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I have been working through the exercises in Tenenbaum's "Introduction to analytic and probabilistic number theory" book, and I am stumped here (Exercise I.1.6). This is not a homework assignment, but just rather for my own edification. I know Tenenbaum's exercises are usually considered quite hard, so I don't feel as embarrassed to ask this question!

For those not aware (I suspect most experts know this exercise and book inside and out), the homework problem reads as follows:

Set $d_n = p_{n+1} - p_n$.

  1. Show that $p_n \sim n \log n \qquad (n \to \infty)$

  2. $\sum_{1 < n \leq x} d_n / \log x \sim x \qquad (x \to \infty)$

  3. $\liminf_{n \to \infty} \frac{d_n}{\log n} \leq 1 \leq \limsup_{n \to \infty} \frac{d_n}{\log n}$

  4. For each $\alpha > 0$ there exists a sequence of integers $\{n_1, n_2, \dots \}$, increasing in the weak sense, such that $p_{n_j} \sim \alpha j \qquad (j \to \infty)$.

  5. The set of rational numbers of the form $p/p'$, where $p$ and $p'$ are prime is dense in $[0, \infty)$.

My questions are two-fold:

How does part 5 of this exercise follow from part 4?

Are there different proofs (elementary or not) of question 5 that do not follows this route?

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From question 4 you can easily show that for any $\alpha > 0$, there exists a sequence $(q_n, q'_n)$ of prime numbers such that $q_n/q'_n$ converges to $\alpha$.

You just have to take the sequence $\{n_1, n_2, \dots\}$ and use the subsequence of the terms with prime index. So you can build this sequence $\{(p_1, p_{n_{p_1}}), (p_2, p_{n_{p_2}}), \dots \}$. Each pair in this new sequence in composed of prime numbers, and from the fact that $p_{n_j} / j \rightarrow \alpha$ you get that $p_{n_{p_i}}/p_i \rightarrow \alpha$ as $i \rightarrow \infty$.

Therefore $\alpha$ is in the closure of the rational numbers that are quotient of primes. Since it is true for any $\alpha > 0$, this set is dense in $[0, \infty)$.

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  • $\begingroup$ Thank you chandok. Chandru, sorry for the repeat question. $\endgroup$
    – JavaMan
    Feb 1 '11 at 17:53
  • $\begingroup$ @DJC: No problem. $\endgroup$
    – anonymous
    Feb 1 '11 at 17:58

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