Prove: if $a\cdot b = 0$ then, $a =0$ or $b =0$ or $a=b = 0$.

Question

$a$ and $b$ are any 2 numbers. Prove that if $a\cdot b = 0$ then, $a =0$ or $b =0$ or $a=b = 0$.

I have proved it when $a = 0$ and $b\ne0$ or $b = 0$ and $a\ne0$ using inverse property:

$a\cdot b = 0\\ \Rightarrow a\cdot b \cdot a^{-1} = 0 \cdot a^{-1} \\ \Rightarrow(a\cdot a^{-1})\cdot b = 0 \\ \Rightarrow 1\cdot b = 0 \\ \Rightarrow b= 0 $

But this solution doesn't work if $a\ne 0$ because if $a = 0$ then $0\cdot 0^{-1}$ is undefined. However, the equation is true when $b=a= 0$ so, what is the right solution?

Answer 1

Assume $a$ and $b$ are both not $0$. Thus, $d = a^{-1}b^{-1}$ exists and if $ab = 0$, then $1 = abd = 0\cdot d = 0$, a contradiction.