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.

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.)