This normed group of rationals $|\cdot|:\Bbb{Q}^{\times} \to \Bbb{Z}$, $|\cdot| = $ the minimum number of generators needed, might be a homomorphism. Easy background lemma: The image set of the function $f(x) = \dfrac{x -1}{x + 1}$ generates $\Bbb{Q}$ multiplicatively?
The subgroup generated by $S_a = \{\dfrac{n - a}{n + a}: n \in \Bbb{Z}, n \neq \pm a \}$ is the whole space $\Bbb{Q}^{\times}$.  Define a norm on $\Bbb{Q}^{\times}$, for each $a \in \Bbb{Z}, a \neq 0, \ |x|_a = \min_{k \geq 0} \{ s_1, s_2, \dots s_k \in S_a : s_1 \cdots s_k = x\}$.
Then $|x y|_a \leq |x|_a + |y|_a$.  Proof. Suppose that $x = s_1 \cdots s_k, \ y = s'_1 \cdots s'_r$.  Then clearly $xy = s_1''\cdots s_{r+k}''$ at most.

Clearly $|x|_a = 0 \iff x = 1$ by definition of empty product.  And clearly we have $|-x|_a = |x|_a$ for all $x  \in \Bbb{Q}^{\times}$.
My question is can we redefine so that $|\cdot|$ is a signed quantity and also a homomorphism into $\Bbb{Z}$, that is the above inequality is actually an equality?
I ask because that would make sense of why it seems that $|\dfrac{x}{y}|_a = |x|_a - |y|_a$ at times.  Take the case $a = 1$, odd $x,y$:
         x-4 x-3 x-2 x-1  x  x+1 x+2
    ...  x-2 x-1  x  x+1 x+2 x+3 x+4  ...

So that $\dfrac{-1}{x} = \dfrac{x -2}{x} \dfrac{x-4}{x-2} \cdots \dfrac{-1}{1}$ and so $|x|_1 \leq \dfrac{x-1}{2}$ and so $|y|_1 \leq \dfrac{y - 1}{2}$ but say $y \gt x$.  Since both are odd, we have that $y = x + 2k$ and $\dfrac{x+2k}{x+2(k+1)}$ occurs in the list after $\dfrac{x}{x+2}$ in fact it occurs $2k$ steps ahead so that $|\dfrac{x}{y}|_1 = k = \dfrac{y - x}{2}$ but then $|\dfrac{x}{y}|_1 = \dfrac{x - y}{2} = \dfrac{x-1}{2} - \dfrac{y - 1}{2} = |x|_1 - |y|_1$.
How can we prove in general that it is indeed a group homomorphism from $\Bbb{Q}^{\times}_{\gt 0} \to \Bbb{Z}$?
 A: $$
\dfrac{n}{n + 2a} = \dfrac{n + i}{n + 2a + i} \iff n^2 + 2an + ni = n^2 +2an + ni + 2ai \iff \\ ai = 0 \iff i = 0
$$
since we assumed $a \neq 0$.  Therefore the generators are unique. The only way to generate a free abelian group on a countable number of generators is to do so isomorphically with $\Bbb{Q}^{\times}_{\gt 0}$.  Of course we can extend this isomorphism to all of $\Bbb{Q}^{\times}$ by setting $\varphi(-x) =-\varphi(x)$.
Therefore let $\varphi : \pm p \mapsto \pm\dfrac{n_{p} - a}{n_{p} + a}$ be an assignment of each $\pm p \in \Bbb{P}$.  I don't know of any natural one other than maybe $p_n \mapsto \dfrac{n - a}{n+a}$ but it doesn't matter for here.
Then since unique factorization (with respect to primes) happens in $\Bbb{Q}^{\times}$ as long as the fraction is reduced, we also have unique factorization for reduced fractions (with respect to generators in $S_a$).  Therefore there is only one way of multiplying to each $x \in \Bbb{Q}^{\times}$ up to sign and we have that $|x|_a = \Omega(\varphi^{-1}(x))$.
The sign though is very confusion.  It is safer to say we have a homomorphism $|x|_a : \Bbb{Q}^{\times}_{\gt 0} \to \Bbb{Z}$ from the positve rationals to the integers that equals the exact number of generators comprising $x$'s numerator minus the exact number comprising its denominator.
