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Re: The naive test for equality

From: Paul <paul_at_test.com>
Date: Sat, 06 Aug 2005 20:30:06 +0100
Message-ID: <42f50fc1$0$24006$ed2619ec@ptn-nntp-reader01.plus.net>


Marshall Spight wrote:
>>I've kind of lost track of what started this thread in the first place
>>now! I think it was just to say I didn't think there was any real
>>difference between equality and equivalence relations. Each one defines
>>the other.

> 
> Equality is a particular type of equivalence relation. It is the kind
> where every value is its own equivalence class. Put another way,
> in equality, the equivalence classes all have cardinality 1.

That's not how I interpret it. 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.

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

Paul. Received on Sat Aug 06 2005 - 14:30:06 CDT

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