Proving that there is no total ordering of $\mathbb{C}$ such that it is an ordered field Posting for proof verification and tips!
Ordered field axioms used: 
1. $\forall_{a,b\in\mathbb{F}}(0\leq a \land 0\leq b \implies 0\leq ab)$
2. $\forall_{a,b\in\mathbb{F}}(a\leq b \implies a+c\leq b+c)$

Theorem 1: Let $\mathbb{F}$ be a field. If $x\in\mathbb{F}$, then  $\forall_{x\neq0}(0<x^2)$.
Proof:
Let $0\leq x$, then $ 0\leq x \land 0\leq x \implies 0\leq x^2$ by axiom (1). 
We also know that $x^2=(-x)^2$, so that $ 0\leq x \land 0\leq x \implies 0\leq x^2$  and $ 0\leq x \land 0\leq x \implies 0\leq (-x)^2$.
As $0\leq x$, $-x \leq 0$, so then $\forall_x(0\leq x^2).$
Let $x\neq0 \iff x^2\neq0$, then $[x^2\neq0 \land \forall_x(0\leq x^2)] \implies [\forall_x(0\lt x^2)]$$\tag*{$\Box$}$ 

With this fact, we can move on to the second part; there is no total orering of $\mathbb{C}$ such that it is an ordered field.
Proof:
We know that $i^2=-1<0$, which contradicts the just proven property: $\forall_{x\neq0}(0<x^2)$.
This proof was originally a bit longer, but I saw that some steps were unnecessary or just stating the same thing repeatedly. However, having it reduced to this one line, I can't really believe this would pass as a valid proof... Maybe it's wrong to use the fact that $i^2=-1?$
 A: Proof (2):
We know that $1\neq 0$ and that $1^2=1\times1=1$ by axioms of a field. Thus, by theorem (1) we know that $0\lt 1^2 \iff 0<1$. Then, by axiom (2), we know that $0+(-1) < 1 +(-1)$ and by axioms of a field, we know that this is equivalent to $-1<0$.
We also know that $i^2=-1$ by its definition, and therefore that $-1=i^2<0$. This is a contradiction, as we have already proved in theorem (1) that a square of any number in a field must be positive. 
Thus, there is no total ordering for $\mathbb{C}$ such that it is an ordered field.
A: 
Let 0≤x, then 0≤x∧0≤x⟹0≤x2=(−x)2 by axiom (1)

You haven't proven that $x^2 = (-x)^2$.  
To prove that you need to prove $(-x)*y = x*(-y)=-(x*y)$ (and you can't assume $-x = -1*x$; if you use that you must prove that.)

As 0≤x, −x≤0, so then ∀x(0≤x2).

You haven't proven that $0 \le x \implies -x \le 0$.
And you have not actually explicitely stated if $x < 0$ then $x^2 > 0$.  You've implied it by noting $-x \le 0$ and $(-x)^2 = x^2 \ge 0$, but you should put this in terms of $x$, not $-x$ directly, not implicitely.
You have not prove that if $0 \ge x$ that that means $-x \le 0$

We know that $i^2=−1<0$

You have not proven $-1 < 0$.  
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But this are all fixible with very few more lines:
i) $x \ge 0 \implies x + (-x) \ge 0 + (-x)$ by axiom ii) so $x\ge 0\implies -x \le 0$.
ii)  $x*(-y) + x*y = x(-y + y)$ (by distribution) $= x*0$.  We will have to prove that $x*0 = 0$ later.  So after we prove $x*0 = 0$ we know that $x*(-y) + x*y = 0$ we have $x*(-y) = -(x*y)$.
iia) $(-x)^2 = (-x)(-x) = x*(-(-x))$ and if we can prove that $-(-x) = 0$ then we will have proven $(-x)^2 =x*(-(-x)) = x*x = x^2$.
iic)  For $-x$ there is one unique $-(-x)$ such that $x + (-(-x)) = 0$.  That is a field axiom.  $-x + x = 0$.  So $x = -(-x)$.
Likewise $(-x)*y + xy = (-x + x)*y = 0* y = 0$ so $(-x)*y =-(x*y) = x*(-y)$.
iia) $x0 = x(0 + 0) = x0 + x0$  so $0 = x*0 + (-(x*0)) = x*0 + x*0 + (-(x*0)) = x*0$.
iii) $1 = 1^2 > 0$.  
iv) and therefore $-1 < 0$.
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All text are different but I like Rudin's Principals of Mathematical Analysis that specifies the unprovable, taken as a given axioms as:


*

*i) $x < y$ then $x + z < x+y$

*ii) if $x > 0$ and $y>0$ then $xy > 0$.


Then the gazillion provable propositions are:


*

*1a) If $x >0$ then $-x < 0$ and if $x < 0$ then $-x> 0$.

*b) if $x > 0$ and $y<z$ then $xy < xz$

*c) if $x < 0$ and $y< z$ then $xy > xz$.

*d) if $x\ne 0$ then $x^2 > 0$. In particular $1> 0$.

*e) if $0 < x < y$ then $0 < \frac 1y < \frac 1$.


It's tedious but if you have them in order everything will work out.
So when he introduces the complex field, he has for an exercise:
Preve that no order can be defined in the complex field that turns it into an ordered field.  Hint: $-1$ is a square.
So solution would be (in my words):
Suppose there were such an order.  Then $1 > 0$ (prop 1d) so $-1 < 0$ (prop 1a) but $i^2 = -1 > 0$ (prop 1d).  This is a contradiction.
A: We know $i \neq 0$ and so either $i \geq 0$ or $-i \geq 0$ so powers of each must be positive. However $i^3=-i$ and $(-i)^3 = i$ so in either case multiplying positive numbers together results in a negative number, contradicting the first axiom.
