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Re: 1 x 1 ?
Posted:
Sep 17, 1999 5:55 PM


In article <937516347.13527.0.nnrp14.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=(bb) a0=a(bb) =ab+(a)b (Distributive law) =ab+(ab) (Prop 2) =0
 Dale Henderson <mailto://dhenders@cpsgroup.com>
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