Prove that there exists no such total order Prove that there does not exist a total order $\leq$ on $\mathbb{C}$ such that 
(i) for all $x,y,z \in \mathbb{C}$, if $x\leq y,$ then $x+z\leq y + z$;
(ii) for all $x, y \in \mathbb{C}$, if $x \geq 0$ and $y \geq 0$ then $xy \geq 0$. 
edited proof:
By the definition of a total order $\leq$ on a set $X$, $\forall x,y\in X$, we have that $x\geq y$ or $x \leq y$. Assume there exists a total order $\leq$ on $\mathbb{C}$ that satisfies the above conditions. Consider the complex number $i$. We have by the definition of a total order that $i\leq 0$ or $i\geq 0$. Similarly, $-i\in \mathbb{C}$ and $-i \geq 0$ or $-i \leq 0$. Suppose that $i\geq 0$. Then by statement (ii), we have that $i\cdot i\geq 0$. By the definition of multiplication with complex numbers, this can be simplified as $i\cdot i = (0+1i)(0+1i)=(0\cdot 0-(1)(1))+(0\cdot 1+0\cdot1)i = -1$. Thus, we have that $-1 \geq 0$. By statement (ii), then, we have that $1=(-1)(-1)\geq 0$. Since $1\geq 0$, then $0\leq 1$. Adding $-1$ to both sides gives $-1\leq 1+(-1)=0$ by the definition of addition with complex numbers, which is a contradiction because $-1\geq 0$ and $-1 \neq 0$. 
So we have that $i\leq 0$. By the definition of addition in complex numbers and statement (i), adding $-i$ to both sides gives $0 \leq -i$, so we have that $-i \geq 0$. Then by statement (ii), $(-i)(-i) = -1\geq 0$. By statement (ii), then, we have that $(-1)(-1)=1\geq 0$. By statement (i), since $0\leq -1$ and $1 \in \mathbb{C}$, we have that $0+1\leq (-1)+1 = 0 \Rightarrow 1\leq 0$, which is a contradiction as $1\geq 0$ and $1 \neq 0$. Thus there exists no such total order since two elements $i,0\in \mathbb{C}$ will never be comparable.
Some questions:
In (ii), $x$ and $y$ do not have to be distinct, right?
Also, is my edited proof valid? Any advice would be appreciated!
 A: Hint/sketch: Suppose there is some such total order.  Then show that $z^2\geq 0$ for any complex $z.$  From this, show $-1, 1 \geq 0$ and show why this gives a contradiction. 
A: Well, the first question I'd ask is, is $i > 0$ or is $i < 0$? (we know $i\ne 0$.)
If $i>0$ then $i^2 = -1 > 0$.   That's fine but weird.  Then $(-1)^2 = 1 > 0$.  And then $0 < -1 + 1 < 0$ that's not good.
So $i < 0$, I'd like to do that if $a < 0$ and $b <0$ then $ab > 0$ to get a contradiction that $i^2 = -1 > 0$, but I'm too lazy to prove $a< 0$ and $b<0$ then $ab> 0$.  I mean, maybe if it were a proposition or an exercise that I had to prove....[1]
But if $i < 0$ then $0< i+(-i)< -i$.  And so $(-i)^2 = -1 > 0$.   And $(-1)^2=1 > 0$ so $0 < 1 + (-1) =0$ 
So we can't have $i > 0$ nor $i< 0$ nor $i=0$.  We're doomed.
.....
[1] I'm being tongue in cheek.   Of course $a<0;b<0$ implies $ab > 0$ is a proposition we will have to prove.  And it isn't hard.  $a <0\implies 0 = a+(-a) < 0 + (-a) = -a$ and so $-a> 0$ and $-b > 0$ and so $(-a)(-b) = ab > 0$.
That's the classic way of doing this: i) prove the proposition that $a > 0 \iff -a < 0$ (by adding $-a$ to both sides) and ii) prove that $x^2\ge 0$ (by noting that either $x\ge 0$ so $x^2 \ge 0$ or $x < 0$ so $-x > 0$ and $(-x)^2 = x^2 > 0$) and iii) proving $1 > 0$ and $-1 < 0$. as $1 \ne 0$ and $1 = 1^2 \ge 0$ so $1 > 0$ and $-1 < 0$.  
And once that's done $i^2 =-1 < 0$ and that violates ii) 
