Spivak's proof of $a\cdot 0 = 0$

Question

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 $a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0)$?

Answer 1

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. $$ 0 = a\cdot 0.$$

Answer 2

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 $0=a\cdot 0$.

Answer 3

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.

Answer 4

$$a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$$ by distributivity.

Then we have that $$2(a\cdot 0) = a\cdot 0$$ which implies that $$a\cdot 0 =0$$

Answer 5

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 $a \cdot 0 = 0$.