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Non ordered field (Posted on 2007-06-06) Difficulty: 2 of 5
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.

See The Solution Submitted by Bractals    
Rating: 3.0000 (1 votes)

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Solution Solution Comment 2 of 2 |

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
  Posted by K Sengupta on 2007-06-07 04:57:56

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