
Re: Quasifields
Posted:
Jul 18, 2011 7:36 AM


On Mon, 18 Jul 2011, David Hartley wrote: > <marsh@rdrop.remove.com> writes >> On Sun, 17 Jul 2011, quasi wrote: >>> On Sun, 17 Jul 2011 11:40:25 0500, quasi <quasi@null.set> wrote: >>>>>> <freddywilliams@btinternet.com> wrote: >>>>>> >>>>>>> In Dauben's biography of Abraham Robinson (page 103 of the >>>>>>> first printing) a quasifield is said to be a commutative >>>>>>> field with distribution >>>>>>> >>>>>>> a(b + c) = ab + ac (1) >>>>>>> >>>>>>> replaced by >>>>>>> a(b_1 + b_2 + ... + b_n) = ab_1 + ab_2 + ... + ab_n. (2) >> >>>> Thus, in a quasifield, the law x*0 = 0 holds iff the >>>> distributive law holds. >>>> >> Define as usual for n in N, na = sum(j=1,n) a. >> >> a0 = a * n0 = n(a0); (n1)(a0) = 0 >> >> Thus quasi distributivity (QD). >> a(b + c) = a(b + c + (n2)0) = ab + ac + (n2)(a0) = ab + ac  a0 >> >> Does (1)a = a hold in a quasi field? > > It requires proof. It does hold for n=3: > > a*b = a*(b + b + b) > = a*b + a*b + a*(b) > > so a*b + a*(b) = 0, a*(b) = (a*b) > In general for n = 2k + 1:
a0 = a(kb + k(b) + 0) = k(ab) + k(a(b) + a0
k(a(b)) = k(ab)
> But that's no help for n even.
For n = 2k: a0 = a(kb + k(b)) = k(ab) + k(a(b))
k(a(b)) = a0  k(ab)
Maybe a0 is an infinitesimal.
>> Assume it does. >> a0 = a(1  1) = a1 + a(1)  a0 = a  a  a0 >> a0 + a0 = 0 >> a0 = a * n0 = n(a0) >> >> If n is even, then a0 = 0 and we've a field.

