Question regarding step 8 in appendix of chapter 1 , baby rudin

Question

This question refers to the construction of R from Q using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.

Specifically, I cannot proof (b) in step 8, bottom of pp.20. To be more precise, I'm not able to show that ${(rs)^*\subset r^*s^*}$ when $r>0$ and $s>0$, could somebody prove it for me?

Here are the original texts in Baby Rudin.

Step 8 We associate with each $r\in Q$ the set $r^*$ which consists of all $p\in Q$ such that $p < r$. It is clear that each $r^*$ is a cut; that is, $r^* \in R$. These cuts satisfy the following relations :

(a) $ r^ * + s^* = (r + s)^*$,

(b) $r^*s^* = (rs)^*$,

(c) $r^* < s^*$ if and only if $r < s$.

Answer 1

You want to prove that if $q\in\Bbb Q$, if $q>0$, and if $q<rs$, then $q=q_1q_2$, with $q_1,q_2>0$, $q_1,q_2\in\Bbb Q$, $q_1<r$ and $q_2<s$. Take a rational number $q_1\in\left(\frac qs,r\right)$; this makes sense, since $\frac qs<r$. Now, take $q_2=\frac q{q_1}$. Then $q_2<s$, since\begin{align}q_2<s&\iff\frac q{q_1}<s\\&\iff q_1>\frac qs.\end{align}And, of course, $q=q_1q_2$.