Let $S$ be a set of rational numbers.

For "special" sets $S$, we can ask if $\pi$ can be written as a $\mathbb{Q}$-linear or $\mathbb{Z}$-linear combination of elements from '$\{\tan^{-1}(x): x \in S\}$'.

In particular, let $\hat S$ be the closure of $S$ under both negation and the binary operation $p*q=\frac{(p+q)}{(1-pq)}$.

For any natural number $b \geq 2$, $S_b =\{ \frac{1}{b^k}: k \geq 1 \}$. Prove that $0 \notin \hat {S_b}$.

Note: $1 \notin \hat S$ since any rational number $P/Q$ in $\hat {S_b}$ in non-reduced form satisfies $(P,Q) \in \{(0,1), (0,-1),(1,0),(-1,0)\}$ (mod $b$).

$tan^{-1}(1/2) + tan^{-1}(1/3) = \pi/4$.

Let $b_1,b_2 \geq 2$ be two natural numbers. For which pairs does 1 belong to the closure of $S_{b_1} \cup S_{b_2}$? For which pairs does zero belong to the closure? Are there pairs for which $\pi$ can be written as a $\mathbb{Q}$-linear combination of the arctangents of their negative powers but not as $\mathbb{Z}$-linear combinations?