# Please explain Spivak's proof of $a\cdot 0 = 0$ to me.

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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I think you asked your question backwards. Spivak's proof involves two equalities. The first follows by the property $a b + a c = a(b+c)$. The second follows since by definition of $0$, $0+x = x$. –  Sam Lisi Mar 6 '12 at 16:30
You're both right. I'm a proof noob. :) Cheers! –  jamesbrewr Mar 6 '12 at 16:42
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. –  Michael Hardy Mar 6 '12 at 16:53

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! –  jamesbrewr Mar 6 '12 at 16:43

$$a \centerdot 0 + a \centerdot 0 = a \centerdot (0+0) = a \centerdot 0$$ 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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So, essentially, what Spivak is doing is starting from an expression whose value is 0 (the expression $a \cdot 0 + a \cdot 0$) and proving that it's equal to $a \cdot 0$? –  jamesbrewr Mar 6 '12 at 16:38
@James Brewer: Spivak is starting from an expression which happens to evaluate to $0$, and proving that it does. I would have preferred if he had written the calculation in this order: $a\cdot 0=a\cdot(0+0)=a\cdot 0+a\cdot 0$ and therefore $\dots$. –  André Nicolas Mar 6 '12 at 16:46
@James: Spivak is starting from an expression, $a\cdot 0 + a\cdot 0$, which he doesn't know yet evaluates to $0$ (he hasn't proven that $a\cdot 0$ is $0$ yet), and proving that it is equal to $a\cdot 0$. From this, he concludes that $a\cdot 0$ is $0$; and from that you can conclude that $a\cdot 0+a\cdot 0$ evaluates to $0$. But we don't know that $a\cdot 0+ a\cdot 0$ evauates to $0$ until we prove that $a\cdot 0$ does. –  Arturo Magidin Mar 6 '12 at 16:48
The room just started to spin .. I think what I missed was that this proof is based off of a property previously listed in the text. See @J.D.'s (accepted) answer. –  jamesbrewr Mar 6 '12 at 16:50

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 $a \cdot 0 = 0$.

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