Proof that a root of an integer is an integer or irrational

Proof

We argue by contradiction, with a prime divisor of the denominator. Let $x$ be real, $n \ge 1$ an integer and $m$ an integer with $x^n = m$. Suppose that $x$ is not an integer.

1. Suppose that $x$ is not irrational, that is, $x$ is rational.
2. By the existence of a fraction in lowest terms, $x = a / b$ with integers $a$ and $b$, $b > 0$ and $\gcd(a, b) = 1$.
3. If $b = 1$, then $x = a$ is an integer, which is false. So $b \ge 2$.
4. By the existence of a prime divisor, there is a prime number $p$ with $p \mid b$.
5. From $x^n = m$ we get $a^n / b^n = m$, so $a^n = m b^n$.
6. Since $n \ge 1$, $b$ divides $b^n$, and so $p$ divides $m b^n = a^n$.
7. Claim: if $p \mid a^k$ for an integer $k \ge 1$, then $p \mid a$. We use induction on $k$. For $k = 1$ there is nothing to prove.
8. If $p \mid a^{k+1} = a^k \cdot a$, then Euclid's lemma gives $p \mid a^k$ or $p \mid a$. In the first case, the hypothesis of the induction gives $p \mid a$.
9. By steps 6 to 8, $p \mid a$. Together with step 4, $p$ is a common divisor of $a$ and $b$.
10. By the definition of the greatest common divisor, $\gcd(a, b) \ge p \ge 2$. This contradicts $\gcd(a, b) = 1$.
11. So $x$ is not rational. Since $x$ is real, $x$ is irrational.