Cube root of n is irrational or natural

Question

I am trying to show that $\sqrt[3] n, n \in \mathbb{N}$ is either natural or irrational. I start off by assuming the thesis, that is $\sqrt[3] n = {p \over q} \in \mathbb{Q} \setminus \mathbb{N}$ where $q$ doesn't divide $p$ without a remainder. I rewrite $n = {p^3 \over q^3}$ and conclude that for this to be a natural number, $q$ has to divide $p$ without a remainder, so ${p \over q} \in \mathbb{N}$ which contradicts the assumption.

My question is, how exactly do we know that ${p^3 \over q^3} \in \mathbb{N}$ implies ${p \over q} \in \mathbb{N}$?

Answer 1

A formal answer of your question could be as follows,$(p,q)=1$:

$${p^3 \over q^3}=a, (a \in \mathbb{N})→{p^3 \over q}=aq^2→p^2({p \over q})=aq^2→{p \over q} \in \mathbb{N}$$

This because (in general) if $a({m \over n})=b$ and $(a,n)=1$, then ${m \over n} \in \mathbb{N}$.

(For more detail about the later statement, please see: Suppose $a,b,m,n$ are natural numbers, prove that if $a({m \over n})=b$ and $(a,n)=1$, then ${m \over n} \in \mathbb{N}$)

Answer 2

If $p/q$ is a root of $x^3 - n = 0$, where $p,q$ have no common divisors other than $\pm 1$ then $q$ divides 1 (the leading coefficient) and hence $p/q \in \mathbb{N}$.