Need clarification in proof (P9) from Chap 1 Spivak

[ActionPotential]

I am taking Calculus at the end of the month for the first time but I have started working through Spivak (wanted to get a headstart and didn't have the class textbook yet) and I have been grasping his beautiful description of some the properties of real numbers yet there is a very minor detail in P9 in Chapter 1 in which a step is missing (or as I realized as I thought about it, I was missing something). I just need clarification to make sure I am understanding what happened.

(P9) If a, b, and c are any numbers, thena\cdot(b+c)=a\cdotb+a\cdotc

If a-b=b-a
then (a-b)+b=(b-a)+b=b+(b-a)
hence a=b+b-a
hence a+a=(b+b-a)+a=b+b
Consequently a\cdot(1+1)= b\cdot(1+1)
and therefore a=b

Now does a+a=b+b become a(1+1)=b(1+1) because of p9 a\cdot(b+c)=a\cdotb+a\cdotc?

I know this should be incredibly obvious and once I understand it I will kick myself but I couldn't move on without knowing exactly.

 

[micromass]

Now does a+a=b+b become a(1+1)=b(1+1) because of p9 a\cdot(b+c)=a\cdotb+a\cdotc?

Yes. And you also need $a\cdot 1 = a$.

 

[ActionPotential]

Yes. And you also need $a\cdot 1 = a$.

Of course!

a+a

a\cdot 1 = a
a\cdot 1+a\cdot 1=a(1+1)

Merci beaucoup mademoiselle! I knew it was something minor but opened my eyes more :)