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Math Help - GCD and Cauchy sequences of rationals

  1. #1
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    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?
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  2. #2
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    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 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.
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