Virgil
Posts:
1,119
Registered:
12/6/04
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Re: 0! = 1
Posted:
Aug 10, 2001 11:06 PM
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In article <IY_c7.168$Iw2.8744@petpeeve.ziplink.net>,
"Duane Jones" <gauss@ziplink.net> wrote:
> "Virgil" <vmhjr2@home.com> wrote in message
> news://vmhjr2-08FC87.16133710082001@news1.denver1.co.home.com...
> > In article <3b73c07e.0@katana.legend.co.uk>,
> > "Carl W." <no-one@dev.null> wrote:
> >
> > > Virgil <vmhjr2@home.com> wrote in message
> > > news://vmhjr2-E3F3EE.23043509082001@news1.denver1.co.home.com...
> > >
> > > > (2) if n! = n*(n-1)!, and n = 1, what is (n-1)!?
> > >
> > > This is a slightly dodgy argument in that we could say the same for n =
> 0.
> > >
> > > i.e. if n! = n((n-1)!), and n = 0, what is (n-1)!?
> >
> >
> > Not so. Anyone can solve 1 = 1*x for x, which defines x = 0! but how
> > do you solve 1 = 0*x for x, which is needed to define x = (0-1)! ?
>
>
> How so? In (2) above, 1! can only be defined after knowing 0!. You
> inadvertently assume 1! = 1 to show that 0! = 1.
>
> Cheers,
> Duane
>
>
>
You are not following the thread. Carl W. assumed 1! = 1 but that 0!
was naturally undefined.
Carl W. then said that if 0! could be found from 1! by downward use
of the relation n! = n*(n-1)!, then (0-1)! be found from 0!
similarly.
I was refuting that thesis.
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