Re: The naive test for equality

"Paul" <paul_at_test.com> wrote in message news:42f50fc1$0$24006$ed2619ec_at_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.
Received on Mon Aug 08 2005 - 05:10:46 CEST