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

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 $$r < s$$.

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