Need help with a proof that if $xy=0$ then $x=0$ or $y=0$ I need to prove the following statement:
if $xy=0$   then  $x=0$ $or$ $y=0$.
My first idea was to prove the negation of the original problem
if $x\neq0$ and $y\neq0$ then $xy\neq0$
but I still don't know how to prove it rigorously.
 A: Hint: Suppose that $x\ne 0$ and $xy=0.$ Since $x\ne 0,$ then there is a very special real number closely tied to $x$ with regard to multiplication. Check your axioms, and see if you can figure out what it could be. What will this real number allow us to do, since $1y=y$ and $z\cdot 0=0$ for all $z$?
A: Suppose $xy=0$ and suppose $x\ne0$ and $y\ne 0$. Then there exists $x^{-1}$ and $y^{-1}$. Now, you should be able to derive a contradiction.
A: Break it up into cases:
Case 1:  x equals 0.
You are done proof in this case.
Case 2:  x does not equal 0.
Since x does not equal 0, 1/x exists.
Take equation xy=0 and multiply both sides by 1/x:
(1/x)*xy = (1/x)*0
       .. thus, y=0
So in this case we are also done.
General comments:
Say you have a theorem of the form...
If P then A or B
The way to prove this "directly" (meaning without contradiction) is to split it up into cases either using A or B.
That is if you choose to split it up on cases using A, the cases will be:
1) Assume A true.
...... Boom, we already proved that one of A or B is true (i.e., A).
2) Assume A not true.
.... need to do some work to prove that B is true.
So when you become "pro" you usually don't show case 1), you usually just show the proof of 2), like:
Assume x does not equal 0.
Then 1/x exists, so .... , boom y=0.
Done.
