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?

Suppose that $9^{1/3}=m/n$ with $m$, $n$ integers with ${\rm GCD}(m,n)=1$.
This may be assumed because if $d$ is an integer divisor of both $m$ and $n$, then $m/n=(m/d)/(n/d)$. 
Then
$$
9n^3=m^3
$$
so that $3$ divides $m^3$, hence $m$ since $3$ is prime. Thus $3^3=27$ divides both sides of the equality so that $3$ divides $n^3$, hence $n$ for the same reason as above. This contradicts the assumption that $m$ and $n$ are coprime.
This arguent generalizes immediately to showing that the $n$-th root of an integer which is not an $n$-th power of an integer is not rational.
