Date: Sep 17, 1999 5:55 PM
Author: Dale Henderson
Subject: Re: -1 x -1 ?

In article <937516347.13527.0.nnrp-14.c2debf68@news.demon.co.uk>,

"Guillermo Phillips" <Guillermo.Phillips@marsman.demon.co.uk> wrote:

> Hello All,

>

> Here's something I've always wondered (perhaps in my naivety). Why

> should -1 x -1 = 1?

> I appreciate that lots of nice things come from this, but what's the

> fundamental reason for it?

>

> Guillermo.

>

>

I've seen a few proofs on this thread that make unfounded assumtions.

One assumes the uniqness of inverses and another assumes 0x=0.

Ironically the result (-a)b=-ab is requred to prove 0x=0. I've added

proofs of a few propositions below. The one your interested in is

Proposition 4. I think Propostions 2 and 3 should be combined to read

(-a)b=a(-b)=-ab. I've tried to assume only the axioms of a Ring if

anyone sees an unfoundes assumptions please tell me. I also threw in

the proof that 0x=0 for the fun of it.

Prop1 additive inverses are unique

Proof:

Let b,c be additive inverses of a.

a+b=0

a+c=0

a+b=a+c

b+a+b=b+a+c

0+b=0+c

b=c

Prop 2 a(-b)=-ab

Proof:

a(-b)+ab+ab = a (-b + b + b)

=a (0+b)

=ab

So,

a(-b)+ab+ab=ab

a(-b)+ab=0

a(-b)=-ab

Prop 3 (-a)b=-ab

Proof:

(-a)b+ab+ab=(a+a + -a)(b)

=(a+0)b

=ab

So,

(-a)b+ab+ab=ab

(-a)b+ab=0

(-a)b=-ab

Prop 4 (-a)(-b)=ab

Proof:

let c=-b

(-a)(-b)=(-a)c

=-ac (Prop 2)

=-(a(-b))

=-(-ab) (Prop 3)

= ab

Prop 5 a0=0

Proof:

0=(b-b)

a0=a(b-b)

=ab+(-a)b (Distributive law)

=ab+(-ab) (Prop 2)

=0

--

Dale Henderson <mailto://dhenders@cpsgroup.com>

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