Is this a real contradiction? Where is the mistake?

Question

If $a \cdot b=0$, then either $a=0$ or $b=0$.

Proof: 

Assume $a\neq 0$, we prove $b=0$.

$ a = a \cdot 1 = a (b(1/b)) = a \cdot b (1/b) = 0 (1/b) = 0 $

Since we assumed $a \neq 0 $, isn't this a contradiction? Where is the mistake?

Answer 1

The problem here is you wish to prove b = 0, but you inverted b, so you are assuming b is not 0.

Answer 2

You got a contradiction because you assumed both $a,b$ are nonzero but $a.b=0!$