
Re: 0! = 1
Posted:
Aug 11, 2001 12:14 AM


"Virgil" <vmhjr2@home.com> wrote in message news://vmhjr2A0ECCE.21141910082001@news1.denver1.co.home.com... > In article <IY_c7.168$Iw2.8744@petpeeve.ziplink.net>, > "Duane Jones" <gauss@ziplink.net> wrote: > > > "Virgil" <vmhjr2@home.com> wrote in message > > news://vmhjr208FC87.16133710082001@news1.denver1.co.home.com... > > > In article <3b73c07e.0@katana.legend.co.uk>, > > > "Carl W." <noone@dev.null> wrote: > > > > > > > Virgil <vmhjr2@home.com> wrote in message > > > > news://vmhjr2E3F3EE.23043509082001@news1.denver1.co.home.com... > > > > > > > > > (2) if n! = n*(n1)!, and n = 1, what is (n1)!? > > > > > > > > This is a slightly dodgy argument in that we could say the same for n = > > 0. > > > > > > > > i.e. if n! = n((n1)!), and n = 0, what is (n1)!? > > > > > > > > > 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 = (01)! ? > > > > > > 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*(n1)!, then (01)! be found from 0! > similarly. > > I was refuting that thesis.
I don't recall Carl stating that. Reread his post. You parsed his argument prematurely leaving only (2). It was from that I was arguing.
Cheers, Duane

