Is "in lowest terms" necessary in this definition of rational exponents?

Question

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?

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.