I'm hoping to have my proof reviewed. Thanks in advance.
Problem:
Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: -1 is a square.
Solution
In the complex field we know that (0,0) = 0 and (1,0) = 1, and (0,1)$^2$ = $i^2$ = (-1,0).
Suppose there is some ordering on the complex field. Then either (-1,0) < (0,0) or (-1,0) > (0,0) or (-1,0) = (0,0).
Previous result: for any x $\ne$ 0 in an ordered field that x$^2$ > 0 and that 1 > 0 since 1$^2$ = 1.
Case 1: (-1,0) < (0,0)
Then $i^2$ = (-1,0), but the square of any non zero number should be greater than 0. A contradiction.
Case 2: (-1,0) > (0,0)
Then the square of i satisfies the condition that squares are greater than 0, but we also have the ordered field property that:
If x > 0 then -x < 0, and vice versa.
So, since (-1,0) > (0,0) then (1,0) < (0,0). But as we pointed out above, 1 > 0.
Case 3: (-1,0) = (0,0). This cannot be the case since (-1,0) does not act as an identity element under addition in the complex field.
