Can the product of infinitely many elements from $\mathbb Q$ be irrational?

I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.

Solution 1:

Yes, it can.

Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$ b_n = \frac{a_n}{a_{n-1}} $$ for $n > 1$.

We then have that $$ b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n. $$

We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.

Solution 2:

Yes, every irrational number is an infinite product of rationals.

We can write an infinite sum of rationals as an infinite product of rationals.

$$\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}$$

For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$

$$ 1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac {1.414}{1.41}\times .....$$