```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?>> MarshallMr. Spight, it's about the equivalency or equipollency or equipotencyof 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 standardlymodeled by real functions:	EF(n,d) = n/d, d E N, n->dwith each having those same properties.Then, the co-image is R[0,1] as is the range.Regards,Ross Finlayson
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