when product of irrational numbers = rational number?
I'm not sure what you're hoping for in terms of an answer, but: exactly when one is a rational multiple of the other's reciprocal!
Ex: $\sqrt{2}\times {3\over \sqrt{2}}$ is rational, but $\sqrt{2}\times{\pi\over\sqrt{2}}$ is irrational.
You might find this unsatisfying - unfortunately, I'm not really sure a satisfying characterization exists!
A somewhat abstract necessary condition can be stated using the language of fields: $\mathbb{Q}(x)=\mathbb{Q}(y)$. However, this is definitely not sufficient, since e.g. $\pi^2$ isn't rational.
For each irrational number, a, there exists a countably infinite number of irrational numbers, b, such that a ⋅ b is rational. These numbers are exactly the rational numbers except zero divided by a. Call this set B.
PROOF -- SUFFICIENT
Let q be a rational number other than zero. Then b = q / a is irrational. However, a * b equals a * q / a which is just q, a rational number. Therefore, every number in B is an irrational that produces a rational number when multiplied by a.
PROOF -- NECESSARY
Consider an irrational number, b, which, when multiplied by a, produces a rational number. Call the result q. Then b = q / a. But q is rational so q / a is in set B.
Q.E.D.
How about this?
$ab$ is rational iff either of $a,b$ is zero or the ratio $a+b : \frac1a+\frac1b$ is rational, for any reals $a,b$.
No products in sight! Heheh...