Spivak's proof of $a\cdot 0 = 0$ 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 $a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0)$?
 A: $$a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$$ by distributivity.
Then we have that $$2(a\cdot 0) = a\cdot 0$$
which implies that $$a\cdot 0 =0$$
A: 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. $$ 0 =  a\cdot 0.$$
A: Since $0=0+0$, then $a\cdot 0 = a\cdot (0+0)$.
Since $\cdot$ distributes over $+$, whenever we have $x\cdot (b+c)$, this is equal to $x\cdot b + x\cdot c$. Applying this to $a\cdot(0+0)$, we get
$a\cdot (0+0) = a\cdot 0 + a\cdot 0$.
So: 
$$\begin{align*}
a\cdot 0 + 0 &=a\cdot 0 &\text{(because }x+0=x\text{ for all }x\text{)}\\
&= a\cdot (0+0) &\text{(because }0=0+0\text{)}\\
&= a\cdot 0 + a\cdot 0 &\text{(because }\cdot\text{ distributes over }+\text{)}
\end{align*}$$
So we have $a\cdot 0 + 0 = a\cdot 0 + a\cdot 0$. Cancelling one $a\cdot 0$ (or adding $-(a\cdot 0)$ to both sides) we conclude that $0=a\cdot 0$.
A: That $a\cdot 0 + a\cdot 0 = a\cdot (0+0)$ is just using the distributive law. (From here you get $a\cdot (0+0) = a\cdot 0$ and then you just subtract $a\cdot 0$ on both sides.
A: You don't.  He's not assuming $a \cdot 0 = 0$.  He's proving that.  So, you don't get from that to $a \cdot 0 + a \cdot 0 = a \cdot (0 + 0)$.  How you do get that last statement is the distributive property.  He then uses the property that 0 plus anything is that anything.  In particular, $0 + 0 = 0$.  Once you have that, we have $a \cdot 0 + a \cdot 0 = a \cdot 0$.  Now, we use the fact that every number has an additive inverse, so we can add that to both sides, which is the same as subtracting $a \cdot 0$ from both sides.  This leaves us with $a \cdot 0 = 0$.
