Prove that if a is a 0 divisor of some ring, then for any b of the ring, ax=b cannot have one x that satisfies the equation (i.e. cannot have one solution)
The Attempt at a Solution
It seems to me this isn't even right. Say a is a right 0 divisor but not a left. Then ax=0 has only one solution (namely x=0) so unless all right 0 divisors are left 0 divisors (which I didn't think is true) isnt' this a contradiction of the theorem I'm trying to prove?