Prove that the field of complex numbers cannot be ordered.
An ordered field F is a field having a subset P satisfying the following:
1) For all x,y in P, x+y in P
2) For all x,y in P, x*y in P
3) For all x in F, exactly one of the statements
(x in P, x = 0, -x in P) is true.
At the outset, if i= 0, then i^2= i.i = 0, which is a contradiction
as i^2 = -1.
Thus, i!= 0.
Now, for any given ordered field F and for all x(- F, we observe that if x! = 0, then either x(- P or, -x (- P, giving: x^2 (- P, where P corresponds to the positive half of F.
Since i^2 = -1 is negative, it follows that i^2 cannot belong to P.
This violates the trichotomy axiom.
Consequently, it follows that the field of complex numbers cannot be ordered.
Edited on May 9, 2008, 4:41 pm