From: gadde@sm.luth.se (Erland Gadde)
Newsgroups: sci.math
Subject: Re: Is there order among imaginery numbers? i.e. Is 2i > i?
Date: Thu, 03 Dec 1998 16:13:08 +0100

In article <nnevels-0312980012510001@dialip3-207.mvp.net>, nnevels@mvp.net
(Bonafide Nupe) wrote:

> Please provide a simple explanation

If such an ordering relation < shall satisfy the conditions
(for arbitrary complex x,y, and z):

1) Exactly one of  x<y, x=y, or y<x  holds
2) If x<y, then  x+z<y+z
3) If x<y and 0<z, then xz<yz

then the answer is no, there is no such ordering relation on
the set of complex numbers.

Proof: If there is such an order, then, by 1), either 0<1 or 1<0 
(since certainly 0=/=1). If 0<1, then 2) gives -1<0 (with z=-1 in 2)). 
Similarly, if 1<0, then 0<-1. Hence, either 0<1 or 0<-1 must hold.
If 0<-1, then 3) gives 0*(-1)<(-1)*(-1), that is, 0<1.
We have now proved that 0<1 must hold. Hence, by above, -1<0.
Now, if 0<i, then, by 3), 0*i<i*i, that is, 0<-1.
If, instead i<0, then 0<-i, by 2), and 0*(-i)<(-i)*(-i), that is, 0<-1
in this case too. Since i=/=0, 1) gives 0<-1. But -1<0.
This contradicts 1). Hence, there is no such ordering relation.

-- 
Erland Gadde
Department of Mathemetics
Luleå University of Technology
Sweden