# Why are the complex numbers unordered?

first post on here!

I have been learning about complex numbers, and how they do not satisfy the trichotomy like real numbers do.

For example, there is no way to say $i<3$, $i>3$, or $i=3$.

But consider this: If $x<y$, then $\sqrt x < \sqrt y$. So let $x=-1$ and $y=9$. Then $-1<9$, so $\sqrt {-1} < \sqrt 9$. In other words, $i<3$.

Can someone explain the flaw in this reasoning?

I imagine that it has something to do with $\sqrt {-1}$ not being real, and so the x,y statement doesn't apply, but I can't think of a way to say why it doesn't apply. It seems like it should, since -1 is real, even if its square root is not.

Thanks!

• So would you say that $-i$ is also $<3$? – lulu Oct 26 '17 at 19:49
• – carmichael561 Oct 26 '17 at 19:51
• Well it depends, do you want only to oder them a set ? Choose any bijection between reals and complexes and you automatically have a total order, however if you hope that your ordering is compatible with arithmetical operations than the answer is you can't (see Peter answer) – Renato Faraone Oct 26 '17 at 19:56
• Lulu, I would say that $$-i = i^3 = (\sqrt {-1})^3 = \sqrt {{-1}^3} = \sqrt {-1} < \sqrt {9} = 3$$ – Sam Oct 26 '17 at 20:05
• I think you already answered your question when you said the statement $x<y \Rightarrow \sqrt{x} < \sqrt{y}$ is a statement about the real numbers not the complex numbers. By the way i think there probably are useful orderings of the complex numbers (for example set inclusion on the complex numbers would give you a partial ordering --but not sure how useful this is). – vkan Oct 28 '17 at 0:37

Suppose, $i>0$ , then since multiplying with a positive number does not change the sign, we have $i^2>0$, but this is false because of $i^2=-1$
Suppose, $i<0$, then since multiplying with a negative number changes the sign, we have $i^2>0$, again a contradiction.
So, neither $i>0$ nor $i<0$ is consistent with the rules of ordered numbers.
• Peter, thanks, this makes sense. Because also if you suppose $i=0$, then $i^2 = -1$, but also $i^2 = 0^2 = 0$, so $-1=0$, which is false. – Sam Oct 26 '17 at 20:08
• @Sam True, but $i=0$ is already a contradiction. – Peter Oct 26 '17 at 20:11