# How to teach a High school student that complex numbers cannot be totally ordered?

I once again need your precious knowledge! I am not sure which is the best pedagogic way to teach a High school student about why complex numbers cannot be totally ordered. When I was in High school we were simply told that we cannot order complex numbers. When we asked why the answer was because there is no total order in $\mathbb C$. But we weren't taught a thing about order in general. Is there a crystal clear way to do this in High school? I am thanking you all in advance!

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You can totally order $\mathbb{C}$. You cannot make it an ordered field, and that's easily seen by the fact that in ordered fields nonzero squares are positive. But $-1 = i^2$ is a square, as is $1 = 1^2$. –  Daniel Fischer Jun 26 '13 at 15:08

Without going into exactly what an ordered field is, you can tell your students that if you want to order $\mathbb{C}$, two things that should be true of the order is

1. The order preserves the ordering of $\mathbb{R}$.
2. For $a,b,c$ with $b<c$, then $ab<ac$ if $a>0$ and $ab>ac$ if $a<0$.

Then ask: Where is $i$ in the ordering compared to $0$?

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Condition 1 can be replaced with $a>b$ implies $a+c>b+c$ and condition 2 with $a>0,b>0$ implies $ab>0$. –  Hagen von Eitzen Jun 26 '13 at 15:22
I find it more fun to ask students to compare $i$ and 1, as it then becomes less obvious and they have to do more manipulations, and they all get excited that $i^2 = -1$. Often, after enough time, they come up with comparing $i$ and 0 by themselves. –  Calvin Lin Jun 26 '13 at 15:47

Algebraic. The condition for a field to be orderable is that sums of nonzero squares are nonzero. The complex numbers are designed to violate that.

Geometric. Take a regular $2n$ sided polygon in the complex plane, with center at $0$. Vertices $2k$ steps apart (moving from vertex to neighboring vertex) differ by a factor $(w^2)^k$ for $w$ a root of unity, which must be positive as a power of a nonzero square, so that all the even numbered vertices have the same sign in any ordering. But the sum of all the even vertices is zero by symmetry.

Quadratically closed. Every nonzero complex number is a square, which should make it positive, but then $z$ and $-z$ cannot have opposite signs.

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