# The Complex Field

1. Oct 15, 2009

### 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!

2. Oct 15, 2009

### pbandjay

You have i2 = -1 > 0, but -1 < 0.

3. Oct 15, 2009

### jgens

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

4. Oct 15, 2009

### Staff: Mentor

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.

5. Oct 15, 2009

### 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?

6. Oct 16, 2009

### 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.

7. Oct 16, 2009

### 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?

8. Oct 16, 2009

### Ben Niehoff

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?

9. Oct 16, 2009

### 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.

10. Oct 16, 2009

### Ben Niehoff

Ah, yes, that does change things a bit.

11. Oct 22, 2009

### 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.

12. Oct 22, 2009

### HallsofIvy

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.