Show that it is impossible to order C (complex numbers) so that it becomes an ordered field.

Question

Hint: Assume (C, +, *, <) is an ordered field for some order relation <. Then consider i is not equal to 0. One has either i>0 or i<0. In either case, find a contradiction.

Answer 1

Hint: what happens when you multiply both sides of the inequality by $i$?

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Clarification:

To begin, suppose that $i>0$. Then since $i>0$, we can multiply both sides of the inequality by $i$ to get $i^2 >0$ (the inequality doesn't flip because $i>0$). Why is this a contradiction?

Now, what happens when we assume $i<0$?

Answer 2

Assume to the contrary that one can define a total ordering $>$ on $\mathbb{C}$. It is obvious that $i \neq 0$. So we need check if $i>0$ or $0>i$ are possible.

Case 1: Assume that $i>0$. By definition and our assumption, we have that

$i\cdot i =-1> i \cdot 0 = 0$

which is impossible.

Case 2: Assume that $0>i$. Then by definition and our assumption,

$0+(-i)=-i>i+(-i)=0$

But then we notice that

$0\cdot (-i)=0>i\cdot(-i)=1$

which is impossible.