[Haskell-cafe] Re: Num instances for 2-dimensional types
Daniel Fischer
daniel.is.fischer at web.de
Wed Oct 7 21:55:08 EDT 2009
Am Donnerstag 08 Oktober 2009 03:05:13 schrieb Felipe Lessa:
> On Wed, Oct 07, 2009 at 08:44:27PM -0400, Jason McCarty wrote:
> > Daniel Fischer wrote:
> > > Am Mittwoch 07 Oktober 2009 23:51:54 schrieb Joe Fredette:
> > > > I generally find semirings defined as a ring
> > > > structure without additive inverse and with 0-annihilation (which one
> > > > has to assume in the case of SRs, I included it in my previous
> > > > definition because I wasn't sure if I could prove it via the axioms,
> > > > I think it's possible, but I don't recall the proof).
> > >
> > > 0*x = (0+0)*x = 0*x + 0*x ==> 0*x = 0
> >
> > This proof only works if your additive monoid is cancellative, which
> > need not be true in a semiring. The natural numbers extended with
> > infinity is one example (if you don't take 0*x = 0 as an axiom, I think
> > there are two possibilities for 0*∞).
It was a proof for a ring (with or without unit), which Joe stated above he didn't recall.
There your additive monoid is cancellative since it's a group :D
>
> Given that
>
> x = 1*x = (0+1)*x = 0*x + 1*x = 0*x + x
>
> we can show that
>
> x = x + 0*x (right)
> x = 0*x + x (left)
>
> so, by definition of 'zero', we have that 0*x is a zero.
Not necessarily, we don't know 0*x + y = y for arbitrary y yet.
If the additive monoid isn't cancellative, that needn't be the case.
In Jason's example, you can indeed set 0*∞ = ∞.
> But we can easily prove that there can be only one zero: suppose we have
> two zeros z1 and z2; it follows that
>
> z1 = z1 + z2 = z2
>
> So 0*x = 0. Any flaws?
>
> --
> Felipe.
More information about the Haskell-Cafe
mailing list