Re: Proposal: 6NF

From: vc <boston103_at_hotmail.com>
Date: 19 Oct 2006 10:42:44 -0700

JOG wrote:
> vc wrote:
> > Jan Hidders wrote:
> > > vc wrote:
> > > > Jan Hidders wrote:
> > > > [...]
> > > >
> > > > A much simpler example. Let {0, 1, 2, 3} be a set of four integers
> > > > with addition modulo 4. Then, none of its subsets, except {0} and
> > > > {0, 2}, retains the addition mod 4 operation which makes the idea of
> > > > 'subtype as subset' utterly silly, [....].
> > >
> > > You keep on making the same mistake. The expression a +[mod 4] b has a
> > > well defined result if a and b are from any subset of {0, 1, 2, 3}.
> >
> >
> > Consider the subset {2, 3}. What is the result of (2+3) mod 4 ? If
> > you say it's '1', what is '1'? There is no such element in {2, 3}.
> > [snip]

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> Hi vc. PMFJI, but does this argument not rely on the assumption of the
> set needing to satisfy the closure property in respect to the
> operation?
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> Mathematically, surely the modulo addition operation as described above
> above can be applied to the subset {2, 3}, while happily having a
> closure of {0,1,2} for instance?

It's not a 'closure'. The closure property, in this context, means that the result of a binary operation must belong to the same set the operands come from What are '0' and '1' ? They do not exist in {2, 3}, so {2, 3} is not closed under '+ mod 4'. Very much in the same fashion, the natural numbers are not closed under subtraction, or odd numbers under addition, etc.

> After all, any operation just maps one
> set of values to another - why the self-closure requirement?

>
> I stand ready for correction if I have made a misinterpretation.
Received on Thu Oct 19 2006 - 19:42:44 CEST

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