The image set of the function $f(x) = \dfrac{x -1}{x + 1}$ generates $\Bbb{Q}$ multiplicatively? Define $f(x) = \dfrac{x - 1}{x+1}$.
Compute the composition $f\circ f(x) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x - 1}{x+1} +1} =\dfrac{\dfrac{x - 1 -x -1}{x+1}}{\dfrac{x-1 + x + 1}{x+1}} =\dfrac{-2}{2x} = \dfrac{-1}{x}$
$$
f^3(x) = \dfrac{-1}{f(x)} \\
f^4(x) = \dfrac{-1}{f^2(x)} = x
$$
Thus $f : Z= \Bbb{Z}\setminus\{-1\} \hookrightarrow \Bbb{Q}^{\times}$ is an injection of sets, because $f$ is invertible.  Its image space is such that for all $x,y \in f(Z)$ we have that $xy = 0$ or $xy \notin f(Z)$.
I was wondering, can we generate $\Bbb{Q}$ multiplicatively using $f(Z)$?
$$
\dfrac{0}{1}, \dfrac{1}{3}, \dfrac{1}{2}, \dfrac{3}{5}, \dfrac{2}{3}, \dfrac{5}{7}, \dfrac{3}{4}, \dfrac{7}{9}, \dfrac{4}{5}, \dfrac{9}{11}, \dots
$$
are the elements for non-negative $x \in \Bbb{Z}$.

Let $H$ be the subset of $\Bbb Q$ which can be written as a product of finitely many elements of $f(Z)$. We want to show that $H = \Bbb Q$.
We have:

*

*$0 \in H$ because $0 = f(1)$;

*$-1\in H$ because $-1 = f(0)$;

*$\frac 1 n \in H$ for any positive integer $n$ because $\frac 1 n = f(2n - 1)f(2n - 3)\cdots f(3)$.

*$n \in H$ for any positive integer $n$ because $n = f(1 - 2n)f(3 - 2n)\cdots f(-3)$.

These obviously imply the result.
