Prove there is no rational number whose cube is $13$

Question

I used the idea that if $13$ divides $n^3$ then $13$ divides $n$.

I begin with assuming that there is a rational number x such that $x^3$ = $13$. Since $x$ is rational, we rewrite as $m/n$ where gcd(m,n)=1.

$x^3$ = $13$

$m^3$/ $n^3$ = 13.

Then $m^3$ = 13 $n^3$.

Using the fact that if 13 divides $n^3$, then 13 divides $n$. What would the next step be?

Answer 1

A simpler proof (by contradiction):

Assume $\left( \frac{a}{b} \right)^3 = 13$, and thus $13 b^3 = a^3$.

Write $a$ as a product of its primes, and $b$ as a product of its primes. Then there are $3k + 1$ primes on the left-hand side of the equation (for some unknown $k$), and $3 m$ primes on the right hand side (for some unknown $m$). But there are no values for $k$ and $m$ for which $3k + 1 = 3m$.

By the fundament theorem of arithmetic (unique factorization into primes), there must be such an equality if the assumption were true. Therefore the original assumption is false.

QED

Answer 2

Then m is divisible by 13. Then let $m=13k$, we get $13^3k^3 =13n^3$. Therefore $n^3=13^2k^3$. Therefore n is divisible by 13. Therefore $\gcd(m,n)\ge 13$, causing a contradiction.