roma at ro-che.info
Thu Jan 1 15:06:19 UTC 2015
You misunderstood. Complex numbers can be ordered in a way that is
compatible with the equality — the lexicographic ordering.
(That said, I don't understand why this discussion is relevant at all.
The fact that the ordering exists doesn't mean that one would want to
declare the Ord instance, like with complex numbers.)
On 01/01/15 16:52, Atze van der Ploeg wrote:
> If we do not require that (a <= b) && (a >= b) ==> a == b (where <= is
> from the total ordering and == is from the equality relation) then it is
> trivial, take
> the total ordering forall x y. x <= y that i mentioned earlier.
> So the compatiblity with equality (you say field structure) is not
> besides the point, in fact
> antisymmetry means that the ordering corresponds to the equality relation.
> Clear now or did I misunderstand?
> 2015-01-01 15:39 GMT+01:00 Tom Ellis
> <tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk
> <mailto:tom-lists-haskell-cafe-2013 at jaguarpaw.co.uk>>:
> On Thu, Jan 01, 2015 at 03:37:09PM +0100, Atze van der Ploeg wrote:
> > This boils down to the question whether on each set with an equality
> > relation defined on it a total ordering (consistent with the equality
> > relation) can also be defined.
> I agree with the essence of this restatement.
> > One counterexample is the complex numbers.
> This is what I don't understand. The complex numbers can be totally
> ordered. (Not in a way compatible with the field structure, but that's
> beside the point).
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