Impossibility of ordering the complex numbers 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 $i < 0$?
 A: If $\mathbb{C}$ is considered as an additive Abelian group, or even as a $2$-dimensional real vector space, then it can be totally ordered, in a way that is compatible with the operations of addition and multiplication by real scalars. Quoting from Ordered vector space - Wikipedia, the free encyclopedia:

$\mathbb{R}^2$ is an ordered vector space with the $\leq$ relation defined in any of the following ways $\ldots$
Lexicographical order: $(a, b) \leq (c, d)$ if and only if $a < c$ or ($a = c$ and $b \leq d$). This is a total order. The positive cone is given by $x > 0$ or ($x = 0$ and $y \geq 0$) $\ldots$

Therefore, in order to prove the non-existence of a compatible total ordering of $\mathbb{C}$, you will have to adopt at least one postulate concerning its multiplicative structure.
A: Axioms 6, 7 and 8 are not sufficient for excluding the possibility to order the complex numbers.
If you define $a+bi\prec c+di$ ($a,b,c,d\in\mathbb{R}$) when either $a<c$ or $a=c$ and $b<d$, you get an order relation satisfying those axioms.
The contradiction will show up only if you add another axiom:

if $x<y$ and $0<z$, then $xz<yz$

First step: proving that $0<1$.
There are two cases: $1<0$ or $0<1$. Suppose $1<0$; then $1-1<0-1$, so $0<-1$. Hence $0(-1)<(-1)(-1)$, that is $0<1$: a contradiction.
Second step: proving that $-1<0$
Since $0<1$, we have $0-1<1-1$.
Third step: getting a contradiction
Suppose $0<i$; then $0i<i^2$, that is, $0<-1$, a contradiction.
Suppose $i<0$; then $i-i<0-i$ and $0<-i$; then $0(-i)<(-i)^2$, that is, $0<-1$, a contradiction.
Conclusion
Axiom 6 cannot hold for $x=0$ and $y=i$.
A: Hint: prove (it is not given, you must prove it) that for $x \ne 0$ then $x^2 > 0$.
Then you have $i^2 > 0$ and $1^2 > 0$.
As to why $1 < 0$ isn't an immediate contradiction?  Why should it be?  Were you ever given an axiom that $1 > 0$? You were not. (but you can prove it.)
A: Re: your first question, is "$-1<0$" explicitly one of your axioms? If not, you have to prove it, and "$-1>0$" isn't immediately a contradiction.
Re: your second question, the case where we assume $i<0$ is similar. You can prove (and probably have done so already as previous exercises) that your axioms imply that $-a>0$ whenever $a<0$, and that $(-a)(-a)=a^2$. So we have that $-i>0$ and $(-i)^2=-1$, so we can run the proof above with $-i$ in place of $i$. 
(Also, no need for all caps.)
A: Why would it be impossible to create an ordering for complex numbers? As you know, a complex number can be described as a point in the complex plane, and as a result, it can be written in polar coordinates (a1*cos(x1),a1*sin(x1)). Let's now say that two complex numbers A1=(a1*cos(x1),a1*sin(x1)) and A2=(a2*cos(x2),a2*sin(x2)) need to be ordered, then we can state following ordering rule:
A1 < A2 if:
  a1 < a2, or:
  a1 = a2 and x1 < x2

What would be wrong with that?
A: You may have forgotten writing the theorem, if for any two numbers $a$ and $b$, $a>b$ then $a\cdot c> b\cdot c$ for all $c>0$.
I think you have used this theorem to start with the inequality assuming $i> 0$ and since you have assumed $i$ to be positive applying the above stated theorem, we claim $i\cdot i >0\cdot i \Rightarrow i^2 >0 \Rightarrow -1>0$ and again since we have (-1) as positive (pretend for a moment you don't know it's nature) we again perform $-1\cdot -1 >0\cdot -1 \Rightarrow 1>0$ and then you face the contradiction.
