I think the problem with making -i=i goes a little deeper than 2*i=0. Rather than addition, start with the multiplication by i i=-i i*i=-i*i -1=1 Having -1=1 makes a mess out of the complex plane. 1+1=1-1 2=0 The argument below doesn't show i is a divisor of zero, 2*i=0 follows from the fact that 2 is 0. Further, 2*k=0 and 2*k+1=1 so the integers degenerate to 0,1. The rationals to 0,1, or undefined. And so a+bi can take on only 4 values 0, 1, i, or 1+i (with a dense set of undefined values where a or b have a 2 in the denominator) Addition works fine for that set of 4 and forms a group that makes sense, but multiplication... i(1+i)=1+i (1+i)(1+i)=0 Hmm... That makes the set of 4 surviving values isomorphic to Z_4 with the identification of 0 0 1 1 2 1+i 3 i Who'd of thought! mark -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Marc LeBrun Sent: Saturday, November 24, 2007 10:56 AM To: math-fun Subject: Re: [math-fun] The Axiom of Choice for roots of z^2 + 1 Either -i is different from i or it is the same. If we choose to make it different we get the usual system. If instead we choose to make it the same i = -i then we can add i to both sides obtaining i + i = -i + i i + i = 0 2 * i = 0 making i a divisor of zero. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun