Date: Nov 29, 2012 10:18 AM
Author: ross.finlayson@gmail.com
Subject: Re: Cantor's first proof in DETAILS
On Nov 28, 4:58 pm, Marshall <marshall.spi...@gmail.com> wrote:

> On Monday, November 26, 2012 11:33:02 PM UTC-8, Virgil wrote:

>

> > I find a citation from r 9/22/99 In which Ross states, what may well be

> > Ross' original "definition" of his alleged "Equivalency Function" which

> > as any mathematician can plainly see is not a function at all, and is

> > only equivalent to nonsense::

>

> > " Consider the function

> > f(x, d)= x/d

> > for x and d in N. The domain of x is N from zero to d and the domain of

> > d is N as d goes to

> > infinity, d being greater than or equal to one.

> > I term this the Equivalency Function, and note it EF(x,d), also EF(x),

> > assuming d goes to

> > infinity."

>

> >http://groups.google.com/group/sci.math/msg/af29323d694cf89e1999 -

> > "Equivalency Function"

>

> Um, so EF is a restriction of division?

>

> The domain of x depends on the value of d. I don't recall having seen

> that sort of thing before, but I guess I do know what that means.

> But I can't figure out what the domain of d is. It sorta looks like the

> domain of d depends on what d is, but what the heck would that mean?

>

> And it's just a name, but what about EF has anything to do with

> equivalency?

>

> Marshall

Mr. Spight, it's about the equivalency or equipollency or equipotency

of infinite sets.

EF(n) = n/d, d->oo, n->d.

Properties include:

EF(0) = 0

EF(d) = 1

EF(n) < EF(n+1)

The domain of the function is of those natural integers 0 <= n <= d.

It's very simple this. Then, not a real function, it's standardly

modeled by real functions:

EF(n,d) = n/d, d E N, n->d

with each having those same properties.

Then, the co-image is R[0,1] as is the range.

Regards,

Ross Finlayson