Let $\mathcal F$ be a field. Suppose that there is a set $P \subset \mathcal F$ which satisfies the following properties:
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For each $x \in \mathcal F$, exactly one of the following statements holds: $x \in P$, $-x \in P$, $x =0$.
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For $x,y \in P$, $xy \in P$ and $x+y \in P$.
If such a $P$ exists, then $\mathcal F$ is an ordered field.
Define $x \le y \Leftrightarrow y -x \in P \vee x = y$.
Exercise: Prove that the field of complex numbers $\mathbb C$ cannot be given the structure of an ordered field.
My Work So Far: (Edit 1 note: This section and the Question is at the beginning, simply leaving this up for reference as to where I started)
Let $i$ be such that $i \in P, i \ne 0 \Rightarrow i > 0$. But $i^2 = -1 \notin P$.
My Question: I am not sure how much I need to redefine, and how I go about rigorously making this patchwork argument airtight. I am aware that I have not addressed how I assumed that $-1 \notin P$, but I'm not sure how to distinguish between $1$ and $i$ in this proof.
Edit #1
1st Step: Showing that $-1 \notin P$, observe that $(-1)(-1) = 1$ therefore if $-1 \in P$, both $x, -x \in P$, a contradiction.
2nd Step: To show $i \notin P$, we have that if $i \in P \Rightarrow i^2 \in P$, but $i^2 = -1 \notin P$, so $i \notin P$.
3rd Step: To show $-i \notin P$, we have $(-i)(-i) = i^2 \notin P$, so $-i$ cannot be in $P$.
Conclusion: Since $i \ne 0$, and $i, -i \notin P$, there is no set $P \subset \mathbb C$ that satisfies the above properties, thus $\mathbb C$ is not ordered.
Thank you André Nicolas and Eric Stucky for your help!
Best Answer
To show that $-1$ is not in $P$, note that if $-1\in P$ then $(-1)(-1)\in P$, which contradicts the fact that if $x \ne 0$ exactly one of $x$ and $-x$ is in $P$.
Next we show that $i\notin P$. Suppose to the contrary that $i\in P$. Then $i^2\in P$, which contradicts the fact that $-1\notin P$.
The same argument shows that $-i\notin P$. This contradicts the fact that if $x\ne 0$, then exactly one of $x$ and $-x$ is in $P$.