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


Daniel W. Johnson wrote:
> 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.
>
> 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.

Mea culpa. I wasn't paying attention; I was assuming that 0.a = 0 had
been proven for all a, from which it follows that 0.(-a) = 0.

My brain's still on vacation.

--- Christopher Heckman

.

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