>>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.
>>
>I'll buy that, but...
>Ordered fields must obey a few certain properties. There are several reasons why C is not an ordered field, but basically the problem is that i is not defined in R (though i**2 is).

I can't follow you here. What's the connection between the posibility to order C and i?

When I say "order relation", it has nothing to do with the usual order in R. An order relation is a relation (on a set) that is reflexive, tranzitive and antisymetric. This is the definition. The usual order in R is just an example of order relation.

>One of the properties for an ordered field that results from order axioms is that for any number, say x, in a set, one of these statements is true: 'x = 0', 'x is > 0', '-x > 0'. Clearly, this is not true for i. Thus C is not an ordered field.

You can't use the order from R in C. According to my order on C, (0,0) <= (0,1). Which means 0 <= i. You can replace here (and in the definition I gave) "<=" with "M". Thus, "z1 M z2" means "z1 is in the relation M with z2", where M is my order relation. "<=" is just a notation, is not the same relation as the usual order in R.

>On another and perhaps easier track, it's my understanding that an isomorphism must maintain a 1-1 correspondence between sets. Intuitively, you can see that C and R are not 1-1. Proof left as exercise :~)

I never said that C and R are isomorphic. I said RxR is isomorphic with C. (An element in RxR is an order pair: (x,y).)

Vlad