Is every subproduct of rational product is rational?

Question

Assume we have a $$\prod_{p \in P}a_p=R$$ where R is rational. Is it possible to have a $P_1\subset P$ for which: $$\prod_{p \in P_1}a_p=I$$ where I is not rational.

ADDITION 1: where $a_p$ is rational.

ADDITION 2: and what if $0<a_p<1$

Answer 1

The answer is still yes. Consider $$\zeta(2) = \prod_{p \textrm{ prime}}\frac{1}{1 - p^{-2}} = \frac{\pi^2}{6}.$$ Then $\tfrac{\pi^2}{6}$ is a subproduct of $$\prod_{p \textrm{ prime}}\frac{1}{1 - p^{-2}}\prod_{p \textrm{ prime}}(1 - p^{-2}) = 1.$$

For an example with $0 < a_p < 1$, consider

$$\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\frac{8}{9}\cdot\frac{10}{9} = \frac{\pi}{2}.$$ Dividing by $2$, grouping consecutive pairs of terms together and squaring gives $$\frac{8^2}{9^2} \cdot \frac{24^2}{25^2} \cdot \frac{48^2}{49^2} \cdot \frac{80^2}{81^2} \cdots \prod_{n=1}^{\infty}\frac{((2n + 1)^2 - 1)^2}{(2n + 1)^4} = \frac{\pi^2}{16}.$$ So then $$\prod_{p \textrm{ prime}}(1 - p^{-2})\prod_{n=1}^{\infty}\frac{((2n + 1)^2 - 1)^2}{(2n + 1)^4} = \prod_{p \in P}a_p = \frac{3}{8}$$ is rational and $0 < a_p < 1$ but there are two obvious irrational subproducts.