# GCD and Cauchy sequences of rationals

Let $P$ be the set of all prime numbers. For any $a,b\in \mathbb{Q}$ with $a = \prod_{p\in P}p^{a_p}, a_p \in \mathbb{Z}$ and $b=\prod_{p\in P}p^{b_p}, b_p \in \mathbb{Z}$, I will use the standard GCD $(a,b) = \prod_{p\in P}p^{\min(a_p,b_p)}$. Let $a: \mathbb{N} \to \mathbb{Q}$ be an arbitrary Cauchy sequence. For notation, $a(n) = \prod_{p\in P}p^{a_p(n)}$. We will call the sequence $a$ GCD-Cauchy if it is Cauchy and $\lim_{n \to \infty} a(n) = \lim_{m\to \infty} \left( \lim_{n\to\infty} {\left(a(n),a(m) \right)} \right)$ (all limits must exist, obviously). Given an irrational number $x$, does there exist a GCD-Cauchy sequence of rationals whose limit is $x$?
I am thinking the answer is yes, and if I can show that for any pair of distinct primes $p,q$, the set $\left\{p^aq^b \mid a,b\in \mathbb{Z} \right\}$ is dense in the reals, then let $P = \{p_1,\ldots , p_n, \ldots \}$. Now, we define $a(0) = 1$ and for each $n>0$, choose $i,j\in \mathbb{Z}$ such that $\left| p_{2n-1}^i q_{2n}^j - \dfrac{x}{a(n-1)} \right| < \dfrac{1}{n \cdot a(n-1)}$. Then, let $a(n) = a(n-1) \cdot p_{2n-1}^i q_{2n}^j$. From there, I would know how to show that this is a GCD-Cauchy sequence whose limit is $x$. But, this construction relies on that density property I mentioned. I have been trying to think of an argument for it (or against it), but so far I have yet to come up with anything substantial.