# Proof that Multiplicative inverse of a Dedekind cut is a Dedekind cut

Let $$\alpha$$ be a Dedekind cut. Then its inverse is constructed as $$\gamma= 0^* \cup \{0\} \cup \{q\in \mathbb{Q}\mid\text{ there exists }r\in \mathbb{Q}\text{ such that }r>q\text{ and }1/r \notin \alpha\}$$

I am looking for proof that $$\gamma$$ is a Dedekind cut. It is clear that $$\gamma \neq \emptyset$$ because $$0 \in \gamma$$.

However I struck at proving that $$\gamma \neq \mathbb{Q}$$.

That definition applies only to positive cuts $$\alpha$$; if $$0\notin\alpha$$, so that $$\alpha\le 0^*$$, then $$\gamma$$ is all of $$\Bbb Q$$.
If $$\alpha$$ is positive, however then there is a $$p\in\alpha$$ such that $$p>0$$. Let $$q=\frac1p$$; if $$q, then $$\frac1r<\frac1q$$, so $$\frac1r\in\alpha$$. That is, $$\frac1r\in\alpha$$ for every rational $$r>q$$, so $$q\notin\gamma$$, and therefore $$\gamma\ne\Bbb Q$$.
Here you’ll find a proof that $$\alpha\gamma=1^*$$, assuming that you’re working with the same definition of multiplication as the author of that question.