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- Apr 14, 2013

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Let $K$ be a field. Suppose that there is an element $u\in K$ such that $u^2+1=0$. Show that there is no order relation on $K$ that would make $K$ an ordered field.

We have to show that no relation $<$ can exist that satisfies the order axioms, i.e.:

- Only one of $a < b$, $a = b$, or $a > b$ is true
- If $a < b$ and $b < c$, then $a < c$
- If $a < b$ and $c < d$, then $a + c < b + d$
- If $a < b$ and $c < d$, then $a c < b d$

Suppose that there is an order relation on $K$ that would make $K$ an ordered field.

In $K$ there is an element $u$ such that $u^2+1=0\Rightarrow u^2=-1<0$.

According to the $4$th axiom we have that $u^2<0$ and $u^2<0$ then $u^2\cdot u^2=\left (u^2\right )^2=\left (-1\right )^2=1>0$, which is a contradiction since it should be $u^2\cdot u^2 <0$.

That means that there cannot be an order relation on $K$ that would make $K$ an ordered field.

Is the proof correct and complete? Could I improve something?