This is something I've been struggling to understand since the past few days. Let's take an example:
Prove/Disprove: $(x+1)^2 = (x+1)^3$ for all real values of $x$.
Proof:
Let us assume the opposite, i.e., $(x+1)^2 \neq (x+1)^3$.
Now, we can split this into two inequalities:
$(x+1)^2 > (x+1)^3$ or $(x+1)^2 < (x+1)^3$
Multiply both sides by $0$ in both inequalities:
$0 > 0$ or $0 < 0$
which are both false, therefore ($x+1)^2 = (x+1)^3$ for all real values of $x$, which is b
While you can multiply both sides of an equation by the same thing and it remains an equality, $$x = y \quad \Rightarrow \quad ax = ay,$$ the same is not true for inequalities, that is, $$x < y \quad \nRightarrow \quad ax < ay.$$ What happens depends on the sign of the thing you're multiplying by. For example, if $x < y$, then $2x < 2y,$ but $-2x > -2y$ (think about it), and, coming to the point of the question, $0x = 0y$ (obviously).