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An ordered field is a field $F$ which is also an ordered set, such that:
(i) $x+y < x+z$ if $x$,$y$,$z \in F$ and $y<z$
(ii) $xy > 0$ if $x \in F$, $y \in F$, $x > 0$, and $y>0$


Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: -1 is a square

My approach

To prove that there can be no order defined in the complex field, I used the second point of the definition.

Let $x=(0,1)$ and $y=(0,1)$. I state that $x>0$ and $y>0$ (but I am unsure if I can do that). Thus $xy$ should be strictly positive, yet $xy= i^2=-1 <0$, thus this disproves the fact that it is possible to define an order in the complex field.

Would this be an acceptable proof? Can I really state that $x=(0,1)>0$?


marked as duplicate by user228113, Did real-analysis May 28 '17 at 15:34

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  • $\begingroup$ you can say: suppose that there is an order in $\Bbb C$ such that $i>0$ and $\Bbb C$ is an ordered field. But you can see that this order cannot be extended to the standard order on $\Bbb R$ because $i^2<0$. The proof is not complete or sufficient. You can complete the proof showing that when $i<0$ it doesnt define either an ordered field in $\Bbb C$. $\endgroup$ – Masacroso May 28 '17 at 14:15

You need to either prove that $x > 0$ or assume that and then assume that $x < 0$. For example, you would write:

Assume $x > 0$ then $x^2 = xx > 0$ by (ii). Now assume that $x < 0$. Then $-x > 0$ (Prop. 1.18(a)) and again $x^2 = (-x)(-x) > 0$. In both cases we see that if $x \ne 0$ then $x^2 > 0$.


$1=1\times 1$

Hence $1$ is a square so it's greater than $0$,

So $-1+0<-1+1$, so $-1<0$ but $i^2=-1$, so $-1$ must be greater than $0$, hence a contradiction.

Note that a square is greater than zero can be easily verified from the axioms.


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