Let $\mathcal F$ be a field. Suppose that there is a set $P \subset \mathcal F$ which satisfies the following properties:

  • For each $x \in \mathcal F$, exactly one of the following statements holds: $x \in P$, $-x \in P$, $x =0$.

  • 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!


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 $P$.

  • $\begingroup$ Why doesn't $0$ work here? If $z\epsilon P$ such that $z=0$, then doesn't that satisfy these properties? Also, who is to say that there isn't another subset out there where we could choose different numbers and make it work? I just don't get the point of choosing -1 and showing that since that doesn't work, then nothing will work. Any help explaining this? $\endgroup$
    – user23793
    Aug 26 '13 at 3:12
  • $\begingroup$ The square of $0$ is not $-1$. There are many ways to put a total order on $\mathbb{C}$. However, no such ordering can "play nice" with the addition and multiplication, which is what the definition of ordered field requires. $\endgroup$ Aug 26 '13 at 3:24
  • $\begingroup$ "The square of 0 is not $−1$" but what does that have to do with finding a subset of $C$? $0+0 \epsilon P, 0*0 \epsilon P$ So doesn't this work? $\endgroup$
    – user23793
    Aug 26 '13 at 3:36
  • $\begingroup$ In the definition of ordered field, we need for any $x$ that $x\in P$ or $-x\in P$ or $x=0$. So we have to decide whether $i$ is in $P$. If it isn't, then $-i\in P$. In either case, since $i^2=(-i)^2=-1$, we have $-1\in P$. But then $(-1)(-1)\in P$. Impossible, since by definition $-1$ and $1$ can't be both in $P$. $\endgroup$ Aug 26 '13 at 3:43
  • 1
    $\begingroup$ No problem, making a mistake of this type probably ensures you will remember the structure of the definition forever. $\endgroup$ Aug 26 '13 at 3:50

I think the definition of $P$ you have is slightly off: if $x=0$ then all three conditions are satisfied. A possible fix is "Either $x\in P$ or $-x\in P$, with both holding iff $x=0$." On the other hand, I'm not convinced that it's important that $0\in P$; you should check that before making things complicated.

For an arbitrary field, $-1\notin P$ because then $(-1)(-1) = 1\in P$, which is impossible.

From there, you assume that $P$ exists and begin a proof by contradiction. Using your work in the OP you can therefore show that $i\notin P$. However, since $i\neq 0$, we also need to show that $-i\notin P$ before we continue. The proof is essentially identical to the one you gave in the OP.

This will contradict the fact that either $i$ or $-i$ is in $P$. Therefore, there cannot be such a set $P\subset\mathbb{C}$.

  • $\begingroup$ I think the definition is OK. 0 is not meant to be an element of P. For example, if we were talking about the reals, then P would be the positive reals, not including 0. $\endgroup$
    – user13618
    Aug 12 '12 at 17:30

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