A rational number that is a infinite product of distinct irrational numbers?

Question

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.

Answer 1

In this product, the integers in the radicals are primes:$$\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{5}}\cdots$$ It converges to $0$. Any finite subproduct is the reciprocal of the square root of a square-free integer, and therefore irrational. None of the factors are rational multiples or inverses of another, as specified.

Answer 2

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.

Answer 3

Let $c$ be any rational number you like. The below describes a construction of a sequence whose product converges to $c$, with its finite subsets meeting your criteria. It is not explicit, because it asks you to pluck numbers from the complement of a countable set within an uncountable set.

Start with some irrational number, $x_1$ in $(c-1,c+1)$. Note that you have a set $\{x_1\}$ and all of its sub-products are irrational, and the product of its elements is within $1$ of $c$. Inductively, we will grow this set.

Assume you have $n-1$ irrational numbers $\{x_1,\ldots,x_{n-1}\}$ with no sub-product rational, and the product of its elements is within $\frac{1}{n-1}$ of $c$. Let $p_{n-1}$ be that product. Consider the interval $I_n=\left(\frac{cn-1}{np_{n-1}},\frac{cn+1}{np_{n-1}}\right)$. (Or reverse the order of those endpoints if $p_{n-1}$ is negative.) Find an $x_n$ in that interval that is algebraically independent over $\mathbb{Q}$ from $\{x_1,\ldots,x_{n-1}\}$. This is possible because the algebraic extension of $\mathbb{Q}$ by $\{x_1,\ldots,x_{n-1}\}$ is countable, and $I_n$ is uncountable.

Then you have $n$ irrational numbers $\{x_1,\ldots,x_{n}\}$ with no sub-product rational, and the product of its elements is within $\frac{1}{n}$ of $c$. That is, you have the same features of a set that is one element larger. So by induction, you have an infinite sequence where every finite subset meets your criteria. And since $\frac{1}{n}\to0$, the infinite product of this sequence converges to $c$.

Answer 4

Let $c$ be any nonzero rational number.

Let $P_n$ be the $n^\text{th}$ prime number.

Define a sequence of nonzero rational numbers $r_n$ recursively so that $$r_1r_2\cdots r_n=\frac c{\left\lfloor\sqrt{P_1P_2\cdots P_n}\right\rfloor}.$$

Let $a_n=r_n\sqrt{P_n}$. Then $$\lim_{n\to\infty}a_1a_2\cdots a_n=\lim_{n\to\infty}c\cdot\frac{\sqrt{P_1P_2\cdots P_n}}{\left\lfloor\sqrt{P_1P_2\cdots P_n}\right\rfloor}=c.$$

Thus the infinite product $\prod_{n=1}^\infty a_n$ converges to $c$. The product of any nonempty finite set of distinct terms is irrational, being a nonzero rational multiple of the square root of a product of distinct primes.