Re: x*x-1=0

From: Vadim Tropashko <vadimtro_at_yahoo.com>
Date: Tue, 23 Jan 2001 19:23:22 GMT
Message-ID: <94klms$bov$1@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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http://www.deja.com/ Received on Tue Jan 23 2001 - 13:23:22 CST