I have to exposition about the impossibility of ordering the complex numbers:

Axioms $6$: Exactly one of the relations $x = y$, $x < y$, $x > y$ holds.

Axioms $7$: If $x < y$, then for every z we have x + z < y + z.

Axioms $8$: If $x > y$ and $y > z$, then $x > z$

As yet we have not defined a relation of the form $x < y$ if $x$ and $y$ are arbitrary complex numbers, for the reason that it is impossible to give a definition of $<$ for complex numbers which will have all the properties in Axioms $6$ through $8$. To illustrate, suppose we were able to define an order relation $<$ satisfying Axioms $6$, $7$, and $8$. Then, since $i \neq 0$, we must have either $i > 0$ or $i < 0$, by Axiom 6. Let us assume $i > 0$. Then taking, $x = y = i$ in Axiom $8$, we get $i^2 > 0$, or $-1 > 0$. Adding 1 to both sides (Axiom $7$), we get $0 > 1$. On the other hand, applying Axiom $8$ to $-1 > 0$ we find $1 > 0$. Thus we have both $0 > 1$ and $1 > 0$, which, by Axiom $6$, is impossible. Hence the assumption $i > 0$ leads us to a contradiction. [Why was the inequality $-1 > 0$ not already a contradiction?] A similar argument shows that we cannot have $i < 0$. Hence the complex numbers cannot be ordered in such a way that Axioms $6$, $7$, and $8$ will be satisfied.

But Why was the inequality $-1 > 0$ not already a contradiction? and it is true for $i < 0$?

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    $\begingroup$ 1 > 0 is not a given axiom. It has to be proven. Hint: prove $x \ne 0$ then $x^2 > 0$ and therefore $1 = 1^2 > 0$. This can get you where you are going. $\endgroup$ – fleablood Aug 25 '16 at 17:31
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    $\begingroup$ Depends on the rest of axioms (and what you have proven with them). If it was already proven starting from the axioms that $-1<0$, then you, indeed, reached a contradiction at that point. If it was not proven, then you need to cater for the possibility that may be there exists an ordering of the complex numbers that does not coincide with the usual ordering of integers. $\endgroup$ – Jyrki Lahtonen Aug 25 '16 at 17:31
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    $\begingroup$ Axiom 8 does not let you infer $i^2>0$ from $i>0$, you must be using a different axiom. $\endgroup$ – Andrea Aug 25 '16 at 17:33
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    $\begingroup$ You are missing the most important axiom. $x > 0$ and $a < b$ then $xa < xb$. Can't do this without it. $\endgroup$ – fleablood Aug 25 '16 at 17:35
  • $\begingroup$ You've proven -1> 0 by considering i > 0 or i < 0. Do the same thing for 1. consider 1 > 0 or 1 < 0 and conclude $1^2 = 1 > 0$. $\endgroup$ – fleablood Aug 25 '16 at 17:52

Axioms 6, 7 and 8 are not sufficient for excluding the possibility to order the complex numbers.

If you define $a+bi\prec c+di$ ($a,b,c,d\in\mathbb{R}$) when either $a<c$ or $a=c$ and $b<d$, you get an order relation satisfying those axioms.

The contradiction will show up only if you add another axiom:

if $x<y$ and $0<z$, then $xz<yz$

First step: proving that $0<1$.

There are two cases: $1<0$ or $0<1$. Suppose $1<0$; then $1-1<0-1$, so $0<-1$. Hence $0(-1)<(-1)(-1)$, that is $0<1$: a contradiction.

Second step: proving that $-1<0$

Since $0<1$, we have $0-1<1-1$.

Third step: getting a contradiction

Suppose $0<i$; then $0i<i^2$, that is, $0<-1$, a contradiction.

Suppose $i<0$; then $i-i<0-i$ and $0<-i$; then $0(-i)<(-i)^2$, that is, $0<-1$, a contradiction.


Axiom 6 cannot hold for $x=0$ and $y=i$.

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If $\mathbb{C}$ is considered as an additive Abelian group, or even as a $2$-dimensional real vector space, then it can be totally ordered, in a way that is compatible with the operations of addition and multiplication by real scalars. Quoting from Ordered vector space - Wikipedia, the free encyclopedia:

$\mathbb{R}^2$ is an ordered vector space with the $\leq$ relation defined in any of the following ways $\ldots$

Lexicographical order: $(a, b) \leq (c, d)$ if and only if $a < c$ or ($a = c$ and $b \leq d$). This is a total order. The positive cone is given by $x > 0$ or ($x = 0$ and $y \geq 0$) $\ldots$

Therefore, in order to prove the non-existence of a compatible total ordering of $\mathbb{C}$, you will have to adopt at least one postulate concerning its multiplicative structure.

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Re: your first question, is "$-1<0$" explicitly one of your axioms? If not, you have to prove it, and "$-1>0$" isn't immediately a contradiction.

Re: your second question, the case where we assume $i<0$ is similar. You can prove (and probably have done so already as previous exercises) that your axioms imply that $-a>0$ whenever $a<0$, and that $(-a)(-a)=a^2$. So we have that $-i>0$ and $(-i)^2=-1$, so we can run the proof above with $-i$ in place of $i$.

(Also, no need for all caps.)

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Hint: prove (it is not given, you must prove it) that for $x \ne 0$ then $x^2 > 0$.

Then you have $i^2 > 0$ and $1^2 > 0$.

As to why $1 < 0$ isn't an immediate contradiction? Why should it be? Were you ever given an axiom that $1 > 0$? You were not. (but you can prove it.)

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Why would it be impossible to create an ordering for complex numbers? As you know, a complex number can be described as a point in the complex plane, and as a result, it can be written in polar coordinates (a1*cos(x1),a1*sin(x1)). Let's now say that two complex numbers A1=(a1*cos(x1),a1*sin(x1)) and A2=(a2*cos(x2),a2*sin(x2)) need to be ordered, then we can state following ordering rule:

A1 < A2 if:
  a1 < a2, or:
  a1 = a2 and x1 < x2

What would be wrong with that?

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  • $\begingroup$ It is possible to create an order, but not an order that preserves the arithmetic. Your suggestion is problematic, because it has $-1>1$ but does not have $0>2$ that should be a consequence of the former (add $1$ to both sides). Anyway, you should ask this as a new question (and this is not exactly an answer to Yobani's question, so will likely attract negative attention soon). $\endgroup$ – Jyrki Lahtonen Aug 26 '16 at 12:54

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