# Question regarding step 7 in appendix of chapter 1 , baby rudin

I have proved step 1 to 6. However, I am stopped by step 7. Here's the quote to Baby Rudin

We complete the definition of multiplication by setting $$\alpha 0^*=0^*=0^*\alpha$$ and by setting

$$\alpha\beta= \begin{cases} (-\alpha)(-\beta),\text{ if }\alpha<0^*,\beta <0^*\\ -[(-\alpha)\beta],\text{ if }\alpha < 0^*,\beta > 0^*\\ -[\alpha(-\beta)],\text{ if }\alpha > 0^*,\beta < 0^* \end{cases}$$

My question is, can these be proved?

I think no. So for example, let's take the first case as a demonstration.

Addition axiom holds for positive and negative reals for now, which is basically the entire real number set. So it suffices to prove $$\alpha\beta=-[(-\alpha)\beta]$$ if $$-\alpha\beta+\alpha\beta=-\alpha\beta+-[(-\alpha)\beta]$$ and $$\alpha <0^*,\beta >0^*$$.

To me, I find this extremely easy if distributive axiom holds in the set of reals, because I can just factor out the common factor negative alpha and apply addition inverse as well as product includes 0 will be 0. However, both distributive axiom and the product includes 0 will be 0 warrants the distributive axiom, which is yet defined in the set of reals for now. In fact, it will be proved afterwards according to baby rudin.

For now, I can't prove these settings. Let say I take it for granted and prove multiplication axioms hold in the set of reals. But how about these settings? How can I prove them?

The above were my efforts in proving according to established lemmas in the book. the following are some links that I have researched. Since it is possible for me to have overlooked some post which may possibly answer my question, for those who have a duplicate post please attach after checking duplicates.

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

Step 6 Real construction from Rational Baby Rudin