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?
 A: Assume the cube root of $9$ is rational, then it can be written as $p/q$ for integers $p$ and $q$ such that $p$ and $q$ share no common divisor. 
$$\begin{align*}(p/q)^3 &= 9\\
&\implies (p^3)/(q^3) = 9\\
&\implies p^3 = 9 * (q^3)
\end{align*}$$
Now, $p$ must be a multiple of $3$, else $p^3$ wouldn't be a multiple of 3 (or 9).
Let $r$ be the integer such that $(3r)^3 = p^3$
So, 
$$\begin{align*}
(3r)^3 &= 9(q^3)\\
&\implies 3 * 3 * 3 * r^3 = 9(q^3)\\
&\implies q^3 = 3 r^3
\end{align*}$$
hence $q$ is also a multiple of $3$.
If $p$ and $q$ are both multiples of $3$, then $p/q$ is not the simplest way to express the cubed root of $9$ and so we have a contradiction.
Therefore, the cubed root of $9$ must be irrational
A: This is essentially the same proof I gave in my answer here.
Suppose $9^{\frac{1}{3}}$ is rational. Then $3^2n^3 = m^3$ for some natural numbers $n$ and $m$. On left side of the equation, the power of $3$ is of the form $3k + 2$ and on the right side it is of the form $3l$. This is a contradiction, because each integer greater than one has a unique prime factorization by the fundamental theorem of arithmetic. Thus $9^{\frac{1}{3}}$ is not rational.
This same proof also works for a more general case. Let $p$ be prime and $n \geq 2$ an integer. Then $\sqrt[n]{p^k}$ is irrational when $n$ does not divide $k$. Just like before, assuming that $\sqrt[n]{p^k}$ is rational leads to a situation where we have a number with two different prime factorizations. One factorization has $p$ of power divisible by $n$, while the other has $p$ of power not divisible by $n$.
A: 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.
A: 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). 
A: I typically use Bézout’s theorem http://mathworld.wolfram.com/BezoutsTheorem.html. And here is an example of how to use it http://rtybase.blogspot.com/2011/06/regarding-square-root-of-n.html which can be easily extrapolated to a more general case.
