Re: x*x-1=0

Vadim Tropashko wrote:
> In article <94rr74$l4j$1_at_news.tue.nl>,
> hidders_at_win.tue.nl (Jan Hidders) wrote:
> > Vadim Tropashko wrote:
> > > In article <94mhpf$jdh$1_at_news.tue.nl>,
> > > hidders_at_win.tue.nl (Jan Hidders) wrote:
> > > > Note that if you have ordered tuples then the projection is also
> > > > likely to be slightly different. If you say PROJ[#1,#2](R) then
> > > > that will mean something else than PROJ[#2,#1](R).
> > >
> > > What are unordered tuples: sets or bags?
> >
> > Neither.
>
> Do tuples have *internal* structure in terms of set/bags theory?

Yes, named tuples are defined as functions that map a name to a value. And, as you probably know, functions can again be defined as a set of pairs, and a pair can also be defined in terms of sets.

> > > > But things get, from an algebraic perspective, a little more
> > > > complicated because the cartesian product does not commute as
> > > > the join does.

>

Yes, but note that he talks about the algebra defined on sets of *named* tuples (and, therefore, needs the rename operator) where I was talking about the algebra defined on sets of *ordered* tuples. The operations on these algebras are similar but have different algebraic properties.

> I agree that renaming could be viewed as algebraic operation. I wonder
> 1. if this is a productive definition

Well, for starters, it is definitely something that you need to be able to do, and it is not expressible with the other operators. So by that definition of "productive" it is not only productive but even inescapable.

> 2. how would I define relational algebra without renaming (renaming,
> then, goes into metamodel). The fact that in classic algebras we never
> consider renaming as part of algebra confuses me

Sorry, renaming has to be there. That it confuses you is not the algebra's problem. :-)

> 3. how algebra with renaming operation relates to the one without it

The expressive power becomes greater.

-- 
  Jan Hidders