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NNTP-Posting-Date: Sun, 07 Aug 2005 22:10:46 -0500
From: "VC" <boston103@hotmail.com>
Newsgroups: comp.databases.theory
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Subject: Re: The naive test for equality
Date: Sun, 7 Aug 2005 23:10:46 -0400
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"Paul" <paul@test.com> wrote in message 
news:42f50fc1$0$24006$ed2619ec@ptn-nntp-reader01.plus.net...
>... The way I see it, an equivalence relation
> *defines* what we mean by equality with respect to a given structure.
>
> So for example you start with expressions of the form "x/y", with x and
> y integers (y!=0)
>
> Now to begin with, "1/2" != "2/4"
>
> But you create an equivalence relation as VC described, which is
> basically grouping certain integer pairs together to create a different
> structure. And you use this equivalence relation to *define* what you
> mean by "equality" on your new structure. So [1/2] = [2/4]. But
> conventionally you drop the square brackets indicating the equivalence
> class and write 1/2 = 2/4, which maybe confuses things though.

Right...

>
> So for the rational numbers, you have equality but the corresponding
> equivalence classes on ZxZ *don't* have cardinality 1

Not quite right.  In the case of rationals equality,  you treat the 
equivalence class,  as a whole, as a single element. E.g,  for integers 
you'd say 2=2;  for rationals you'd say [5/10] = [1/2],  no difference 
really since both [5/10] and [1/2] is the *same* element.  In other words, 
your *equality* relation pair would be, say,  for integers (1,1) and for 
rationals (E_half, E_half), where E_half = {1/2, 2/4,, 5/10, ..} etc.


>
> Paul. 


