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.
Yes, the proof is correct. But why do you represent complex numbers as pairs of real numbers? There's no need for that.