# Need clarification in proof (P9) from Chap 1 Spivak

1. Aug 11, 2013

### 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 text book 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, then

$a$$\cdot$$(b+c)=a$$\cdot$$b+a\cdot$c

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$$\cdot$$b+a\cdot$c?

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.

Last edited: Aug 11, 2013
2. Aug 11, 2013

### micromass

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

3. Aug 11, 2013

### ActionPotential

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 :)