Does this completed metric space form a semi-ring? Consider the natural numbers with $0$. Let $\hat{d}(a,b) = \log( \frac{ab}{\gcd(a,b)^2})$ for $ab \neq 0$ which is a metric on the natural numbers without $0$ (see Encyclopedia of Distances). Consider the Steinhaus tranform of this metric at the point $p=1$:
$$d(a,b) = \frac{2\hat{d}(a,b)}{\hat{d}(a,1)+\hat{d}(b,1)+\hat{d}(a,b)} = \frac{2\hat{d}(a,b)}{\log(a)+\log(b)+\hat{d}(a,b)}$$
The following are easy to prove:
1) If $m|n$ then $\hat{d}(m,n) = \log(n/m)$
2) $\hat{d}(a,1) = \log(a)$
3) For all $a,b,m$ $\neq 0$: $\hat{d}(ma,mb) = \hat{d}(a,b)$
4) If $\gcd(a,b) = 1, a\neq b$ then $d(a,b)=1$
5) If $a \neq 0$ then the sequence $(a^n)_n$ is a Cauchy sequence and $\lim_{n \rightarrow \infty} d(a^n,a^{n+1}) = \lim_{n \rightarrow \infty} \frac{1}{n+1} = 0$
6) By the triangle inequality we have: $d(a,b) \in [0,1]$
7) We have $\hat{d}(a_1 b_1, a_2 b_2) \le \hat{d}(a_1,a_2)+\hat{d}(b_1,b_2)$
Let $a \oplus b=\gcd(a,b)$. Then $c(a\oplus b) = ca \oplus cb$, hence the natural numbers with zero form a semiring with $\oplus$ as addition and $\cdot$ as multiplication.
My question is this:

If $x_n,y_n$ form Cauchy sequences with respect to $d$. Do the sequences $$x_n \oplus y_n, x_n \cdot y_n$$ also form Cauchy sequence with respect to $d$?
How do I extend $d$ from $\mathbb{N}^2$ to $\mathbb{N_0}^2$?

 A: There are actually no nontrivial Cauchy sequences under this metric. Let $(a_n)_{n=1}^\infty$ be a Cauchy sequence. Then in particular there is some $N$ such that $d(a_m, a_n) < 1/2$ whenever $m, n \geq N$, so $d(a_N, a_n) < 1/2$ for all $n \geq N$. For all such $n$, we have 
$$d(a_N, a_n) = \frac{2\hat{d}(a_N, a_n)}{\log(a_N) + \log(a_n) + \hat{d}(a_N, a_n)}  = \frac{2\log(a_N) + 2\log(a_n) - 4 \log(\gcd(a_N, a_n))}{2\log(a_N) + 2\log(a_n) - 2 \log(\gcd(a_N, a_n))} < 1/2$$
hence after rearranging,
$$\log(a_n) < -\log(a_N) + 3\log(\gcd(a_N, a_n)) \leq 2\log(a_N)$$
(where the last inequality holds since $\gcd(a_N, a_n) \leq a_N$) from which it follows that $a_n < a_N^2$ for all $n \geq N$. Most importantly, this means that the sequence $(a_n)_{n=1}^\infty$ takes only finitely many values. 
Let the set of all values taken by the sequence be $A = \mathrm{range}((a_n)_n)$, and let $D = \min_{i, j \in A, i \neq j} d(i, j)$ be the minimum distance between distinct points in $A$, so necessarily $D > 0$ since $A$ is finite. Then since we can take $N'$ such that $d(a_m, a_n) < D$ for all $m, n \geq N'$, it follows that the sequence must be constant at this point, i.e. all $a_n$ for $n \geq N'$ are equal. 
We've shown any Cauchy sequence is eventually constant, so it trivially follows that if $(x_n)_n$ and $(y_n)_n$ are Cauchy, then the sequences given by $x_n \oplus y_n$ and $x_n \cdot y_n$ are Cauchy. This also means that completion does not change the space.
