Completion of a prime-factor-based metric space over $\mathbb{Q}$

Question

For fun, I am considering the completion of metric space defined using prime factorizations which is somewhat similar to the $p$-adic construction.

Consider any two positive rationals: $$ \begin{aligned} x &= p_1^{n_1}p_2^{n_2}... \\ y &= p_1^{m_1}p_2^{m_2}... \end{aligned} $$ expressed in terms of their unique prime factorizations ($p_k$ is the $k^{th} $ prime).
Then, assign a distance function such that $$ D(x,y) = \sum |n_k-m_k|\log (p_k), $$ where the $\log$ base lies in the interval $(1,2)$ to ensure positivity of the log term.
This may also be expressed with the convention that $ x • y = \prod p_k^{|n_k-m_k|}$ so that $$D(x,y) = \log(x•y).$$

Note that, since logs of different rationals are linearly Independent, it follows that $$ x\neq y \implies D(x,y)\neq 0,$$ which satisfies Axiom 2 of definition of a metric space. With Axioms 1 and 3 being obvious, we should only prove the triangle inequality.

Let $ x,y,z \in \mathbb{Q}$ and adopt the convention $ z := \prod p_k^{l_k}$ so that we can show that $ D(x,z) \leq D(x,y) + D(y,z)$ generally holds.

Proof: For any $k$ $$ |n_k-l_k| \leq |n_k-m_k| + |m_k-l_k|$$ by the standard triangle inequality. Therefore, multiplying by $\log p_k$ and summing over $k$, we conclude the proof. Note all summands are positive.

Awesome. Thus, $(\mathbb Q^+, D)$ is a metric space...
Except that nonpositive numbers are not defined here!

Question 1) We could just use $\mathbb{Q}^+$, but I wonder if the definition can be extended to include zero and negatives without contradicting the axioms of a metric space.

Regardless of the answer to Question 1, we also may ask

Question 2) Because it is a metric space, it has a completion $X$. What is $X$? Is it isometric to well-known metric space like $\mathbb{R}$ or is it a different animal? What can we say about its topology?

And finally,

Question 3) Can the definitions of any of the main operations on rationals (such as the four basic operations, exponents, etc.) be extended to $X$?

I have considered the answers to all these questions but have not yet come up with a viable path toward a conclusion. Any ideas or guidance are appreciated!

Answer 1

• Question 1. $\mathbb Q^{\times} \simeq \mathbb Q^+\times\{\pm 1\}$. Here $\times$ can be considered as Cartesian product of sets. So given any metric on $\{\pm 1\}$ (e.g. $d(1, -1)=1$), we can equip $\mathbb Q$ with any product metric.

• Question 2. For any distinct $x,y\in\mathbb Q^+$, we have $d(x,y)\ge \log 2$, topologically it's just discrete, and any Cauchy sequence must eventually stabilize, so the completion is simply itself.

• Question 3. It's a discrete topology, so do whatever you want, it's going to be continuous, as long as it makes sense algebraically. However, it's perhaps better understood $\mathbb Q^{\times}$ as the free abelian group with infinitely but countably many generators. That is, $(\mathbb Q^{+}, \cdot)\simeq \oplus_{i=1}^\infty\mathbb Z$, and the metric is just a product metric of componentwise metrics $d(a,b)=|a-b|$. And you really don't need to use the linear independence of $\log p_i$'s when they are all positive and the coefficients $|n_k-m_k|\ge 0$. Anyway, this looks fancy but not very interesting, unlike the super interesting $p$-adic stuff.