Re: Another stupid Cantor's "proof"

From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 09/25/04 

Date: 25 Sep 2004 16:04:35 -0700

poespam-trap@yahoo.com (Randy Poe) wrote in message news:<df76407e.0409250518.7aaddfad@posting.google.com>...
> Keckman <keckman@welho.com> wrote in message news:<opsevcj7mi3uk9lu@cs81133.pp.htv.fi>...
> > On Fri, 24 Sep 2004 21:32:54 -0500, Ryan Reich <ryanr@uchicago.edu> wrote:
> >
> > > Of course, you'll never see a function which
> > > can't be defined by a formula because if you know what it is you have a
> > > formula....
> > >
> >
> > This is a function; f(n) = n.
>
> This is a function defined by a formula. It tells you what f(n)
> is for every n.
>
> > prime number
>
> Huh? Where does primeness come into it?

OK, after looking at your response, I see what you intended.
You meant f(n) = n-th prime number.

> > f(1)=1
> > f(2)=2
> > f(3)=3
> > f(4)=5

1 is not ordinarily defined to be a prime number, but that's
a small point.

> > But you don't have a an algebraic formula to tell me what is
> > f(99^999^99999999)

Right. We don't know what the 99^999^99999999-th prime number is.
You are correct.

Nevertheless, f(n) is uniquely defined for every natural number
n. We know that there is a value f(n) for every n, no matter how
large a value of n we choose (including yours). Euclid proved
that millenia ago.

So what point did you want to make about this f? It has already
been stated that not every mapping has an algebraic rule defining
it.

               - Randy