In our new books in freshman highschool,it is mentioned that :
When $a<0$ then, the expression
$$a^{m \over n} m,n\in \mathbb Z $$
even if the $n$ in the denominator is odd.
And my teacher gave the argument that:
$$((-8)^2)^{1 \over 6}=64^{1 \over 6}=2$$
But:$$(-8)^{2 \over 6} =(-8)^{1 \over 3}=-2$$
And he said it resulted into a contradiction that:$$ (a^m)^n=a^{mn}$$
My argument :we cannot make the step that:
$$(-8)^{2\over 6}=(-8)^{1\over3}$$ since there is no meaning ful way to go from the latter to the prior without changing the sign.
What is the correct answer if we take $a\in \mathbb C$or $a\in \mathbb R$?
If you want fractional exponents, then you have to accept that $\frac13 = \frac26$. Otherwise the exponents wouldn't deserve to be called fractions, and I for one do not know what rules such numbers would follow. This, along with the fat that we like the rule that $a^{mn} = (a^m)^n$, means that fractional exponents customarily only are defined for non-negative bases.