when product of irrational numbers = rational number?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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...

Solution 4:

A bit too large for a comment..

For the subset of irrationals being $n$-th roots of rationals:

For (all) primes $p_i$ : consider the numbers being on the form $$q = \prod_{\forall i} {p_i}^{q_i}$$Any root of a rational number could be uniquely represented as $q_i\in\mathbb{Q}$ and then iff the sum of $q_i$ for two numbers $\in \mathbb{Z}$ for all $i$ then their product should be a rational number.

$q_i$ being a negative integer would mean the denominator has a positive power of prime $p_i$ in it's representation and positive would mean the same for numerator.

However there are of course very many irrationals except for the integer roots of rationals so this only expresses a method for a very tiny subset of irrational numbers.

Solution 5:

There cannot be a condition on one of the numbers alone, as for any irrational real $a$, there is at least one $b$ that fits (just take $b=\frac1a$).

Now the necessary and sufficient condition for a pair of numbers is... $a\cdot b\in\mathbb Q$.

You can weaken to get a necessary but not sufficient condition such as $a^2\cdot b^2\in\mathbb Q$.

You can strengthen to get a sufficient but not necessary condition such as $a\cdot b\in\mathbb N$.

This is not of much use.

A vaguely interesting condition is that $a$ and $b$ must be both transcendental or algebraic.