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.