Re: proof of (-1).a= -a

From: panoptes@xxxxxxxxxx (Daniel W. Johnson)
Date: Sun, 4 Sep 2005 01:12:11 -0500

Proginoskes <CCHeckman@xxxxxxxxx> wrote:

> José Carlos Santos wrote:
> > Noh wrote:
> >
> > > 0.a = 0.a + 0
> > > = 0.a + 0.-a
> > > = 0.(a+ -a)
> > > = 0.0
> > > = 0
> >
> > Therefore, when proving that one always has 0.a = 0, you've used the
> > fact that 0.(-a) = 0. Don't you see a problem here?
>
> I don't. -a is in the field if a is; that's one of the axioms.
> --- Christopher Heckman

I suppose you also wouldn't see a problem with using commutativity to
prove commutativity. Or using the Axiom of Choice to prove the Axiom of
Choice.
--
Daniel W. Johnson
panoptes@xxxxxxxxxx
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
.

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- Re: proof of (-1).a= -a
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- Re: proof of (-1).a= -a
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