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

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.

• $a_1a_2\cdot\frac 1{a_1a_2}\cdot\ldots$. You might want to require that no finite subproduct is rational. – amsmath Oct 19 '19 at 3:46
• Let $(x_n)$ be a sequence of irrationals converging to $n$ such that $\frac{x_{n+1}}{x_n}\neq\mathbb Q$ for all $n$. Then $n = x_1\cdot\frac{x_2}{x_1}\cdot\frac{x_3}{x_2}\cdot\ldots$. – amsmath Oct 19 '19 at 3:52
• @amsmath Very clever! Simple but effective. – Don Thousand Oct 19 '19 at 4:01
• @amsmath darn that is clever, how do I discount that type of solution from the product? – yosmo78 Oct 19 '19 at 4:09
• @yosmo78 I don't understand what you mean. – amsmath Oct 19 '19 at 4:12

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.

• This is amazing, and exactly like the product I am thinking of. Can I ask where you found the proof for this? – yosmo78 Oct 19 '19 at 4:41
• You didn't specify that $0$ couldn't be the product, so it was easy to imagine larger and larger irrational denominators doing the job. Square roots of integers are one of the first types of irrational numbers to consider. A little extra thinking goes in to how to make sure they won't have a subset that multiples to a rational number is all. – alex.jordan Oct 19 '19 at 4:43
• Well I am going to have to devise a more rigorous question in the future I guess ;) – yosmo78 Oct 19 '19 at 4:45
• Your infinite product is divergent. It diverges to $0$. By definition, an infinite product is convergent if the sequence of partial products converges to a nonzero finite limit. – bof Oct 19 '19 at 4:57
• en.wikipedia.org/wiki/Infinite_product – bof Oct 19 '19 at 6:01

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.

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 $$c$$.

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.