TITLE: Spivak's proof of $a\cdot 0 = 0$
QUESTION [5 upvotes]: I picked up Spivak's Calculus (3rd Edition) today and it seemed like a good idea to go through the section Basic Properties of Numbers. In this chapter, Spivak proves that 
$$a \cdot 0 = 0$$
The proof looks simple:
$$a \cdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$$
My question is just as simple: How do I get from $a \centerdot 0 = 0$ to $a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0)$?

REPLY [11 votes]: I am reading pp 7 of Spivak's Calculus (3rd Edition) right now.
What he's trying to do is to prove: $$ a\cdot 0 = 0$$ using $$  a\cdot(b+c) = a\cdot b + a\cdot c \tag{P9}$$ 
Let $b,c = 0$. We have: $$ a\cdot 0 = a\cdot (0+0) = a\cdot 0 + a\cdot 0. $$ By adding $(-a\cdot 0)$ to both sides of $ a\cdot 0 = a\cdot 0 + a\cdot 0 $ we get $$ a\cdot 0 + (-a\cdot 0) = a\cdot 0 + a\cdot 0 + (-a\cdot 0) $$ i.e. $$ 0 =  a\cdot 0.$$

REPLY [2 votes]: $$a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$$ by distributivity.
Then we have that $$2(a\cdot 0) = a\cdot 0$$
which implies that $$a\cdot 0 =0$$