On Sun, 17 Jul 2011, quasi wrote: > On Sun, 17 Jul 2011 11:40:25 -0500, quasi <firstname.lastname@example.org> wrote: >>>> <email@example.com> wrote: >>>> >>>>> In Dauben's biography of Abraham Robinson (page 103 of the >>>>> first printing) a quasi-field 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 quasi-field, the law x*0 = 0 holds iff the >> distributive law holds. >> >> To show that a quasi-field need not satisfy the distributive >> law, it suffices to produce an appropriate model, thus >> providing an example of a quasi-field which is not a field. >> >> But even in a quasi-field, 0*0 = 0. > > Actually, is the above necessarily true? > If so, how is it proved? > All I get (see below) is (n-1)(0*0) = 0
>> Thus, if a quasi-field is not a field, there must exist >> nonzero elements a,b of the quasi-field 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); (n-1)(a0) = 0
Thus quasi distributivity (QD). a(b + c) = a(b + c + (n-2)0) = ab + ac + (n-2)(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