Why is it that (negative number) ^ (irrational number) is nonreal?

This is because the most general definition of $x^\alpha$ for complex $x,\alpha$ is $x^\alpha:=\exp(\alpha\log x)$, where $\exp$ is the exponential function and $\log$ is the (principal) natural logarithm. Pictorially, this means that taking $x$ to the power of $\alpha$ means multiplying the argument of $x$ by $\alpha$; that is, rotating $x$ with respect to the origin by an angle of $\alpha$ times the seperation between $x$ and the positive real axis. So if $x$ is a negative real number and $\alpha$ is an irrational real number, then we must rotate $x$ by some irrational multiple of $\pi$ (i.e. $180^\circ$). Obviously, irrational numbers are never integers, so the result after the rotation never ends up on the real axis; that is, $x^\alpha$ is not real.

When we write $x^{\frac{1}{n}}$, this is just a notation. We mean to say that there should be a real number $y$ such that $y^n = x$. This definition is valid only when $n \in \mathbb{Z}$. But, when we want to do the same for irrationals, we do not have any definition. The usual intuition of exponents, which is multiplying the number itself as many times, will not work in case of irrationals. This is why we have to go into complex numbers where $w^z$ is defined for any complex numbers $w$ and $z$.

Every $z\in \mathbb{C}\setminus\{0\}$ can be written as $z=r\cdot e^{i\theta}$, with $r>0$ and $\theta\in\mathbb{R}$.

If $z$ is a negative real number, $r=-z$ and $\theta=(2n+1)\pi$, $n\in\mathbb{Z}$.

So, if $z^x=r^x\cdot e^{ix\theta}=(-z)^x\cdot e^{ix(2n+1)\pi}$ is real, then

$$(-z)^x\cdot e^{ix(2n+1)\pi} = r_0\cdot e^{ik\pi}$$ for some $r_0>0$ and $k\in\mathbb{Z}$. (Remember: $e^{ik\pi}=\pm1$, if $k\in\mathbb{Z}$)

So $ix(2n+1)\pi=ik\pi$ and then $x=\frac{k}{2n+1}\in\mathbb{Q}$.

In particular, if $x$ is irrational, so $z^x$ is not a real number