Let $x$ and $y$ be real numbers. If $xy=0,$ then $x=0$ or $y=0.$

Question

When solving this proof would it be okay to say

Proof: Let $x,y \in \mathbb{R}.$ We will do a proof by case

• Case 1. If $x=0,$ then $xy=0y=0$
• Case 2. If $y=0,$ then $xy=0x=0$

In each case, $xy=0.$

Or would you prove by contradiction saying:

If $x\neq0$ and $y\neq0$ so $x^{-1}$ and $y^{-1}$ must exist.

Then $\frac{1}{xy}xy=0\frac{1}{xy}.$

$1=0$ which is a contradiction.

Answer 1

Proving by cases is a valid way.

Case I: when only $x=0$

Case II: when only $y=0$,

Case III: when both $x=0$ and $y=0$.

Answer 2

For a complete proof we can consider four cases

• $x=0 \land y\neq 0$
• $y=0 \land x\neq 0$
• $x=0 \land y=0$
• $x\neq 0 \land y\neq 0$

to conclude that

$$xy=0 \iff x=0 \quad \lor \quad y=0$$

and then of course

$$xy=0 \implies x=0 \quad \lor \quad y=0$$