Show $xy\neq0$ is the same as $x\neq0 \wedge y \neq0$

Question

I have to show: $$xy\neq0 \Leftrightarrow x\neq0 \wedge y \neq0 $$

I think I can "simplify" it to this: $$xy=0 \Leftrightarrow x=0 \vee y=0 $$

Since $a\cdot0=0$ is an proven theorem, I can show: $$x=0 \vee y=0 \rightarrow xy=0 $$

But that is just one of the directions. How can I use the field axioms to show this theorem?

Answer 1

It helps to spell out what you're trying to show.

Claim: If $xy = 0$, then $x = 0$ or $y = 0$ (or both).

Proof: We'll consider two cases.

If $x = 0$ , then there is nothing to show (we already have one of $x$ and $y$ equal to 0, which is what we want).

If $x \neq 0$ , we hope to establish that $y = 0$ (since the conclusion we desire is that at least one of the two is equal to 0). Since $x \neq 0$ and we are in a field, the multiplicative inverse of $x$ exists.

With $x^{-1}$ in hand, how can you isolate $y$ in the equation $xy=0$ ? What do you learn about $y$ when you do so?

Answer 2

Hint: Suppose that $xy=0$. If $x=0$, then certainly $x=0\vee y=0$. Otherwise, $x\neq0$. What do we know about non-zero elements of a field that would let us conclude that $y=0$?

Answer 3

This is much the same as the other answers, but you do not need to consider cases. If you want to show that $p \implies q$, you can assume $p \wedge \neg q$ is true and derive a contradiction.

Suppose $x \neq 0$, $y \neq 0$, and assume $xy = 0$. Then, $x$ and $y$ have inverses. Hence, $xy$ does too (this needs formal justification). But,...(this is where you come in).