Why ${a}^{\frac 1n}$ $ (a\in \Bbb{R},n\in \Bbb{N}, n\ge2)$ should be undefined unless $a\gt0$?

Question

In the most recent national mathematics textbook for 10th grade in my home country, ${a}^{\frac 1n}$ is defined in the following way:

For every natural number $n \ge2$, positive number $a$ raised to power of ${\frac 1n}$ is defined as: $${a}^{\frac 1n} = \sqrt[n]a$$

($a \in \Bbb{R}$ as stated before definition)

and then adds:

Throught this book if $a\lt0$, ${a}^{\frac 1n}$ is not defined. For example expressions such as ${(-1)}^{\frac 13}$ and ${(-2)}^{\frac 24}$ are undefined.

(Definition and comment above can be seen on page 65 of the pdf version available through the following link (Persian)(PDF): http://chap.sch.ir/sites/default/files/lbooks/96-97/40/C110211.pdf)

What is the reason of defining it this way?

Answer 1

Presumably, the students do not learn about the complex numbers. As a result, there is no meaningful definition for things like $(-1)^{\frac12}$, which under any reasonable definition should be something that, when squared, gives $-1$; but such a number does not exist within the real numbers.

In case $n$ is odd, there is of course a good definition as a negative number, but for simplicity they presumably just decide to restrict radicals to positive numbers, no matter which radical.

Answer 2

That is for two reasons:

First, $a^x$, for $x\in \mathbf R$ is defined as $\;\mathrm e^{x\ln a}$, and $$\ln a$ is defined for $a>0$.

Second, while we may very well speak of $\;\sqrt[3]{\mathstrut -8}$, for instance, the exponential notation $(-8)^{\tfrac13}$ leads to inconsistencies: \begin{align}(-8)^{\tfrac13}&=(-8)^{\tfrac26}=\bigl((-8)^2\bigr)^{\tfrac16}=(64)^{\tfrac16}=2,\\ \text{but}\qquad (-8)^{\tfrac13}&=\Bigl((-8)^{\tfrac16}\Bigr)^2\qquad\text{is not defined!} \end{align}