Help With Basic Proof from Rudin PMA Chapter One From Walter Rudin's Principles of Mathematical Analysis, Third Edition, page 20, Step 8:





I want to complete the proof of (b) by proving the following claim.
Claim: For $r\in{}\mathbb{Q}^+$ and $s\in{}\mathbb{Q}^-$, $r^*\cdot{}s^*\subset{}(rs)^*$


Attempt at Proof: $q\in{}r^*\cdot{}s^*\implies{}q\in{}-[r^*\cdot{}(-s)^*]\implies{}\forall{}r'\forall{}s'(r',s'\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,r'\in{}r^*\,\,\,\wedge{}\,\,\,s'\in{}(-s)^*\implies{}\exists{}a(a\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,-q-a>r's'))$


At this point I am stuck. I need to show $rs>q$ so that $q\in{}(rs)^*$.
This post covers the question, but I think the proof is wrong at the relevant place and I'm stuck.

First prove for all rational $r,s$, we have:
(1) $\quad r^* < s^*$ if and only if $r < s$.
(2) $\quad (-r)^* = -(r^*)$
(3) $\quad r,s \ge 0$ implies $r^*s^* = (rs)^*$.
Then, using the definition of multiplication of reals (Rudin's Step 7), the other cases of (3) follow immediately, unless I'm missing something!
I believe the proofs of (1)–(3) are very straightforward, again unless I'm missing something.
(I'm not sure what the etiquette is on this site, but I feel like I should restrict my contribution to getting past the roadblock, not filling in every detail.  Anyway, in the meantime, I have verified that the proofs of (1)–(3) are straightforward.)
