>>14168114If
(a) 0 is an additive identity,
(b) 1 is a multiplicative identity,
(c) everything has an additive inverse,
(d) addition is associative, and
(e) distributivity holds, then:
0 = g + (-g) = 1*g + (-g) = (0+1)*g + (-g) = 0*g + 1*g +(-g) = 0*g + g + (-g) = 0*g + 0 = 0*g
Then we conclude, if 0*g = 1, then 0 = 1.
You may be able to relax some of (a) - (e) -- there's only so much effort I am willing to exert here.
