I'm trying to prove that $\sqrt[n]{\frac{s}{t}}$ is irrational unless both s and t are perfect nth powers. I have found plenty of proofs for nth root of an integer but cannot find anything for rationals. Also trying to work up from the proofs I have found is rather difficult.
My lecturer wants me to prove this using uniqueness of prime factorisation but I also have no idea where to start with that.
Any help is greatly appreciated.
Best Answer
This is more of a comment than an answer, because it doesn't use the uniqueness of prime factorisation. Instead, it uses the more basic result that if a positive integer $c$ divides the product of two positive integers $a$ and $b$, and $b$ and $c$ are relatively prime, i.e. they have no common factors other than $\pm1$, then $c$ divides $a$.
Suppose that the fraction $\tfrac{s}{t}$ is in its lowest terms, i.e. $s$ and $t$ are relatively prime; and suppose that $\left(\tfrac{u}{v}\right)^n = \tfrac{s}{t}$, where $\tfrac{u}{v}$ also is in its lowest terms, i.e. $u$ and $v$ are relatively prime. Then: \begin{equation} \tag{$*$}\label{eq:simple} tu^n = sv^n. \end{equation}
The hypothesis that $u$ and $v$ are relatively prime implies that $u^n$ and $v^n$ are also relatively prime:
Proof. Suppose that an integer $r > 1$ divides both $u^n$ and $v^n$. Every integer $> 1$ has a prime factor. (This fact is admittedly an ingredient of the proof of uniqueness of prime factorisation, but still, it is a simpler theorem.) Let $p$ be a prime factor of $r$. By the hypothesis that $u$ and $v$ are relatively prime, $p$ cannot divide both $u$ and $v$. But if $p$ does not divide $u$, then by repeated application of the theorem quoted above, $p$ does not divide $u^n$. Similarly, if $p$ does not divide $v$, then $p$ does not divide $v^n$. Therefore, $p$ does not divide both $u^n$ and $v^n$. This contradicts the hypothesis that $p$ divides $r$. Therefore, no such $p$ and no such $r$ can exist. $\square$
Now, four applications of our favourite theorem to \eqref{eq:simple} show that $s$ divides $u^n$, $u^n$ divides $s$, $t$ divides $v^n$, and $v^n$ divides $t$. Therefore, $s = u^n$ and $t = v^n$. $\square$
(Perhaps in an effort to be simple, this has ended up being too laborious? Oh, well.)