Proof: Complex Numbers Don't Form an Ordered Field

[jgens]

Would the following prove that the set of complex numbers do not form and ordered field?

Clearly $i \neq 0$. Therefore, if the complex numbers form an ordered field either $i > 0$ or $i < 0$. Suppose first that $i > 0$, then $i^2 = -1 > 0$, a contradiction. Now suppose that $i < 0$, then $i^2 = -1 > 0$, another contradiction. Thus, the set of complex numbers do not form an ordered field.

This seems awfully fishy and I wouldn't be surprised to find that it's completely invalid. Feedback and suggestions are welcome. Thanks!

 

[pbandjay]

You have i² = -1 > 0, but -1 < 0.

 

[jgens]

Yes, that's why $i > 0$ and $i < 0$ are contradictions as I state in my original post.

 

[Mark44]

Yes, that's why $i > 0$ and $i < 0$ are contradictions as I state in my original post.

I understand the logic behind what you're saying, but you could say it in a way that is clearer.

Suppose that i > 0.
Multiplying by a positive number preserves the direction of the inequality, so i² must be positive. This is incorrect, though, because i² = - 1 < 0. Thus, assuming i > 0 leads to a contradiction.

And similar for the assumption that i < 0.

 

[jgens]

Thanks for the suggestion! It seems like I always need to add details to my proofs in order to make them clearer. Does the method I'm using work?

 

[HallsofIvy]

What you have proven is that the usual order on the real numbers, in which -1< 0, cannot be extended to the complex numbers.

But you can prove more. Can there exist some order? Perhaps one in which 1< 0and -1> 0?

Suppose such an order existed. The i is not 0 (the additive identity does not depend upon the order). If i> 0 then it follow that (i)(i)= -1> 0. From that i(-1)= -i> i(0)= 0. Now that contradicts i>0. If i< 0, the -i> 0 so (-i)(i)= 1< 0. From that i(1)= i< i(0)= 0 and that contradicts i< 0.

 

[qspeechc]

Ok, that proves that there can be no order where 1<0 and -1>0. How does one show that there can not be any order relation defined on the complex numbers? Or R^n for n>1 for that matter?

 

[Ben Niehoff]

What about this one:

Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, where $\theta$ is restricted to be in $[0, 2\pi)$. Then define > as follows:

1. If $r_1 > r_2$, then $z_1 > z_2$

2. If $r_1 = r_2$ and $\theta_1 > \theta_2$, then $z_1 > z_2$

3. If $r_1 = r_2$ and $\theta_1 = \theta_2$, then $z_1 = z_2$

This looks like an order relation to me...it even has a least element, 0. Now, I know that C is not supposed to have an order relation, so what's wrong with this? Have I implicitly used the axiom of choice or something?

 

[Moo Of Doom]

You can certainly put some order on the complex numbers, Ben, but you can't find one that behaves nicely with their algebraic structure. For a field to be considered an ordered field, it must have an order that is compatible with multiplication and addition. That is, we want

a < b ==> a + c < b + c
a < b, c > 0 ==> ac < bc

for all a, b, c. Your order doesn't satisfy either of these properties, so it doesn't make the complex numbers into an ordered field. Sorry.

 

[Ben Niehoff]

Ah, yes, that does change things a bit.

 

[elibj123]

Order of a field is defined via a subset of that field. Let's denote it P. An ordered field F contains a subset P that satisfies the following properties:

1) for any a (which is not 0) in F, Either a or -a are in P, but not both.
2) For any a,b in P, a+b is in P.
3) For any a,b in P, a*b is in P.

Then, we say a>b iff (a-b) is in P.

You can see immediately that the C doesn't contain such a set because
if i belongs P, it implies that -i=i*i*i belongs to P (3) in contradiction to (1).
The same if we assume -i belongs to P.

 

[HallsofIvy]

Ok, that proves that there can be no order where 1<0 and -1>0. How does one show that there can not be any order relation defined on the complex numbers? Or R^n for n>1 for that matter?

As elibj123 said, one of the requirements for an "ordered field", trichotomy, is that one and only one of the following be true of any member of the field, x: (trichotomy)
a) x= 0
b) x> 0
c) x< 0

You cannot have "-1< 0" and "1> 0" in any ordered field (where "1" is defined as the multiplicative identity and "0" is defined as the additive identity).

Your reference to $R^n$ doesn't make sense because it is not a field. What arithmetic operations would you wish to define on it to make it a field?
If you ignore the operations, then the complex numbers can be ordered (any set can be). One such order ("lexicographical") is "If a< c then a+bi< c+ di. If a= c and b< d, then a+bi< c+ di." That's a perfectly good order on C or $R^2$ and can be extended in an obvious way to $R^n$. As long as you don't look at the operations, there is no problem.