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Vadim Tropashko wrote:
> 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
No, we don't. You seem to assume here that {k*<a,b>} is equal to {<k*a,k*b>} and that is obviously nonsense. The bag {<1,2>,<1,2>} is not equal to the bag {<2,4>}.
So the right conclusion would be:
k1*k2 = 1
a11 = 1 a12 = 0 a11 = 0 a12 = 0
Which indeed contains a contradiction. But that the equation does not have a solution, even if we assume bags, is quite obvious. In the beginning you may already assume that
x = { k1*<1,0>, k2*<0,0>, ... }
So then
x*x = { k1*k1*<1,0,1,0>, k1*k2*<1,0,0,0>, k2*k1*<0,0,1,0>, k2*k2*<0,0,0,0>, ... }
If you want this to be equal to { 1*<1,0,0,0> } then you need to find two numbers k1 and k2 such that k1*k1 = 0, k1*k2 = 1, k2*k1 = 0, k2*k2 = 0. That is obviously going to be a bit of a problem. :-)
So your conclusion that this specific equation still doesn't have a solution is correct. But maybe you should first ask yourself the question why <1,0,0,0> should correspond with 1, and why not, for example, <1,1,1,1>.
-- Jan HiddersReceived on Wed Jan 24 2001 - 06:46:37 CST