Linear order on $\mathbb{C}$

Question

I know that the $\mathbb{C}$ -- non-ordered field. And the answer that I have, that $i^2 < 0$. But in axioms of order field I see only this:

1. $\forall a,b,c$, if $a \leq b \Rightarrow a+c \leq b+c$
2. If $0 \leq a$ and $0 \leq b \Rightarrow 0 \leq ab$

I think, that this follows from 2., but how? $i \leq 0$ or $0 \leq i$?

If $0 \leq i \Rightarrow i^2 \leq 0$, if $i \leq 0 \Rightarrow i^2 \leq 0$, but how this axioms follows, that for all numbers in ordered field must be true that: if $a,b \leq 0 \Rightarrow 0 \leq ab$?

Answer 1

The axiom "$0\leq a$ and $0\leq b$ implies $0\leq ab$" together with other ordered field axioms implies this \begin{align}a,b\leq 0 &\implies a+(-a)\leq 0 + (-a)\ \text{and}\ b+(-b)\leq 0+(-b)\\ &\implies -a,-b\geq 0 \implies (-a)(-b) \geq 0 \implies ab\geq 0.\end{align}

Thus, if you take linear order on $\mathbb C$, either $i\geq 0$ or $i\leq 0$, but both imply $i^2\geq 0$.

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So, how does this prove that $\mathbb C$ is not an ordered field? Well, in any ordered field we have $-1\not\geq 0$. Assume the contrary, $-1\geq 0$. Now, it follows that $(-1)^2 \geq 0$ and $-1+1\geq 0 + 1$, i.e. $1\geq 0$ and $0 \geq 1$. Thus, $1=0$. Contradiction.