Ordering on complex numbers

Question

So I just learned that complex numbers cannot be ordered from Total ordering on complex numbers.

Now, does this mean that 4 - 3i < 4 + 4i isn't true? I haven't really seen any concrete examples and I don't even get what the proof means, but I know it makes sense - I just can't relate to it.

In short, what are some concrete examples of this theorem? What is also a walkthrough of the theorem?

And what do you mean by some concrete examples of a theorem that claims something does not exist? How is non-existence demonstrated concretely? — Ittay Weiss, Sep 13 '13 at 19:57
that's the proof of the claim, which is given in the link you provide. What is your question then? — Ittay Weiss, Sep 13 '13 at 19:58
No, $-3i$ is not less than $4i$; $-3<4$, but the statement $-3i<4i$ is meaningless. — Brian M. Scott, Sep 13 '13 at 19:59
and $3i<4i$ is just as meaningless as the previous inequality. — Ittay Weiss, Sep 13 '13 at 19:59
@Jossie In my opinion you need to think about the meaning of $<$. Given, $x,y\in \Bbb C$, what does $x<y$ mean to you? — Git Gud, Sep 13 '13 at 20:01

score 16 · Accepted Answer · edited Apr 13 '17 at 12:20

You can order the complex numbers, just not in a way compatible with multiplication in the way we want it to be.

An ordered field is not just a field that is ordered, it's one that has compatibility with the multiplication and the ordering. One of the conditions is that if $a<b$ and $c>0$, then $ac<bc$, and another is that if $a>0$ and $b>0$, then $ab>0$.

This solution at the problem you linked to gives the elementary details of showing that no total ordering is consistent with the multiplication of $\Bbb C$.

But if you don't care about the compatibility of the order with multiplication, then you can order it as a set however you like. One example ordering could be ordering lexicographically by real and imaginary parts.

score 1 · Answer 2 · answered Sep 13 '13 at 20:09

It appears like you are considering the product ordering in your example, i.e. $a + bi \leq c + di$ if and only if $a\leq c$ and $b\leq d$. This is a perfectly fine partial ordering, but it is (1) not total, i.e. not all complex numbers are comparable: $1-i \not\leq -1+i$ and $-1+i\not\leq 1-i$ and (2) not compatible with multiplication: $i\geq 0$, but $i^2 \not\geq 0$.

Ordering on complex numbers

2 Answers2

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