Let $n$ and $k$ be positive integers, and suppose $x^k = n$ for some $x \in \mathbb{Q}$. Show that $x \in \mathbb{Z}$. In other words, $\sqrt[k]{n}$ is either an integer or is irrational.

Suppose $x \in \mathbb{Q}$,

$$\begin{aligned} x &= \frac{s}{t} \text{ such that } gcd(s,t) = 1 \text{ for some } s \in \mathbb{Z} \text{ and } t \in \mathbb{N} \\ n &= x^k = (\frac{s}{t})^k \\ s^k &= n \cdot t^k \\ \end{aligned}$$

which means $n \mid s^k$.

For each prime $p$, we may define the function $v_p$, mapping non-zero integers to non-negative integers, as follows: for every integer $n \ne 0$, if $n = p^em$, where $p \nmid m$, then $ν_p(n) := e$. We may then write the factorization of $n$ into primes as

$$\begin{aligned} n &= \pm \prod_{p}p^{v_p(n)} \text{ where the product is over all primes }p \end{aligned}$$

From $s^k = n \cdot t^k$,

$$\begin{aligned} \pm \prod_{p}p^{k \cdot v_p(s)} = \pm \prod_{p}p^{v_p(n)} \prod_{p}p^{k \cdot v_p(t)} \\ \end{aligned}$$

Observe that $n \mid s^k \iff v_p(n) \le v_p(s^k) \equiv k \cdot v_p(s)$ for all primes $p$. Thus,

$$\begin{aligned} \prod_{p}p^{k \cdot v_p(s) - v_p(n)} &= \prod_{p}p^{k \cdot v_p(t)} \\ \end{aligned}$$

which shows that if $k \cdot v_p(s) - v_p(n) > 0$ for some prime $p$, $p$ must be a factor of both $s$ and $t$. But this contradicts the fact that $gcd(s, t) = 1$. So the supposition $x \in \mathbb{Q}$ must be false, or $x$ (ie $\sqrt[k]{n}$) must be irrational in such case. On the other hand, if $k \cdot v_p(s) - v_p(n) = 0$ for all prime $p$, $t$ would necessarily be equal to 1, which means $x$ (ie $\sqrt[k]{n}$) must be an integer in such case.

$\Box$

# posted by rot13(Unafba Pune) @ 4:24 PM