Stupid Question on Complex Numbers - Not Ordered

humbert

I'm on the very first chapter of the book and things are going well but then I'm told that real numbers are ordered because a < b OR a = b OR a > b and if x != 0 then x > 0 or -x > 0. That's fine.

Then I'm told that if C were ordered i or -i would be positive. Seems reasonable. But i^2 = (-i)^2 = -1 and this would imply that -1 is positive and that 1 and -1 cannot both be positive .

So simply the square of a number must be positive. However this is not true for complex numbers. So I'm either misunderstanding something or there is assumed knowledge that I don't have. Both are quite possible.

I looked up ordered fields and found this definition:
if a <= b then a + c <= b + c
if 0 <= a and 0 <= b then 0 <= a b

The second condition is much the same as the one given above for real numbers and would fail for complex numbers. Is that simply it?

I think maybe my problem is that if C is unordered then z1 < z2 is undefined but to get the result we have to decide if z1 or z2 are <= 0.

Is it simply the case that it must be positive or negative and since both give -1 then it is not ordered?

Sorry about the awful rambling. I'm not even sure there's a proper question in there. Just don't feel completely comfortable with it.

FoxT

Some quotes below that may help.

In plain english , a complex number is an ordered pair of real numbers (R1,R2). These numbers (pairs) can be represented on a plane. Complex numbers cannot be ordered in a standard way.

You can of course compare complex numbers by comparing say their muduli |z| = |x+iy| = sqrt(x^2 + y^2) - but be aware that the set of complex numbers with a modulus of M has an infinite number of elements, all of which are points on the circumference of a circle in the complex plane of radius M.

There are other ways you can compare them as well, but all of them will bump into this limitation (infinite no. of complex numbers have the same scalar property) .





"Unlike real numbers, complex numbers do not have a natural ordering, so there is no analog of complex-valued inequalities. This property is not so surprising however when they are viewed as being elements in the complex plane, since points in a plane also lack a natural ordering."

http://mathworld.wolfram.com/ComplexNumber.html


or

"Unlike the reals, C is not an ordered field, that is to say, it is not possible to define a relation z1 < z2 that is compatible with the addition and multiplication. In fact, in any ordered field, the square of any element is necessarily positive, so i2 = −1 precludes the existence of an ordering on C."

http://en.wikipedia.org/wiki/Complex_numbers#Field_structure

CJC86

FoxT wrote: »

In plain english , a complex number is an ordered pair of real numbers (R1,R2). These numbers (pairs) can be represented on a plane. Complex numbers cannot be ordered in a standard way.

I'll disagree slightly with FoxT on this point, since the complex numbers can be ordered as a set, but not as a field.

That is, we could compare 2 complex numbers z_1 = a+bi and z_2 = c+di, by saying that z_1 <= z_2 if a<c or if a=c and b<=d. This is the lexicographic ordering of R^2, which is in fact a total ordering, so it must order the complex numbers since C is isomorphic to R^2 as a set.

However, as FoxT pointed out, it is impossible to order C as a field, since the ordering does not preserve multiplication in the field of complex numbers.

FoxT

Thank You, CJC86.

I agree you can order them as described, and I'm sure there are other ways as well to do this.

I guess I should have stated there is not a general way of arranging them that 'is useful' or 'makes sense' for the newcomer.

This isn't very rigorous, but is essentially true as the focus on the study of complex numbers is generally algebra/analysis rather than sets.

- FoxT

ZorbaTehZ

I think I see where your problem is, it can be difficult at first but when thinking about ordering you need to forget completely about the real ordering, because it's not necessarilly the one being talked about, it's easiest to just use another symbol. I'll make a shot at showing it, might not be totally correct.

So suppose [latex]\prec[/latex] is some total ordering of [latex]\mathds{C}[/latex], then we must have [latex]z^2\prec 0[/latex] because consider: we either have [latex]z\prec 0[/latex] or [latex]0\prec z[/latex]. Suppose it's the first case, then we use the fact that for a total ordering we must have [latex]ac\prec bc[/latex] if we have [latex]a\prec b[/latex] and [latex]c\prec 0[/latex] (easy to show) then just let b=0 and a=c=z. Suppose it's the second case, then we use the fact that [latex]a\prec 0 \Rightarrow 0\prec -a[/latex] and set a=-z, then and use the same law again and we get the same result

humbert

Thanks to everyone who took the time to respond.

ZorbaTehZ wrote: »

So suppose [latex]\prec[/latex] is some total ordering of [latex]\mathds{C}[/latex], then we must have [latex]z^2\prec 0[/latex] because consider: we either have [latex]z\prec 0[/latex] or [latex]0\prec z[/latex]. Suppose it's the first case, then we use the fact that for a total ordering we must have [latex]ac\prec bc[/latex] if we have [latex]a\prec b[/latex] and [latex]c\prec 0[/latex] (easy to show) then just let b=0 and a=c=z. Suppose it's the second case, then we use the fact that [latex]a\prec 0 \Rightarrow 0\prec -a[/latex] and set a=-z, then and use the same law again and we get the same result

I think this is exactly what I was looking for. An argument (I wont say proof) that is self contained and explains why [latex]z^2[/latex] must be "greater" than zero (when z is not zero).

I'm not familiar with ordered fields so the only evidence I had for this to be the case was that it is true for reals and I wasn't happy having to maintain an analogy to reals when the definition of the set of real numbers was independent of them.

The lexicographic ordering brought up by CJC86 was also bothering me a little as it was evidence that ordering was of course possible but as far as I could see you had to give priority to one element of the pair which seemed unjustified.

I looked up ordered fields to get a counterpoint. Is a necessary property of an ordered field that each element reduces to a unique scalar...or am I oversimplifying?


PS. I didn't know you could use latex notation on boards, excuse the poor formatting in the first post!

RoundTower

yes, any ordered field is isomorphic to some subset of the real numbers.