Show $xy\neq0$ is the same as $x\neq0 \wedge y \neq0$ 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?
 A: 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 $y=0$?
A: 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?
A: 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).
