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} $.


1 Answer 1


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 $q<r\in\Bbb Q$, 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.


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