Irrationals forming rationals Can we obtain every rational number from the multiplication of two irrational numbers? If not, which ones can we not obtain?
 A: Let $r$ be a non-zero rational number.  Then $\pi$ and $r/\pi$ are irrational, and $\pi \cdot \frac{r}{\pi}=r$ is rational.  To see why $r/\pi$ is irrational, suppose for the sake of a contradiction that it were rational, so $r/\pi=r'$, where $r'\in \mathbb{Q}$; in fact, $r'$ is non-zero since the quotient of two non-zero numbers is non-zero.  But then $r=r'\pi$ and $r/r'=\pi$.  The quotient of two rational numbers is rational, so this would show that $\pi$ is rational, which is not the case.  Thus, $r/\pi$ must be irrational.
However, the same cannot be said about $0$, for if $ab=0$, then at least one of $a$ and $b$ must also be $0$.  The above proof that $r/\pi$ is irrational fails because in this case $r'=0$, and we could not divide by it.
A: Commonly believed thought:
The product of an irrational number and an irrational number is irrational.
But, sometimes this is false for instance, the product of multiplicative inverses like $\sqrt{2}$ and $\frac{1}{\sqrt{2}}$ will be 1).
$\sqrt{2}$ was probably the first number known to be irrational.
Example:
Given Irrational Numbers
$$\frac{1}{\sqrt{2}} = 0.70710678118\text{...}$$
$$\sqrt{2}=1.4142135623730950488016887242096\text{...}$$
$$\frac{1}{\sqrt{2}} * \sqrt{2} = 1$$
$\frac{1}{\sqrt{2}} \text{and} \sqrt{2}$ and both irrational numbers but the product of these two irrational numbers will be a rational number.
Conclusion:
The product of two irrational numbers can be either irrational or rational.
A: Take the square root of a rational.  It's likely irrational.  Multiply that square root by itself, you have a rational.  Thus you have a big set of irrational * irrational = rational numbers.
