Completing a proof Say we are given this:
Impossibility of ordering the complex numbers. 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 $—1 > 0$ yields $1 > 0$. 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. 
So we are given this "passage" and the question is 
By reading the passage: Suppose that $<$ is a relation on $\mathbb{C}$ that satisfies Axioms 6, 7, and 8.  and show that the assumption $i <0$ leads to a contradiction.
Axiom 6 - Exactly one of the relations $x = y$, $x<y$ and $x>y$ holds. Note $x>y$ means the same thing as $y<x$
Axiom 7 - If $x<y$, then for every $z$ we have $x+z <y+z$.
Axiom 8 - If $x>0$ and $y>0$, then $xy>0$.
So this is what I have but I am really confused from that passage. Didn't we already complete the proof because we ended up with a contradiction? Help on this one please.

The passage proves only for the case when $i>0$ was considered, and the $i<0$ doesn't readily follow from this, though it is not hard neither:
By axiom 7., and if $i<0$, we have $0=i+(-i)<0+(-i)=-i$, that is, $-i>0$. But then the passage can be applied again, as $(-i)^2=i^2=-1$.
