Why can't a negative base be raised to a non-integer exponent? Why can't we do this?
Example: $(-1) ^ {1/3}$
Math definitions are based on a definite logic. What is the logic here?
Can you give me some examples where it violates the equation?
I'm just a high school student b.t.w.. It would be greatly appreciated if you could describe it as simply as possible. thanks.
 A: It's not that you can't do it (if you're willing to accept answers that are non-real complex numbers).  It's that there are multiple possible answers and no obvious way to choose among those answers that preserves certain "nice" properties we want those answers to have.
That's not a problem with with positive reals.  When we take a rational power $s=r^{\frac pq}$, we're really saying that $s^q=r^p$.  There are multiple complex numbers $s$ (in fact, $q$ of them) that satisfy this equation, but exactly one of them will always be a positive real number and that's the "obvious" choice.  Importantly, this choice lets us extend the definition of $r^t$, where $t \in \Bbb R \setminus \Bbb Q$, in a "nice" (continuous) way when $r$ is a positive real number.
If you try this when $r$ is a negative real number, there is often no obvious choice because there are no negative real solutions for $s$ unless $q$ is odd.  And it turns out that there's no choice that always lets you extend the definition of exponentiation beyond rational exponents in a "nice" way, although any choice you make will allow you to extend the definition of exponentiation to almost all complex numbers.  These wrinkles are usually taught in advanced undergraduate or graduate classes, so for high school classes by far the more prudent course is to simply say "Don't do it."
A: $-1$ can very well be raised to some rational exponents, namely those with an odd denominator.
E.g. $(-1)^{4/7}=1$ because $(-1)^4=1^7$ and $(-1)^{3/7}=-1$ because $(-1)^3=(-1)^7$.
Irrational exponents do not allow this "trick".
