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

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    He does _not_ go from $a\cdot0=0$ to $a\cdot0+a\cdot0=a\cdot(0+0)$. That's backwards. He goes from $a\cdot0+a\cdot0=a\cdot(0+0)$ to something else, and from that something else eventually to $a\cdot0=0$. You don't go from what you're trying to prove to what you already know. You go in the other direction.2012-03-06

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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.$

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    Since I'm new to these parts I can't vote yet, but I'll accept your answer. Thanks!2012-03-06
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$a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$ by distributivity.

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

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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 $0=a\cdot 0$.

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    The room just st$a$rted to spin .. I think wh$a$t I missed was that this proof is based off of a property previously listed in the text. See @J.D.'s (accepted) answer.2012-03-06
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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.

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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 $a \cdot 0 = 0$.