Irrational Exponents In a maths textbook I have all of the rules defined on exponentiation e.g.
$$
a^ma^n = a^{m+n} 
$$
and
$$
(ab)^n = a^nb^n
$$
defined only for $a,b \in\Bbb R$ and for $m,n \in \Bbb Q$.
Why can't $m,n \in \Bbb Q?$
Also stated:
$0^{-n}$ is not defined for any positive integer $n$. Why just integers? Is it defined for all other real numbers? Why?
 A: Oops.   I completely misread your question.
$(ab)^n = a^nb^n$ is acceptable for $a,b \in \mathbb R$ but only for $n \in \mathbb Z$ because $1 = 1^{\frac 12} = (-1*-1)^{\frac 12} \ne (-1)^{\frac 1n}(-1)^{\frac 12} = [(-1)^{\frac 12}]^2 = -1$.
$0^{-n} = \frac 1{0};$ if $n > 0$ so that is not defined.  But if $x \in \mathbb R; x > 0$ then $0^x = 0$ is defined.
However $0^0$ is not defined.  There is a conflict as to whether $0^x = 0$ so $0^0 = 0$ of $x^0 = 1$ so $0^0 = 1$.  
==== oops; ignore the rest ===
The intuitive and basic definition of $a^n$ is "$a$ multiplied by itself $n$ times" or more formally, recursive by $a^0 = 1; a^{n+1} = a^n*a$.
So it simply hasn't been defined for $n \not \in \mathbb N$. 
....yet. 
But we can extend the definition to $n \in \mathbb Z$ and $n \in \mathbb Q$ and $n\in \mathbb R$ and even $n \in \mathbb C$ if we want to.  We can define them any way we want.  I could say $5^{-7} = \pi$ and $2.8^{\frac 43} = $ "a pink elephant named boo-boo that likes bubblegum" if we wanted to.  We just don't have any reason to define it those ways and if we did, it wouldn't make any sense.
BUT.  If $n\in \mathbb N$ there are two things that become immediately clear.  $a^{n+m} = (a*a*a......*)$[n times]$ * (a*a*a.....)$[m times]$ = a*a*a*......*a$ [n+m times].
So if we could define $a^{-7}$, we should have it that $a^{-7}* a^{7} = a^{-7 + 7}  = a^{0} = 1$.  That would mean $a^{-7} = \frac 1 {a^7}$.  So if we define $a^n$ for $n \in \mathbb Z$ it makes sense to extend the definition as:
$a^0 = 1;$  For $n \in \mathbb N$, $a^{n+1} = a^n*n$.  For $k \in \mathbb Z; k < 0$ then $a^k = a^{-|k|} = \frac 1{a^{|k|}}$.
We also have $(a^n)^k; k, n \in \mathbb N$, we have $(a^n)^k = a^n *a^n * ... *a^n = a^{n+ n+n+n....} = a^{kn}$.  So it makes sense that if we could define $a^{\frac 1n}$ it should make sense that $(a^{\frac 1n})^n = a^{\frac 1n*n} = a^1 = a$.  So that means $a^{\frac 1n}$ should equal $\sqrt[n]{a}$ it would make sense that we can extend the definition so:
$a^{\frac pq} = \frac {q}{a}^p$ for $q = \frac pq \in \mathbb Q; p, q \in \mathbb Z; q > 0; \gcd(p,q) = 1$.
[Caveat: if $q$ is even than $a$ can not be negative. ]
So we have extended the basic definition from only being defined for $n \in \mathbb N$ to $n \in \mathbb Q$.
And we have extended it in a way that: i) makes sense and ii) is consistent.
Now, I'm  probably getting ahead of your book but there are two equivalent ways to define $a^x; a \ge 0; x \in \mathbb R$, that are consistent and make sense.
One is: if $x = \lim\limits_{n\rightarrow \infty} q_n$ where $\{q_n\}$ are a sequence of rational numbers that get infinitely close $x$ then $a^x = \lim\limits_{n\rightarrow \infty} a^{q_n}$.
Ex.  If $\sqrt{2} = 1.4142135623730950488016887242097.....$ then $a^{\sqrt{2}} \approx a^{1.4142135623730950488016887242097}$ taken to "as many decimal places as we like".
The other is: define $\ln (x) = \int\limits_{1}^x \frac 1z dz$; $e = \lim\limits_{n\rightarrow \infty; n\in \mathbb N}(1 + \frac 1n)^n$ and define $exp (x) = \lim\limits_{n\rightarrow \infty; n\in \mathbb N}(1 + \frac xn)^n$.   Then $a^x = exp(x*\ln a)$.
I won't get into how to define  $e^z; z \in \mathbb C$ in this post but it can be done...  Oh, okay, I will .... but I won't explain it:  It is $a^{x + iy} = a^x(\cos (y\ln a)+ i\sin(y\ln a))$.
