I have to exposition about the impossibility of ordering the complex numbers:
Axioms $6$: Exactly one of the relations $x = y$, $x < y$, $x > y$ holds.
Axioms $7$: If $x < y$, then for every z we have x + z < y + z.
Axioms $8$: If $x > y$ and $y > z$, then $x > z$
As yet we have not defined a relation of the form $x < y$ if $x$ and $y$ are arbitrary complex numbers, for the reason that it is impossible to give a definition of $<$ for complex numbers which will have all the properties in Axioms $6$ through $8$. To illustrate, suppose we were able to define an order relation $<$ satisfying Axioms $6$, $7$, and $8$. Then, since $i \neq 0$, we must have either $i > 0$ or $i < 0$, by Axiom 6. Let us assume $i > 0$. Then taking, $x = y = i$ in Axiom $8$, we get $i^2 > 0$, or $-1 > 0$. Adding 1 to both sides (Axiom $7$), we get $0 > 1$. On the other hand, applying Axiom $8$ to $-1 > 0$ we find $1 > 0$. Thus we have both $0 > 1$ and $1 > 0$, which, by Axiom $6$, is impossible. Hence the assumption $i > 0$ leads us to a contradiction. [Why was the inequality $-1 > 0$ not already a contradiction?] A similar argument shows that we cannot have $i < 0$. Hence the complex numbers cannot be ordered in such a way that Axioms $6$, $7$, and $8$ will be satisfied.
But Why was the inequality $-1 > 0$ not already a contradiction? and it is true for $i < 0$?