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The question at hand is stated in the title.

Typical proofs that I have read involves taking a complex number i, then reaching some contradiction with i2 > 0.

I feel like these proofs bypass the problem of what it means to define an order in the complex field. In other words, what does it mean to have 1 complex number larger/smaller than the other?

Intuitively, taking the norm is the obvious way to compare, but that violates 'if x>y, y-x < 0'. Nonetheless, the cause to my confusion is there seems to be many ways that an 'order' can be defined, and until each one of them is explored, one cannot prove if a complex field can or cannot be ordered.

Keen to hear your comments on my amateurish view.

Kenneth Or
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    Welcome to Mathematics Stack Exchange. Complex numbers can be ordered, but not in a way compatible with field operations – J. W. Tanner Oct 18 '20 at 03:36
  • Acknowledging J.W. Tanner's comment, the way my complex function theory book specifically defines order is that field $F$ must have a subset $P$ such that (1) $(a \in P \wedge b \in P) \Rightarrow ([a+b] \in P \wedge [a\times b] \in P).$ and (2) $0$ not in $P$ and $a \neq 0$ implies that exactly one of $[a, -a]$ are in $P$. Without some definition of ordering, any proof becomes meaningless. Re your instinct about linking ordering to (for example) $|z|$, don't you want ordering to provide that $z_1 \neq z_2$ implies that they don't have the same order? – user2661923 Oct 18 '20 at 04:15
  • Thanks for the welcome. – Kenneth Or Oct 18 '20 at 09:24
  • Tanner: may I have a proof/reference to the statement? Or was the proof I listed the one.

    user2661923: I understand that there must be a definition for ordering for a proof to function. According to your textbook's def, it seems complex numbers are ordered? (I know they are not) In mine (Rubin), order is defined with: For 2 elements, one of x>y, x=y, x<y is true. But I don't see how 2 complex numbers can be compared.

    Thank you for the responses and I apologize in advance for the difficulty in communication due to my lack of rigorous mathematical knowledge.

     – Kenneth Or Oct 18 '20 at 09:37
  • With respect to proving that $\mathbb{C}$ is not an ordered Field, based on the definition of an ordered Field, in my earlier comment, I will (again) excerpt "An Introduction to Complex Function Theory" 1991 (Appendix A) - [Bruce Palka]. Suppose that $P$ is a subset of $\mathbb{C}$, with the properties re my previous comment. Since $i \neq 0$ exactly one of ${i, -i}$ is in $P$. In either case, $(i)^2 = (-1) = (-i)^2$ must then be in $P$. Therefore, $(-1)^2 = (1)$ is in $P$. This is a contradiction, since $P$ can not contain both $(-1)$ and $(1)$. – user2661923 Oct 18 '20 at 10:07
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    It was only an accident (re my random browsing) that I saw your comment. The following link shows how to address a comment so that the user is automatically notified: https://math.stackexchange.com/help/privileges/comment. – user2661923 Oct 18 '20 at 10:09
  • @user2661923 thanks for clarifying I see that this proof does not involve the '<' '>' symbols, and I am more comfortable. – Kenneth Or Oct 18 '20 at 11:11
  • Using $\mathbb{R}$ as an example of an ordered field, with $\mathbb{R^+}$ serving as $P$ re the proof in one of my previous comments, you then define the relation $(a > b)$ by ${[a > b] \iff [a + (-b) \in \mathbb{R^+}]}$. – user2661923 Oct 18 '20 at 12:25
  • @J.W.Tanner " Complex numbers can be ordered, but not in a way compatible with field operations" : Is the following an example of what you have in mind. If $z = (x + iy), w = (u + iv)$ then ${[z > w] \iff [(x > u) \vee (x = u \wedge y > v)]}$? – user2661923 Oct 18 '20 at 12:43
  • Cf. this question and this question and this question and this question ... – J. W. Tanner Oct 18 '20 at 14:16
  • and this question and this question and this question and this question and this question – J. W. Tanner Oct 18 '20 at 14:21
  • and this question – J. W. Tanner Oct 18 '20 at 14:29

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