Let $\alpha$ be a Dedekind cut and define $\alpha^* :=\{x\in\mathbb{Q}|\exists r>0\space \text{such that} -x-r\notin\alpha\}$. I need to show that $\alpha^*$ is a Deddekind Cut and the additive inverse of $\alpha$. Note the the additive identity is defined as $0^*:=\{x\in\mathbb{Q}|x<0\}$. Any help is greatly appreciated and needed. Thanks in advance.
For the $1$st part I can show that $\alpha^*$ is downwardly closed as if $y\in\mathbb{Q}$ and $x\in\alpha^*$ with $y<x$ then for some $r>0$, $-x-r>a$, $\forall a\in \alpha$, but as $-y>-x$, we have $-y-r>-x-r>a$$\forall a \in\alpha $ i.e. $-y-r\notin \alpha$ so $y\in\alpha^*$. But I am stuck showing that $\alpha^*$ has no top element and that $\alpha^*\neq\emptyset$ and $\alpha^*\neq\mathbb{Q}$, so any help please.
For the second part I can show that $\alpha+\alpha^*\subset 0^*$ as if $w\in\alpha+\alpha^*$ then $w=u+v$ where $u\in\alpha$& $v\in\alpha^*$. Therefore for some (rational) $s>0$, $-v-s>u$, that is $u+v<-s<0$. But I am stuck with the reverse inclusion. So please any help will be greatly appreciated and please let me know if what I've done is correct and on the right track. Thanks