Proving that the $k$-th power of no rational equals a natural number that is not a perfect $k$-th power It is easy to show that if a natural number is not a perfect $k$-th power, then there is no rational $q$ with $n=q^k$.
However, in my book this is a corollary to the following proposition: if $\gcd(a,b)=1$ and $ab=c^k$ where $a,b$ are natural numbers, then $a=d_1^k$ and $b=d_2^k$ for some natural numbers $d_1,d_2$.
Using the above-mentioned proposition to prove the former, I could not check that the coprimality hypothesis holds. Is there something missing?
 A: This is the proof:
Suppose that there is a rational $a/b$ where $\gcd(a,b)=1$ with $n=(a/b)^k$. Then $n.b^k=a^k$. It follows that $n$ divides $a^k$ having no nontrivial common divisors with $b^k$, since otherwise $\gcd(a,b)$ would not be $1$.
From the above-mentioned proposition, it follows that $n$ is a perfect $k$-th power, contrary to the assumption.
A: Suppose there is a non-natural rational $q$ such that $n = q^k$ for some some natural numbers $n$ and $k > 1$. We write $q$ as the ratio of two coprimes $a$ and $b > 1$. Among all possible values of $q$, we choose one for which $b$ is the least possible denominator. Writing $n = a^k/b^k$, we see that $n$ divides $a^k$, so that we may write $b^{2k}$ as the product of two coprimes, $a^k/n$ and $b^k$. From the proposition we conclude that there is a natural number $c$ such that $b^k = c^{2k}$. We now have $nc^k = (a/c)^k$, and since $b > 1$, we also have $b > c > 1$. This contradicts the fact that $b$ is the least possible denominator.
Edit: To reach the contradiction we need $a$ and $c$ to be coprimes. This follows from $b^k = c^{2k}$, then $b = c^2$, and $a$ and $b$ being coprimes.
Also, it is worth noting that the proposition being proved is almost equivalent to a third proposition, stating that $\text{gcd}(a, b) = 1$ implies $\text{gcd}(a^k, b^k) = 1$. This is because if $a^k/b^k$ is a natural number $n$ and its numerator and denominator are coprimes, then $b^k = 1$, making $n = a^k$.
