# Spivak Example

1. Sep 30, 2013

### Bashyboy

Hello,

I am embarking to read Spivak's book on Calculus, and have come across some difficulty with something that is perhaps rather trivial. In the third edition, there is a section entitled Basic Properties of Numbers. Near the end of page 7, the author begins discussing how he will use property 9 to derive the fact that, when negative numbers are multiplied, the result is a positive number. Before this, he must show that (-a) x b = -(a x b) is true. To prove this, we must show that
(-a) x b + a x b = [(-a) + a] x b, which I can follow. Next, he says that, because this is true, then we can add -(a x b) to both sides of the equation:

$(-a) \cdot b + a \cdot b + [-(a \cdot b)] = [(-a) + a] \cdot b + [-(a \cdot b)]$

Clearly, the first term on the RHS of the equation will yield zero.

$(-a) \cdot b + a \cdot b + [-(a \cdot b)] = -(a \cdot b)$

$[-a + a - a] \cdot b = -(a \cdot b)$

$(-a) \cdot = -(a \cdot b)$

But are these steps truly valid; how I can know that, when factoring the b, the negative symbol isn't appended to the b; isn't this what I am trying to prove, that the negative symbol is appended to the a?

Last edited: Sep 30, 2013
2. Sep 30, 2013

### gopher_p

Are those the actual steps listed in the text? The next-to-last line looks a bit suspect. It seems like it should be

$(-a)\cdot b+a\cdot b=(-a+a)\cdot b$ by distributivity of multiplication over addition
$(-a)\cdot b+a\cdot b=0\cdot b$ by definition of additive inverses
$(-a)\cdot b+a\cdot b=0$ from results related to the definition of the additive identity and distributivity of multiplication over addition
$[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=0+(-(a\cdot b))$ because addition is well-defined as a binary operation (?)
$[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=-(a\cdot b)$ by definition of additive identity
$(-a)\cdot b+[a\cdot b+(-(a\cdot b))]=-(a\cdot b)$ by associativity of addition
$(-a)\cdot b+0=-(a\cdot b)$ by definition of additive inverse
$(-a)\cdot b=-(a\cdot b)$ by definition of additive identity

It's possible that some of those steps were left out, but that's what is going on as far as I can tell.

3. Sep 30, 2013

### Bashyboy

Oh, yes, I see now. Thank you very much.