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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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