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