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Forum:
Visual FoxPro
Catégorie:
Autre
Divers
Thread ID:
00017699
Message ID:
00018777
Vues:
62
>>>A general isomorphism cannot exist since R is an ordered field but C is not...
>>
>>Why isn't C ordered?
>>
>The order axioms aren't satisfied. For example, is i greater than -i?

We can set an order defined as follows:

z1 <= z2 iif ((y1 < y2) or ((y1 = y2) and (x1 <= x2)))

Where z=x+y*i (or zk=xk+yk*i). (k is an indice)

This relation is reflexive, tranzitive and antisymetric. So, C can be ordered.

Anyway, the fact that we don't have an order on a set doesn't mean it cannot be isomorphic with a set for which we have an order.

As an example, we can have an isomorphism between a circle and R. R is ordered, but we don't have (usual) an order on a circle.

Vlad
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