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I have heard that the Complex Numbers do not form an ordered field, so you can't say that one number is larger or smaller than another. I've always been fine with this, but I recently wondered what happens if one describes size as the distance along a space-filling curve? If the real numbers can be ordered, and you can map a curve through all points in 2D space, then what prevents you from ordering the points purely by how far along the curve the points are? Each point should be a unique distance, and the distance continuously increases (as far as my understanding goes). Is there perhaps some difference between being "ordered" and being an "ordered field" that I am missing? I know the basics about why they differ, but I lack formal math training above Algebra 2... Everything else (up to most of Real Analysis) I have studied on my own, so my knowledge is spotty in areas like this.

Brevan Ellefsen
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    The distance would generally be infinite, and space-filling curves must be self-intersecting, so you'd have to deal with that, too, but you could probably get an ordering relation out of it if you tried. It's just that "ordered field" means more than "a field with an order on it"; the ordering and the field structure have to be compatible in a specific way that won't hold here. – user2357112 Jan 21 '16 at 01:47
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    @user2357112 that comment actually cleared things up for me more than the answers... Thank you!! – Brevan Ellefsen Jan 21 '16 at 02:40

3 Answers3

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You can of course put a total order on the complex numbers (an example easier than your space-filling curve: simply take the lexicographic order on $\mathbb{C}\cong\mathbb{R}^2$). However, if you are studying the complex numbers as a field, then you'd like your order to be in some sense compatible with sum and multiplication. For example, you want

  • if $a<b$, then $a+c<b+c$
  • if $a<b$ and $c>0$, then $ac<bc$

and some others. However, this is impossible to obtain in $\mathbb{C}$. Sadly, I don't know the proof of this fact. I would be interested in seeing it though.

Daniel Robert-Nicoud
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    The easy proof of the fact is to ask whether $i \lt 0$ or $i \gt 0$. In either case, you should have $i^2=-1 \gt 0$ – Ross Millikan Jan 20 '16 at 19:55
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    Ah right, you show that a square is always $>0$ and derive a contradiction, right? Thanks! – Daniel Robert-Nicoud Jan 20 '16 at 20:03
  • @RossMillikan using the above definition 2, we assume $0<i \implies 0(i)<i(i)$ this leads to $0<-1$, a clear contradiction. However, what choice of $c$ can we use if we assume $i<0$? I can't seem to find a $c>0$ that yield contradiction without using something like $-i$, which I'm not satisfied is justified here. I tried using $i^4=1$ but found no success – Brevan Ellefsen Jan 21 '16 at 02:44
  • @Brevan Ellefsen: there is a counterpart to the second bullet for $c \lt 0$, then $ac \gt bc$ and $i^2$ fails – Ross Millikan Jan 21 '16 at 03:45
  • @BrevanEllefsen: Suppose $i < 0$. Using the first bullet with $a=i$, $b=0$ and $c=-i$, conclude $0 < -i$. Then apply the second bullet with $a=0$, $b=c=-i$ and get $0 < (-i)(-i) = -1$. – Nate Eldredge Jan 21 '16 at 04:27
  • @NateEldredge yes, I already proved it this way but note in my comment above I wished to avoid referencing any other complex number than $i$ (including $-i$). I did however manage to prove the counterpart Ross mentions using the two properties above (about 15-20 steps formally) and then used that to solve my problem. The only problems I ran into were assuming that $a=a$ and that you can make a substitution $u=-c$. I have no idea how to formally prove that these are valid, and I feel doing so would lead me down that endless loop of lower fundamentals (although set theory might suffice) – Brevan Ellefsen Jan 21 '16 at 04:36
  • @RossMillikan thank you!! Using that hint I was able to prove the counterpart formally using the two properties above, and used that to finish my proof without directly referencing any other complex numbers other than $i$. Your help was most appreciated! – Brevan Ellefsen Jan 21 '16 at 04:38
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    @BrevanEllefsen: You need the field axioms. A basic property of the complex numbers is that they are a field, so the axioms listed on that page are satisfied. Among those properties is that every element $z$ has an additive inverse, which we might denote $-z$. So if there is an $i$ then there is a $-i$, and they satisfy $i + (-i) = 0$, which is really all that I used. – Nate Eldredge Jan 21 '16 at 04:39
  • @NateEldredge I'm aware it's a valid proof... My intentions here were somewhat unrelated to the main question I posted (hence why I asked in comment section). It's not that I disregard the proof you use in any sense; I merely wished to gain insight into the problem by working it another way (which I was perhaps not clear on above!!) – Brevan Ellefsen Jan 21 '16 at 05:00
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You can definitely put an order $\prec$ on the complex numbers. In fact, the order can be a well-order, by the well ordering theorem. The question is whether such an order "is useful."

Here are some things we probably want in the order $\prec$:

  • if $0 \prec \alpha, \beta$, then $0 \prec \alpha \beta$.
  • if $\alpha, \beta, z, w$ are complex with $\alpha \prec \beta$ and $z \prec w$, then $\alpha + x \prec \beta + y$.

In fact, an order on a field (such as the complex numbers) that satisfies these properties gives us an ordered field. See Ross Millikan's comment for why you cannot put such an order on $\mathbb C$.

Community
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Snow
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  • The ordering of an ordered field is never a well ordering, so "finding" a well ordering (using the axiom of choice) is rather beside the point. Also defining a (totally) ordered field is not the same as defining a "total order on a field", so using that phrase is not very appropriate here. – Marc van Leeuwen Jan 21 '16 at 05:58
  • Thanks for pointing that out. "Total order" was misleading. – Snow Jan 21 '16 at 13:22
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In an ordered field the order has to be compatible with the algebraic structure; in particular it should be translation invariant (if $a<b$ then $a+c<b+c$).

Justpassingby
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