
Re: Quasifields
Posted:
Jul 18, 2011 1:18 AM


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. >> >> To show that a quasifield need not satisfy the distributive >> law, it suffices to produce an appropriate model, thus >> providing an example of a quasifield which is not a field. >> >> But even in a quasifield, 0*0 = 0. > > Actually, is the above necessarily true? > If so, how is it proved? > All I get (see below) is (n1)(0*0) = 0
>> Thus, if a quasifield is not a field, there must exist >> nonzero elements a,b of the quasifield such that a*0 = b. >> Hmmm ... >> Does that mean b/a = 0? > Yes, and in addition a*0/b = 1.
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? 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.
Defining 2 = 1 + 1, does 2a = a + a in a quasi field? Assume it does. Then
a0 + a0 = 2a0 = a2*0 = a(0 + 0) = a0 + a0  a0 = a0
Thusly a0 = 0 and we've a field.

