I have questions about one proof.

It concerns numbers with rational exponents.

Having in mind that for all $\displaystyle r,s \in Q,a \in R$ is $\displaystyle (a^r )^s = a^{rs}$ then author of book wants to show that $\displaystyle a$ must be positive real number.

Proof is:

If $\displaystyle r$ is rational number then $\displaystyle \frac{r}{2}$ is also a rational number, so:

$\displaystyle a^r = a^{\frac{r}{2} \cdot 2} = (a^{\frac{r}{2}} )^2 \ge 0$

Similarly, if $\displaystyle r$ is rational number then $\displaystyle -r$ is rational number, so:

$\displaystyle a^{ - r} = a^{r \cdot ( - 1)} = (a^r )^{ - 1} $ which gives us that $\displaystyle a^r \ne 0$. Then is $\displaystyle a^r > 0$

If $\displaystyle r \ne 0$ then is $\displaystyle \frac{1}{r} \ne 0$.

Based on above, we get that $\displaystyle a = a^{r \cdot \frac{1}{r}} = (a^r )^{\frac{1}{r}} $ which means that $\displaystyle a$ must be positive number because $\displaystyle a^r $ is positive number so is $\displaystyle (a^r )^{\frac{1}{r}} $ positive number.

My question is why $\displaystyle a^r$ must be positive number?

If $\displaystyle r$ is rational number then $\displaystyle \frac{r}{3}$ is also a rational number, so:

$\displaystyle a^r = a^{\frac{r}{3} \cdot 3} = (a^{\frac{r}{3}} )^3$ which means that $\displaystyle a^r$ can be also negative number.

Based on that, then $\displaystyle a = a^{r \cdot \frac{1}{r}} = (a^r )^{\frac{1}{r}} $ can mean that $\displaystyle a$ can be also negative number.