Wednesday 10 October 2018

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 $xSo 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!

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...