Show that it is impossible to order C (complex numbers) so that it becomes an ordered field.


Hint: what happens when you multiply both sides of the inequality by $i$?


To begin, suppose that $i>0$. Then since $i>0$, we can multiply both sides of the inequality by $i$ to get $i^2 >0$ (the inequality doesn't flip because $i>0$). Why is this a contradiction?

Now, what happens when we assume $i<0$?


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Updated on July 21, 2022


  • arielle
    arielle less than a minute

    Hint: Assume (C, +, *, <) is an ordered field for some order relation <. Then consider i is not equal to 0. One has either i>0 or i<0. In either case, find a contradiction.

    • Hagen von Eitzen
      Hagen von Eitzen over 8 years
      Yes, go ahead. You are almost there.
  • arielle
    arielle over 8 years
    what do you mean by this?
  • Ben Grossmann
    Ben Grossmann over 8 years
    I've added to my answer in an attempt to make what I meant clearer. Let me know if you're still confused.