GCD and Cauchy sequences of rationals

Let $\displaystyle P$ be the set of all prime numbers. For any $\displaystyle a,b\in \mathbb{Q}$ with $\displaystyle a = \prod_{p\in P}p^{a_p}, a_p \in \mathbb{Z}$ and $\displaystyle b=\prod_{p\in P}p^{b_p}, b_p \in \mathbb{Z}$, I will use the standard GCD $\displaystyle (a,b) = \prod_{p\in P}p^{\min(a_p,b_p)}$. Let $\displaystyle a: \mathbb{N} \to \mathbb{Q}$ be an arbitrary Cauchy sequence. For notation, $\displaystyle a(n) = \prod_{p\in P}p^{a_p(n)}$. We will call the sequence $\displaystyle a$ GCD-Cauchy if it is Cauchy and $\displaystyle \lim_{n \to \infty} a(n) = \lim_{m\to \infty} \left( \lim_{n\to\infty} {\left(a(n),a(m) \right)} \right)$ (all limits must exist, obviously). Given an irrational number $\displaystyle x$, does there exist a GCD-Cauchy sequence of rationals whose limit is $\displaystyle x$?

Re: GCD and Cauchy sequences of rationals

I am thinking the answer is yes, and if I can show that for any pair of distinct primes $\displaystyle p,q$, the set $\displaystyle \left\{p^aq^b \mid a,b\in \mathbb{Z} \right\}$ is dense in the reals, then let $\displaystyle P = \{p_1,\ldots , p_n, \ldots \}$. Now, we define $\displaystyle a(0) = 1$ and for each $\displaystyle n>0$, choose $\displaystyle i,j\in \mathbb{Z}$ such that $\displaystyle \left| p_{2n-1}^i q_{2n}^j - \dfrac{x}{a(n-1)} \right| < \dfrac{1}{n \cdot a(n-1)}$. Then, let $\displaystyle a(n) = a(n-1) \cdot p_{2n-1}^i q_{2n}^j$. From there, I would know how to show that this is a GCD-Cauchy sequence whose limit is $\displaystyle x$. 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.