In the Appendix to Chapter 1 of Rudin's Principles of Mathematical Analysis, the additive inverse of a cut $\alpha$ is shown to be
$$-\alpha = \{p\in \mathbb{Q} \mid \exists r \in \mathbb{Q}^{+} \text{ with } (-p-r) \in \alpha\}$$
when reading this section, I found that
$$-\alpha = \{x\in \mathbb{Q} \mid \forall y \in \alpha, x+y<0 \}$$
which seems to be not only valid but easier to prove that it is a cut and that it is the additive inverse of $\alpha$. Is there anything wrong with my reasoning?
Here are the details of the proof:
To show $-\alpha$ is a cut:
Since $\alpha \neq \emptyset$, $\exists z \in \alpha \Rightarrow (\forall x \in -\alpha) \space x<-z \Rightarrow -\alpha \neq \mathbb{Q} $ and if $m$ is an upper bound of $\alpha$, then $-m \in -\alpha$, so $\alpha \neq \emptyset$.
If $x \in -\alpha$, $w \in \mathbb{Q}$, and $w<x$, since for all $y \in\alpha$, $x+y<0$, for all $y \in\alpha$, $w+y<0$, so $w \in -\alpha$.
For each of the y in $\alpha$, $\exists y_{0} \in \alpha$ with $y_{0}>y$, and if $x \in -\alpha$, then $x+y_{0}<0 \space \forall y_{0} \in \alpha$, so $(x + (y_{0} - y)) + y < 0 \space \forall y \in \alpha$ with $(x + (y_{0} - y)) > x$. Therefore, if $x \in -\alpha$, then $x<x_{0}$ for some $x_{0} \in -\alpha$.
Then, directly from the definition, $\alpha -\alpha$ is the set of negative rational numbers.
Suppose that $\alpha=\{q\in\Bbb Q:q<-1\}$. Then $1+q<0$ for each $q\in\alpha$, so your definition would make $-\alpha=\{q\in\Bbb Q:q\le 1\}$, which is not a Dedekind cut: it has a largest element.