Re: x*x-1=0
Date: Tue, 23 Jan 2001 19:23:22 GMT
Message-ID: <94klms$bov$1_at_nnrp1.deja.com>
In article <94gv2l$c38$1_at_news.tue.nl>, hidders_at_win.tue.nl (Jan Hidders) wrote:
>
> And that also tells how a join can be defined on bags because it will
> be similar to the cartesian product on bags. So if a tuple occurs n
> times in a table and is joined with another tuple in another table
that
> appears there m times, then the joined tuple will appear n*m times in
> the result. And that gives an idea of what will happen if you also
> assume that your equation has a solution: you will get "complex bags"
> that may contain a tuple i times where i is the complex number such
> that i^2 = -1.
Let's solve some equations:
x*x = { 1*<1,0,0,0> }
Assume
x = { k1*<a11,a12> , k2*<a21, a22> }
(Here my decision to guess a solution with 2 tuples only is somewhat arbitrary.) If we try to match
k1*k2*<a11,a12,a11,a12> = 1*<1,0,0,0>
then we have
k1*k2*a11 = 0 &
k1*k2*a11 = 0
- contradiction. On the other hand if we match
k1*k2*<a11,a12,a21,a22> = 1*<1,0,0,0>
then
k1*k2*a11 = 1 k1*k2*a12 = 0 k1*k2*a21 = 0 k1*k2*a22 = 0
has no solutions as well.
For equation x*x=a, where 'a' has tuples that are of not square arity I don't even know how to approach. I assume that they don't have solutions even on the domain of complex number weighted tuple bags. Am I multiplying bugs correctly?
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Received on Tue Jan 23 2001 - 20:23:22 CET