# Re: Can relvars be dissymetrically decomposed? (vadim and x insight demanded on that subject)

From: erk <eric.kaun_at_gmail.com>
Date: 17 Jul 2006 11:28:35 -0700

Cimode wrote:
> [snip]
> > > [snip]
> > > For instance, if you consider the function F(X) = A(X) and you add
> > > the value B to F(X), you are basically doing a translation making a new
> > > function T(X) = F(X) + B = A(X) + B In this case, the result of adding
> > > B to the function can be expressed (characterized) as a mathematical
> > > translation (jumping from F(X) to T(X)). It says a lot about the
> > > function F(X) behavior and may help describe it better over time.
> >
> > But F isn't changing. You're creating a new function, not changing an
> > existing one. Assuming the expression F(X)+B makes sense, T has the
> that would mean that once an INSERT is done, there would be necessarily
> a new relation resulting from the INSERT operation performed?

> > > The
> > > challenge of this thread is to do a similar effort on relations. For
> > > instance, an interesting question raised would be: is an INSERT
> > > operation equivalent to a translation in relational perspective?
> I rephrase then : is an INSERT operation equivalent to a translation in
> relational perspective resulting in a new relation?

Any relational expression produces a relation value. If that's what you mean by the "new relation," then yes. If it's an UPDATE/INSERT/DELETE, that value is then assigned to the relvar, so the relvar has a new value - maybe that's what you mean.

> > > Overall, how does a relation R1 *changes* when an INSERT, UPDATE,
> > > DELETE operation is made on it? Characterizing such change would help
> > > describe better relations themselves.
> Are operations in general performed on relations create new relations.?

Expressions (relational or otherwise) evaluate to values, which in the case of relational expressions are relations. "Create new" doesn't have much meaning in that context.

• erk
Received on Mon Jul 17 2006 - 20:28:35 CEST

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