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Re: Quasi-fields
Posted:
Jul 18, 2011 1:18 AM
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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 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
a0 + a0 = 2a0 = a2*0 = a(0 + 0) = a0 + a0 - a0 = a0
Thusly a0 = 0 and we've a field.
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