Do we know a non-trivial example of irrational $\times$ irrational $=$ rational?

Question

By non trivial I mean something else than $a \times \frac{b}{a}$ and than $\sqrt{a}\times \lambda \sqrt{a}$.

Answer 1

HINT: use that $$\sqrt[3]{625}\cdot \sqrt[3]{25}$$

Answer 2

$$ (b-\sqrt{a})(b+\sqrt{a}) = b^2-a, $$ perhaps? The two solutions to the quadratic equation with rational coefficients always have rational product: these two numbers are solutions to $x^2-2bx+(b^2-a)=0$.

Answer 3

It is still not known whether $\pi e$ is rational or irrational, so if it was proven that $\pi e$ is rational that could be considered a non-trivial example. In general it's hard to prove irrationality of a number hence it's hard to come up with non trivial examples.

Answer 4

If $$a\times b = r,$$ where $a,b$ are irrational, $r$ is rational,

then one can consider arithmetic mean $m = \dfrac{a+b}{2}$.

According to Vieata's Formulas, there exists quadratic equation $$x^2-2mx+r=0$$ with solutions $x_1=a, x_2=b$. And these values have form $$ a = x_1 = m-\sqrt{m^2-r},\\ b = x_2 = m+\sqrt{m^2-r}. $$

So, you can choose as weird as you wish irrational value $m$, but $a,b$ can be expressed anyway in this simple way via $m$ and $r$ (using radicals).

Examples:

$$ \left(\sqrt[4]{3}-\sqrt{\sqrt{3}-1}\right) \times \left(\sqrt[4]{3}+\sqrt{\sqrt{3}-1}\right) =1; $$

$$ \left(\pi-\sqrt{\pi^2-\dfrac{3}{7}}\right) \times \left(\pi+\sqrt{\pi^2-\dfrac{3}{7}}\right) =\dfrac{3}{7}; $$

$$ \left(2+\sqrt[3]{3}-\sqrt{4\sqrt[3]{3}+\sqrt[3]{9}}\right) \times \left(2+\sqrt[3]{3}+\sqrt{4\sqrt[3]{3}+\sqrt[3]{9}}\right) =4; $$

$$\left(\tan(3\pi/7)-\sqrt{\dfrac{2\tan(3\pi/7)}{\tan(\pi/7)}}\right) \times \left(\tan(3\pi/7)+\sqrt{\dfrac{2\tan(3\pi/7)}{\tan(\pi/7)}}\right)= 1.$$

Hope not all of them are trivial ;)

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Note: since $m$ was chosen irrational, then both $a,b$ are irrational.

(if both $a,b$ are rational, then their arithmetic mean is rational)

(if one of $a,b$ is rational and other is irrational, then their product cannot be rational)