Does there exists a rational number that can be expressed as the product of an infinite amount of distinct irrational numbers without multiplying with an inverse of the irrational numbers or a scalar multiple of them? And no finite sub-product from within the product is rational(a nice suggestion from @amsmath)
Does $\exists n \Bigl(n= \prod_{i=0}^\infty a_i \Bigr)$ and $\frac{1}{a_i}$ is not in the product. Also $k(a_i), k\in \mathbb {Z} $ is not in the product either and each $a_i$ is distinct?
$n \in\mathbb{Q}$
$a\in \mathbb {R}\setminus \mathbb{Q}$
For instance $\sqrt{2}$ and $\frac{1}{\pi}$ could be in the product but not $\pi$ or have $\pi$ but not $\frac{1}{\pi}$. Same thing with $\sqrt{2}$ but not $2\sqrt{2}$ or vice versa or even some other scaled version of $\sqrt{2}$ but not the others.
Best Answer
Take $n=2, a_0=\pi$ then for $i \ge 1, a_i=\sqrt{\frac 2{\prod_{j=0}^{i-1}a_j}}$ In a geometric sense each $a_i$ takes us halfway to $2$. If any of the $a_i$ or one of the partial products were rational we would have a polynomial for $\pi$, which we know is transcendental.