Irrationals forming rationals

Question

Can we obtain every rational number from the multiplication of two irrational numbers? If not, which ones can we not obtain?

Answer 1

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.

Answer 2

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.

Answer 3

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.