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\noindent {\bf Question}

\noindent Prove that there does not exist an order on the complex
numbers ${\bf C}$ so that ${\bf C}$ becomes an ordered field.

\medskip
\noindent {\bf Answer}

\noindent Suppose there were such an order on ${\bf C}$, and
denote it by $<$.  Compare $0$ and ${\rm i}$.  Since $0\ne {\rm
i}$, it must be that either $0 <{\rm i}$ or ${\rm i} <0$.

\medskip
\noindent Suppose that $0 <{\rm i}$.  Multiplying both sides by
${\rm i}$ and remembering that $0 <{\rm i}$, we see that $0\cdot
{\rm i} < {\rm i}\cdot {\rm i}$, which simplifies to $0 < -1$.
Adding $1$ to both sides, we see that $1 <0$.  Again multiplying
both sides by ${\rm i}$ and remembering that $0 <{\rm i}$, we see
that $1\cdot {\rm i} < 0 \cdot {\rm i}$, which simplifies to ${\rm
i} < 0$.  Hence, if $0 < {\rm i}$, then ${\rm i} <0$,
contradicting the second condition in the definition of an order.

\medskip
\noindent Suppose now that ${\rm i} <0$.  Adding the additive
inverse $-{\rm i}$ of ${\rm i}$ to both sides, we get that $0 <
-{\rm i}$.  Multiplying both sides by $-{\rm i}$, we get that
$0\cdot ( -{\rm i}) < (-{\rm i})\cdot (-{\rm i})$, and so $0 <
-1$.  Multiplying both sides by $-{\rm i}$ again, we get that $0 <
(-1)\cdot (-{\rm i}) = {\rm i}$. Hence, if ${\rm i} < 0$, then
then $0  > {\rm i}$, again contradicting the second condition in
the definition of an order.

\medskip
\noindent Hence, since we have that neither $0 <{\rm i}$ nor ${\rm
i} <0$, we see that there cannot exist an order on ${\bf C}$ that
makes ${\bf C}$ into an ordered field.
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