# Confusion about the proof that if $x \in \mathbb{R}$, $x + (-x) = 0$

On page 19 of Rudin's Principles of Mathematical Analysis, the author constructs real numbers using Dedekind cuts. The proof that the sum of a real number $$x$$ and its negation $$(-x)$$ equals $$0$$ is somewhat clever and more involved. I don't exactly follow it?

Rudin defines the cut $$\beta = -\alpha$$ as:

$$\beta := \{p\in\mathbb{Q}:\exists r>0 \text{ such that }(-p-r)\notin \alpha\}$$

In other words, some rational number smaller than $$-p$$ fails to be in $$\alpha$$.

We show that $$\beta \in \mathbb{R}$$ and $$\alpha + \beta = 0^{*}$$.

If $$r \in \alpha$$ and $$s \in \beta$$, then $$-s \notin \alpha$$. Hence, $$r < -s$$, so $$r + s < 0$$. Thus, $$\alpha + \beta = 0^{*}$$.

Question. Why does $$s \in \beta \implies -s \notin a$$ mathemtically?

I mentally picture it as follows:

Suppose for the sake of contradiction $$-s\in\alpha$$ then for every $$r>0$$, $$-s-r< -s$$ and since $$\alpha$$ is closed downward we would have $$-s-r\in\alpha$$ for every $$r>0$$, contradicting the definition of $$\beta$$.