5
$\begingroup$

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

  • 2
    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 $0$, $0+x = x$.2012-03-06
  • 0
    You're both right. I'm a proof noob. :) Cheers!2012-03-06
  • 2
    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

5 Answers 5

11

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

  • 0
    Since I'm new to these parts I can't vote yet, but I'll accept your answer. Thanks!2012-03-06
2

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

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

1

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 $0=a\cdot 0$.

  • 0
    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 $a \cdot 0$?2012-03-06
  • 0
    @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 $\dots$.2012-03-06
  • 0
    @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.2012-03-06
  • 0
    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.2012-03-06
1

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.

1

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 $a \cdot 0 = 0$.