Prove that field of complex numbers cannot be equipped with an order relation

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  • Proof of Non-Ordering of Complex Field

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Solutions Collecting From Web of "Prove that field of complex numbers cannot be equipped with an order relation"

What they are asking is to show that no relation $<$ can exist that complies with the order axioms, i.e.:

  1. Only one of $a < b$, $a = b$, or $a > b$ is true
  2. If $a < b$ and $b < c$, then $a < c$
  3. If $a < b$ and $c < d$, then $a + c < b + d$
  4. If $a < b$ and $c < d$, then $a c < b d$

In this case, if we take $i > 0$ we get $i^2 = -1 < 0$, contradicting (4). So by (1) it must be $i < 0$. But $i^4 = 1 > 0$, again contradicting (4). So no relation $<$ can exist on $\mathbb{C}$ which complies with (1) to (4).

try z = i the square root of -1 for a simple contradiction.