In article <IY_c7.168$Iw2.firstname.lastname@example.org>, "Duane Jones" <email@example.com> wrote:
> "Virgil" <firstname.lastname@example.org> wrote in message > news://vmhjr2-08FC87.email@example.com... > > In article <firstname.lastname@example.org>, > > "Carl W." <email@example.com> wrote: > > > > > Virgil <firstname.lastname@example.org> wrote in message > > > news://vmhjr2-E3F3EE.email@example.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.