I am studying James Stewart's "Algebra and Trigonometry 4th Edition". On page 27, a definition of rational exponents is presented:

For any rational exponent $m/n$ in lowest terms, where $m$ and $n$ are integers and $n > 0$, we define

$a^{m/n} = \left(\sqrt[n]{a}\right)^m$ or equivalently $a^{m/n} = \sqrt[n]{a^m}$

If $n$ is even, then we require that $a \geq 0$.

Is the conditional "in lowest terms" required in this definition?

I can see that $(-1)^{2/6} \neq (-1)^{1/3}$. However, Stewart is also requiring that $a \geq 0$ if $n$ is even which would preclude $(-1)^{2/6}$.

Given that Stewart is requiring that $a \geq 0$ if $n$ is even, the proviso "in lowest terms" seems needlessly restrictive. I don't understand why it is included here.

What breaks if "in lowest terms" is omitted from this definition?

  • 4
    $\begingroup$ The problem is precisely what you state: $\frac{2}{6}$ is the same rational as $\frac{1}{3}$, so we would like any expression that uses $\frac{1}{3}$ to give the same result as that same expression if you replace $\frac{1}{3}$ with $\frac{2}{6}$... if that is allowed. But $(-1)^{2/6}$ is not defined, while $(-1)^{1/3}$ is. So we just say that it's not that the same computation yields different results depending on how you write it, but rather that the first expression is not a well-formed formula. $\endgroup$ Feb 16 at 14:58
  • $\begingroup$ If you restrict $a>0$ then there is no need to assume minimality but the burden is on you to prove that it is well defined. $\endgroup$ Feb 16 at 15:03
  • $\begingroup$ @ArturoMagidin: Shouldn't the included statement "If $n$ is even, we require that $a \geq 0$" be enough to disallow expressions like $(-1)^{2/6}$? If you remove "in lowest terms" from the definition, what issues arise? $\endgroup$
    – ntwk
    Feb 16 at 15:06
  • 1
    $\begingroup$ The point is that then $(-1)^{2/6}$ is not defined, but $(-1)^{1/3}$ is, even though the bases are the same and the exponents are equal as numbers. So instead what we say is that to calculate $(-1)^{2/6}$, first you reduce the exponent to lowest terms, and then you calculate. Note also, as Shinrin-Yoku has noted, that you would have to verify that if $\frac{n}{m}=\frac{p}{q}$, then the $\sqrt[m]{a^n} = \sqrt[q]{a^p}$ when both are defined. So you'll end up having to do work either at the front end or the back end. Stewart wants the work done at the front end. $\endgroup$ Feb 16 at 15:12
  • 1
    $\begingroup$ That's the point: $2/6$ and $1/3$ are the same number. So we would expect the expressions $a^{1/3}$ and $a^{2/6}$ to mean the same thing. But if we define $a^{m/n}$ to mean "$\sqrt[n]{a^m}$, except that when $n$ is even we require $a$ to be positive" then $(-1)^{1/3}$ and $(-1)^{2/6}$ do not mean the same thing. So we have to do something about that: either we say that they do not in fact mean the same thing even though the numbers that show up are "the same", or we say "to calculate $a^{m/n}$, first reduce $\frac{m}{n}$ to lowest terms, and then..." which is what Stewart does. $\endgroup$ Feb 16 at 17:11

1 Answer 1


With the help of @ArturoMagidin I now understand why the definition was stated this way. I've organized and documented my reasoning here in case it helps someone else.

To start:

  1. Avoid thinking of the rational exponent as just being two parameters $m$ and $n$ that you plug into to some operation $\sqrt[n]{a^m}$.
  2. Think of the rational exponent as being one thing: a number.
  3. Think of how that number would be written in lowest terms.

Only once you have the number in lowest terms do you have $m$ and $n$ such that $a^{m/n} = \sqrt[n]{a^m}$ (and if $n$ is even, $a$ must be nonnegative).

For example, $2/6$, $1/3$, and $0.\overline{3}$ are all the same rational number. When a base $a$ is raised to this rational number, we would like it to yield the same result regardless of how this rational number happens to be represented symbolically. So, how do the words "in lowest terms" help us do this?

As an experiment, let's remove "in lowest terms" from the definition and see what happens. Nothing breaks for the base $a \geq 0$. When $a$ is nonnegative, $a^{2/6}$ and $a^{1/3}$ will give the same result. However, if $a < 0$, then $a^{2/6}$ and $a^{1/3}$ will yield different results.

But wait! Stewart is also requiring that if the rational exponent's denominator is even, $a$ can't be negative. With this additional requirement, we might think we've avoided any issues. An expression like $(-1)^{2/6}$ simply remains undefined. However, this still leaves us with a problem.

Remember, $2/6$ and $1/3$ are the same number. Consider $x$. If $x = 2/6$, then $x = 1/3$. So, is $a^x$ defined or undefined? We have a paradox unless "in lowest terms" is included in the definition.


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