I am confused about this piece of text on rational exponents! I was reading the free book by Stitz and Zeager, Pre-requisites for precalculus, Chapter 0 section 0.2 (page 32):
https://www.stitz-zeager.com/
https://www.stitz-zeager.com/ch_0_links.pdf

It presents a case where rational exponents cannot be used with the power rule.
First they state this:
$(a^{\frac{2}{3}})^{\frac{3}{2}}$ where using the power rule we get $(a^{\frac{2}{3}})^{\frac{3}{2}} = a^{\frac{2}{3}.\frac{3}{2}} = a^1 = a $
Then they say suppose a = -1:
$-1^{\frac{2}{3}} = (\sqrt[3]{-1})^2 = (-1)^2 = 1$ thus $(-1^{\frac{2}{3}})^{\frac{3}{2}} = 1^{\frac{3}{2}} = 1$ and therefore $(a^{\frac{2}{3}})^{\frac{3}{2}} \neq a $
but then in the last sentence says that the power rule is true and the reader should show this?
I am therefore confused and would appreciate anyone trying to explain better what the authors mean here? Is it true or is it not true for rational exponents?
 A: They have altered the order of the exponents in the last sentence. It can be shown for $a=-1$ as:
$$a^{\frac 32}=(-1)^{\frac 32}=((-1)^3)^{\frac 12}=(-1)^{\frac 12}=i$$
Now, $$i^{\frac 23}=(i^2)^{\frac 13}=(-1)^{\frac 13}=-1=a$$
The proof for general negative $a$ is of the same flavor. It is interesting to note that the authors seem to ignore the complex roots $-\omega, -\omega^2$. I can't seem to understand why.
A: The rules for integer exponents work also work for rational exponents, provided that $a$ is greater than $0$. What the textbook is warning about is that the rule
$$
(a^n)^m=a^{mn} \tag{*}\label{*}
$$
can fail if $a<0$. For example, if we tried to apply \eqref{*} in the case $a=-1$, then we would get
$$
\left((-1)^{2/3}\right)^{3/2} = (-1)^1=-1 \, .
$$
The above equation is false. We can prove this by noting that $(-1)^{2/3}$ is defined as $\left(\sqrt[3]{-1}\right)^2$. Hence,
$$
\left((-1)^{2/3}\right)^{3/2}=\left(\left(\sqrt[3]{-1}\right)^2\right)^{3/2}=\left((-1)^2\right)^{3/2}=1^{3/2}=1 \, .
$$
It is also not true that
$$
\left((-1)^{3/2}\right)^{2/3}=-1 \, .
$$
The "number" $(-1)^{3/2}$ would seem to equal $\left(\sqrt{-1}\right)^3$. But the function $\sqrt{x}$ is only defined when $x\geq0$, and so the above equation doesn't make any sense. So, to reiterate, exercise extreme caution when dealing with negative bases.
A: With complex numbers, there is a reason why $\sqrt{-1}=i$. This is not just to make a stylistic reason, but rather because the rules for exponents are not always true, for example:
$$i^2=-1=\sqrt{-1}\sqrt{-1}≠\sqrt{-1*-1}=\sqrt1=1=\sqrt{(-1)^2}=|-1|=-1$$
For this reason, there was an amendment, like a constitution, to make the rules of algebra work for imaginary and complex numbers. Even though reals numbers are complex, the normal rule of the “multiplicativity” of radicals and exponents still applies. Please give me feedback and correct me!
As another “answerer” showed, $$(-1)^{2/3}={\left((-1)^{1/3}\right)}^2=1^2=1$$
I will work on this a bit more...
