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Topic: Quasi-fields
Replies: 38   Last Post: Jul 24, 2011 7:21 PM

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 William Elliot Posts: 1,948 Registered: 5/30/08
Re: Quasi-fields
Posted: Jul 18, 2011 7:36 AM
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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 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.
>>>>

>> 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?

>
> 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.

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