesn't understand that $1\ne -1$, that's what he needs to be taught. $\endgroup$ Mar 9, 2020 at 16:08

4 Answers 4


Should we impose that $(a^m)^n=a^{mn}$ only when $a \gt 0$?

Maybe you should tell your student that he/she have discovered by himself/herself the proof that the rule $(a^m)^n=a^{mn}$ cannot be true with negative bases and rational exponents, and this is a great achievement.

Try to explain that he/she proved that

If the said rule is true for negative bases and rational exponents, then $1=-1$.

As we know that $1\neq -1$, the rule cannot be true. This is why the restriction $a>0$ appears in the books and why you will impose it.


  • Maybe I'm wrong but, in view of the title's question ("How should I convince a student..."), I suppose that the question is not about the rule $(a^m)^n=a^{mn}$ itself (which, I believe, the OP knows), but on how to approach it with the student.

  • The essence of my suggestion is: tell to the student what he/she did right (a mathematical discovery) instead of focusing on the mistake (wrong application of a rule that, probably, the student didn't know yet).

  • Of course, the bold text above is not as general as possible. But this is only the first step (see my comments below). Afater that, the professor could guide the student to further investigations in order to discover other forms of the rule.

  • $\begingroup$ Note that most high school/college algebra textbooks I'm aware of do define values of $a^{m/n}$ with $a < 0$ and odd $n$. $\endgroup$ Feb 25, 2020 at 18:21
  • 1
    $\begingroup$ @DanielR.Collins Then, the restriction mentioned in my post have to be replaced by the appropriate one. $\endgroup$
    – Pedro
    Feb 25, 2020 at 18:34
  • $\begingroup$ Your restriction is too restrictive. Consider $a = -2$, $m=2$ and $n=3$. Then you have $((-2)^2)^3=(-2)^{2\times 3} = 64$. Note that we have $(-2)^2 \in R$ and $(-2)^3\in R$. $\endgroup$ Mar 5, 2020 at 19:02
  • $\begingroup$ @DanChristensen You are correct, but generality is not my point here. As I said in my previous comment, the mentioned restriction have to be adapted according to the context of each class. (However, the restriction I used is much less restrictive than the one suggested by the OP (and thus, probably, suitable for his class)). My point is how to explain a "strange mathematical thing" to a child (child = low level of mathematical maturity). Maybe, the bold text is the least important part of my answer. $\endgroup$
    – Pedro
    Mar 6, 2020 at 11:19
  • $\begingroup$ If something in the answer would be better replaced/adapted, then I think we'd encourage editing the answer to reflect/improve that aspect? $\endgroup$ Mar 6, 2020 at 13:54

Obviously the correct mathematical answer is to show how the exponent rules actually work, and when they do not work.

So please don't accept this answer.

Anyway, the educational answer is to see that student is using the fact that $1^2 = (-1)^2$ in the critical middle step. Show them the corresponding fact for cubes, because it's crazy, and might expand their mathematical playground:

$$ 1^3 = \bigg( \frac{-1 + i\sqrt{3}}{2} \bigg)^3 $$

See what they can "prove" next. your student has an imagination; show them the rules and when they work -- but also keep that spark alive by showing them neat things.


When I asked the What are the Laws of Rational Exponents? question on SE Mathematics, I was largely thinking about this context; teaching at the level of high school or early (remedial) college math. While it wasn't the top-voted, my answer there represents my best thinking about the status of this issue in classes at that level. As I wrote:

Regarding the example in the question, most everyone agrees that $(-1)^{2 \cdot \frac{1}{2}} \ne ((-1)^2)^\frac{1}{2}$, if both sides are simplified in the standard order of operations; and this highlights the fact that the identity $(a^r)^s$ = $a^{rs}$ is not true unrestrictedly. Exactly what restrictions need to be honored depend on the definitions in use in a particular textbook.

Just to expand on the last line there; part of the conversation around these questions usually dredges up the fact that there is a standard definition for fractional exponents in real-numbers, and another one in complex-numbers, and that these two definitions actually disagree with each other (e.g.: the value of $(-8)^{1/3}$, on which numerous academic articles have been written). As a result, real-number theorists and complex-numbers theorists have a tendency to start arguing with each other on these questions.

So: I assume you are working from some standard textbook. That textbook must have some restrictions around the $(a^m)^n = a^{mn}$ identity (although they vary between books and contexts). I would recommend you read your book very carefully, note the restriction used there, and apply it conscientiously in your classroom practice.


(My new and hopefully improved answer)

Should we require that $(a^m)^n=a^{mn}$ only when $a \gt 0$ ?

That might "solve" the problem in some sense, but we do have legitimate cases with negative bases.

Example: $((-2)^2)^3 =(-2)^{2\times 3}=(-2)^6=64$

According user "Gae. S." at Math SE, $a^m, a^n \in \mathbb{R}$ is a sufficient condition for $(a^m)^n=a^{mn}$ when using real-number arithmetic.

So the Power of a Power Rule (on $\mathbb{R}$) can be restated: $(a^m)^n = a^{mn}$ if we have $a^m, a^n \in \mathbb{R}$.

This may provide a more satisfying resolution for this precocious student: Since $(-16)^{1/4} \notin \mathbb{R}$, we cannot infer that $((-16)^2)^{\frac{1}{4}}=(-16)^{\frac{1}{2}}$.


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