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Topic: -1 x -1 ?
Replies: 29   Last Post: May 13, 2009 9:01 AM

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 Dale Henderson Posts: 1 Registered: 12/12/04
Re: -1 x -1 ?
Posted: Sep 17, 1999 5:55 PM
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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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Date Subject Author
9/16/99 Guillermo Phillips
9/16/99 Ian A. Mason
9/16/99 Peter Percival
9/16/99 Dave Seaman
9/17/99 Steve Leibel
9/17/99 Dale Henderson
9/17/99 Chuck Cadman
9/18/99 Peter Percival
9/18/99 Bill Taylor
9/18/99 John Savard
9/19/99 carel
9/20/99 Jon Haugsand
9/20/99 HH
9/20/99 Jon Haugsand
9/20/99 Jonathan Hoyle
9/20/99 Jim Hunter
1/21/09 Tay
1/21/09 mensanator
1/21/09 lwalke3@lausd.net
1/22/09 Henry
1/22/09 David R Tribble
1/22/09 Dave L. Renfro
1/22/09 mensanator
5/12/09 Kayama
5/12/09 mensanator
5/12/09 hagman
5/12/09 Bill Dubuque
5/12/09 Bill Taylor
5/13/09 Cobra
5/13/09 Cobra

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