The product of a Dedekind cut and its inverse equals one

Exercise 1.11 of the textbook Real Mathematical Analysis by Pugh asks what the inverse of a cut $x=A|B$ is ($x$ being positive).

The first step is to show that $x^*=C|D$ where $C=\{1/b: b\in B \space\& \space b\neq \min B\}$ and $D=C^c$ is a cut. Suppose we already know this fact.

The last step is to verify that $x.x^*=1$. Here is my proof:

Let’s denote $x.x^*$ by $E|F$. Let $m\in E$. So $m=a.1/b$ for some $a,b\in A,C$ respectively. But $b\gt a$ so $m\lt 1$.

Conversely, let $m<1$. Choose some $b\in B$. There is a natural number $n$ for which $m^nb\in A$. Choose the smallest such $n$. If $m^{n-1}b= \min B$, choose a sufficiently large natural number $s$ and take $b^*=b+1/s$ instead. Now $m=\frac{m^nb}{m^{n-1}b}\in AC$, namely, $m\in E$.

Is my proof correct?