Proof that the nth root of a rational s/t is irrational unless s and t are perfect nth powers 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.
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 $r &gt; 1$ divides both $u^n$ and $v^n$. Every integer $&gt; 1$ 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.)
To build off Vasya and Calum's helpful comments:

Let $\frac{s}{t}$ be an irreducible fraction (to address Vasya's comment).

Continuing with the main proof, recall Vasya's comment:


  $\sqrt[n]{\frac{s}{t}} = \frac{\sqrt[n]{s}}{\sqrt[n]{t}}$


We will quickly prove this.

Obviously $(\sqrt[n]{\frac{s}{t}})^n = \frac{s}{t}.$

Notice $\frac{s}{t} = \frac{(\sqrt[n]{s})^n}{(\sqrt[n]{t})^n} = (\frac{\sqrt[n]{s}}{\sqrt[n]{t}})^n.$

It follows that $\sqrt[n]{\frac{s}{t}} = \frac{\sqrt[n]{s}}{\sqrt[n]{t}}$ since the $n$th root is either a bijection or always positive.

Without loss of generality, let $s$ have a prime factorization $p_1^{a_1}p_2^{a_2}...p_i^{a_i}.$ Let $\sqrt[n]{s}$ be an integer with prime factorization $q_1^{b_1}q_2^{b_2}...q_j^{b_j}.$

Clearly $p_1^{a_1}p_2^{a_2}...p_i^{a_i} = q_1^{nb_1}q_2^{nb_2}...q_j^{nb_j}.$
We know the prime factorization of $s$ is unique, so we know that each power of $p$ corresponds to some power of $q.$
It's not clear which $p$ corresponds to which $q,$ but it is clear that all the prime factors are $n$th powers.