I'm studying the proof for theorem 1.19 from Principles of Mathematical Analysis by Walter Rudin (public online copy here).
The lengthy proof for this theorem is contained in the appendix to chapter 1 on page 17.
I am working on Step 4:
If $\alpha \in R$ and $\beta \in R$ we define $\alpha + \beta$ to be the set of all sums $r + s$, where $r \in \alpha$ and $s \in \beta$.
We define $0^*$ to be the set of all negative rational numbers. It is clear that $0^*$ is a cut. We verify that the axioms for addition (see Definition 1.12) hold in $R$, with $0^*$ playing the role of $0$.
I am specifically working on step 4, (A5), on page 19:
Fix $\alpha \in R$. Let $\beta$ be the set of all $p$ with the following property:
There exists $r > 0$ such that $-p - r \not\in \alpha$.
In other words, some rational number smaller than $-p$ fails to be in $\alpha$.
We show that $\beta \in R$ and that $\alpha + \beta = 0^*$.
For this question, I am focusing on the $\beta \in R$ part.
I encountered the following proposition in Step 4, (A5), on page 19:
If $q \in \alpha$, then $-q \not\in \beta$.
It wasn't immediately clear to me how this was true, and Rudin didn't give an explanation, so I set out to prove it myself. Since we had the keyword "NOT" in the conclusion, I decided that it would be best to use proof by contradiction. The proof is as follows.
A (Hypothesis): $q \in \alpha$, where $\alpha \in R$.
B (Conclusion): $-q \not\in \beta$.
NOT(B): $-q \in \beta$.
I now work forward from the hypothesis and the negation of the conclusion.
NOT(B) 2: There exists $r > 0$ such that $-(-q) - r \not\in \alpha$
$\implies q - r \not\in \alpha$
NOT(B) 3: $q \in Q$ and $r \in Q$, which means that $q - r \in Q$.
A1: $q \in R$, where $R \subset Q$.
NOT(B) 4: $q - r \not\in \alpha \in R \subset Q$, where $q - r \in Q$. But this is a contradiction.
$Q.E.D.$
I would greatly appreciate it if people could please take the take to review my proof for correctness and comment as to whether I went about this correctly.