- Show that it is impossible to define a total ordering on C. In other words, one cannot find a relation between complex numbers so that:
For any two complex numbers z, w, one and only one of the following is true: \(z \succ w , w \succ z \ \mathrm { o r } \ z = w .\)
For all \(z _ { 1 } , z _ { 2 } , z _ { 3 } \in \mathbb { C }\) the relation \(z _ { 1 } \succ z _ { 2 }\) implies \(z _ { 1 } + z _ { 3 } \succ z _ { 2 } + z _ { 3 }\)
Moreover, for all \(z _ { 1 } , z _ { 2 } , z _ { 3 } \in \mathbb { C }\) with \(z _ { 3 } \succ 0\) , then \(z _ { 1 } \succ z _ { 2 }\) implies \(z _ { 1 } z _ { 3 } \succ z _ { 2 } z _ { 3 }\)
[Hint: First check if \(i \succ 0\) is possible.]
Solution
Assume such an ordering exists. For every nonzero \(z\in\mathbb C\), trichotomy gives either \(z\succ0\) or \(-z\succ0\). In either case, compatibility with multiplication implies \[ z^2\succ0. \] Applying this to \(z=1\) and \(z=i\) gives \[ 1\succ0, \qquad i^2=-1\succ0. \] Adding \(1\) to the inequality \(-1\succ0\) yields \[ 0\succ1, \] contradicting \(1\succ0\). Therefore no total ordering on \(\mathbb C\) can satisfy the ordered-field axioms (i)–(iii).