The common argument against the existence of a total order on complex numbers that observes the known axioms on real numbers is that 0 < i or 0 < -i. But why? 0 is not even an imaginary number.

There's a possibility of treating ni < -oo for all n in the set of natural numbers, such that we have the real line on the right and the imaginary line on the left, where a comparison operation compares the imaginary part before the real part.

The thing is, both negative numbers and imaginary numbers are not scalars that satisfy the usual axioms of multiplication. Why must imaginary numbers satisfy what even negative numbers cannot satisfy? In order to establish the 2-D complex plane, we just have to see the set of complex numbers as a cross product of 2 sets of real numbers, where the imaginary part comes before the real part, the opposite way of writing a complex number conventionally, namely, ai + b. Well, we've been using cross products to establish multiple dimensions for a long time. I don't see a problem here.

------------ Appendix ------------

Recall that imaginary numbers are derived to be more negative than negative numbers in the first place, namely, i i = -1 and (-1)(-1) = 1. So, it makes intuitive sense to treat them as numbers less than negative infinity. However, I forgot that they also cancel out like real numbers, namely, i + -i = 0 i = 0. The key is to not treat 0 i = 0. The problem is that they are designed to be like real numbers, except for the i sign, which is similar to the - sign.

In any case, I was trying to define an INumber interface, but not being able to compare numbers is really inconvenient. It means that they cannot participate in many known formal verification and optimization techniques. The reason why I ask this question here is that if mathematicians can agree on this definition, it will be easy to adopt it in ISO, IEC and IEEE standards. Of course, I can define anything I want, but who will use my definition? If the new definition works, we will have safer and faster computer programs. It makes perfect economic sense.

I know I can simply try not to use complex numbers at all, but why invent them in the first place? Plus, it's really convenient to express wave functions in terms of complex numbers. Nowadays, what is not a wave in physics and engineering? By the way, we mostly deal with equations and not inequalities or any sort. So, it's not as meaningful as I pitch it here. Anyways.

I will work out the addition requirements before posting another question. Thank you all for your participation.

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    $\begingroup$ Not quite sure what you're saying. The well-ordering theorem is equivalent to the Axiom of Choice, so assuming that there does exist an ordering on the complex numbers, just not one that makes it an ordered field in the sense that the product of two positives is positive, etc. If it's an ordered field, either i>0 or i<0, and either possibility leads to a contradiction. $\endgroup$ – John Brevik Mar 10 '15 at 23:31
  • $\begingroup$ But, both i and -i are not positive numbers. $\endgroup$ – David Mar 10 '15 at 23:37
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    $\begingroup$ 0 is a complex number $\endgroup$ – pre-kidney Mar 10 '15 at 23:37
  • $\begingroup$ You can concatenate two rays but not two lines. $\endgroup$ – N. F. Taussig Mar 10 '15 at 23:40
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    $\begingroup$ "0 is not even an imaginary number." Yes it is. $0 = 0i$ is a purely imaginary number. It is the only number that is both purely imaginary and purely real. $\endgroup$ – fleablood Sep 4 '18 at 18:40

Note that the proof does not prove that there is no total ordering of the complex numbers. It is easy to order the complex numbers totally.

What it does prove is that there is no total ordering of the complex numbers which is compatible with their algebraic structure.

You say you want to have both $i<0$ and $-i<0$, but that is not compatible with the algebraic structure. Namely, if we have $i<0$ then adding $-i$ to both sides gives $0<-i$. Since you're asserting $-i<0$ which contradicts this, your ordering doesn't obey the rule $$ a<b \implies a+c<b+c $$ which is one of the axioms for an ordered field.

  • $\begingroup$ You're right. I totally forgot about addition. Thanks! $\endgroup$ – David Mar 11 '15 at 2:47
  • $\begingroup$ I give up for today. Do you have any thought on the appendix? If not, I can still do it the old school way. No worries. $\endgroup$ – David Mar 11 '15 at 5:37

Your argumenation is somewhat weird. You want to order the set of complex numbers and complain that $0$ is not an imaginary number. This already being questionable, $0$ is certainly a complex number.

Next you come up with comparisons against $-\infty$ - and that is not a complex number.

Lastly, it is possible to impose a total order on $\mathbb C$ (for example by taking lexical ordering as you seem to secribe in the last paragraph). However, none that is compatible with the field operations (which the proof you mention in the first paragraph is about).

  • $\begingroup$ Well, 0 is a real number and not an imaginary number. So, it's definitely a complex number. $\endgroup$ – David Mar 10 '15 at 23:38
  • $\begingroup$ Well, -1 is not a natural number, but I can compare it with 2, which is a natural number. $\endgroup$ – David Mar 10 '15 at 23:40
  • $\begingroup$ For me, both i and -i are less than 0, a condition that doesn't meet the assumption of the proof. $\endgroup$ – David Mar 10 '15 at 23:44
  • $\begingroup$ The proof that there cannot exist a total order on complex numbers (proofwiki.org/wiki/Complex_Numbers_cannot_be_Totally_Ordered) indicates that either 0 < i or 0 < -i, but we can have both 0 > i and 0 > -i. More, I want -oo > ni > -i for all n in natural numbers. $\endgroup$ – David Mar 11 '15 at 0:36

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