How can one prove that the cube root of 9 is irrational? Of course, if you plug the cube root of 9 into a calculator, you get an endless stream of digits. However, how does one prove this on paper?

By the rational root test, if
$$P(x) = a_nx^n+\cdots +a_0$$
is a polynomial with integer coefficients, and $\frac{u}{v}$ is a rational number with $\gcd(u,v)=1$ such that $P(\frac{u}{v})=0$, then $u$ divides $a_0$ and $v$ divides $a_1$. 
(This holds in any GCD-domain, even if there is no unique factorization into primes)
(If you don't understand the last parenthetical comment, don't worry; it's a nod towards those who know some more abstract algebra)
Look at $x^3-9$. If $\frac{u}{v}$ is a root, then 
$$\begin{align*}
\frac{u^3}{v^3}-3&=0\\
\frac{u^3}{v^3} &=9\\
u^3&=9v^3\\
v^3 &\text{divides }u^3\\
v &\text{divides }u^3\\
v &\text{divides }1 &&\text{(since }\gcd(u,v)=1\text{)}
\end{align*}$$
hence any rational root must be an integer. 
But if $a$ is a positive integer, $a\leq 2$ implies $a^3\leq 8$, and $a\geq 3$ implies $a^3\geq 27$. So there are no integers with cube $9$. 
More generally, if you look at $x^n-b$ with $b$ an integer, it has rational roots if and only if it has integer roots, if and only if $b$ is a perfect $n$th power. 
Even more generally, rational roots of monic polynomials with integer coefficients ("monic" means "leading coefficient equal to $1$) are necessarily integers (and must divide the constant term). 
