Ordering on complex numbers

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?

• the inequality $4-3i<4+4i$ is, indeed, meaningless. – Ittay Weiss Sep 13 '13 at 19:56
• 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
• @IttayWeiss But $-3i < 4i$? – Don Larynx Sep 13 '13 at 19:57
• 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

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 $\Bbb C$.
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 $i^2 \not\geq 0$.