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In the literature is there a terminology for the following property of rational numbers;

A rational number in $\mathbb{Q}$ whose reduced form is in the form $\frac{p}{q}$ for two primes $p,q$.

By prime we mean $\pm p$ for a positive prime $p$.

Lets call a rational number with this property, a pure rational number.

Is the set of all pure rational numbers a dense subset of $\mathbb{R}$?

So if twin prime conjecture is true then "1" is an accumulation point for the set of pure rationals.

Let $\alpha$ be an irrational number is the following set a dense subset of $\mathbb{R}$? $$\{ p+q\alpha \mid p ,q \quad\text {are prime integers}\}$$

Edit: I realized that my question is similar to the following MO post but I did not check if the second part of my post can be answered by this MO post:

Using Quotient of Prime Numbers to Approximation Reals

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    $\begingroup$ No twin prime conjecture needed, the Prime Number Theorem is all you need to prove the "pure" rationals are dense in the reals. I'm pretty sure the question has been asked & answered on mathstack. See also mathoverflow.net/questions/477989/… $\endgroup$ Commented 21 hours ago
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    $\begingroup$ The two questions you ask are very different. The former is a fairly standard exercise, the latter seems much harder (and largely unrelated to pure rationals...). It will probably follow from existing results on prime exponential sums, but I'm not well versed in those. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ @GerryMyerson Thank you for sharing the interesting post by Stefen Kohl. I am not sure that post gives an answer to my second question. $\endgroup$ Commented 18 hours ago
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    $\begingroup$ @GerryMyerson your suspicion is correct: math.stackexchange.com/questions/1013414/… $\endgroup$ Commented 17 hours ago
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    $\begingroup$ Given the responses and closure all addressing the first question, I would suggest asking the second question separately in a new question. I believe that one to be considedably less trivial, although it is likely to exist in the literature, so I'd suggest doing a more thorough search before asking. $\endgroup$ Commented 11 hours ago

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As pointed out by Wojowu and Emil Jeřábek, the prime number theorem implies that the OP's set is dense in $(0,\infty)$. Equivalently, if $(p_n)$ denotes the sequence of prime numbers, then the fractions $p_m/p_n$ with $m<n$ form a dense subset of $(0,1)$. In fact more can be proved along these lines: if $n$ is large, then the fractions $p_m/p_n$ with $m<n$ are approximately equidistributed in $(0,1)$. For a proof, see my response here.

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The set of pure rationals is indeed dense in the reals. Suppose we seek a pure rational between $rs^{-1}$ and $rs$ with $r>0, s>1$. Choose integers $m,n,k$ with $$rs^{-1/3} < \frac{m^3}{n^3} < rs^{1/3}$$ $$\frac{(k+1)^3}{k^3}<s^{1/3},$$ $$k>e^{e^{34}}.$$ Choose a prime $p$ between $(mk)^3$ and $(mk+1)^3$, and choose a prime $q$ between $(nk)^3$ and $(nk+1)^3$, which must exist by a result of Ingham. Then $p/q$ will indeed be in the desired range.

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    $\begingroup$ Ingham's results are way overkill. It gives you prime gaps which are $O(p_n^{2/3})$, while all you need is $o(p_n)$ which is easy from PNT. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ @Wojowu, I take your point. On the other hand, the Prime Number Theorem was proved in 1896 and Ingham’s result in 1937, so one could say with similar timeframes (en.wikipedia.org/wiki/Ballpoint_pen#Origins) that using a quick-drying ballpoint pen like Biro’s is overkill when all you need is the earlier versions that wrote well on leather. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ An amusing analogy, but I bring that up as a matter of accessibility so to speak - PNT is much easier and more well known than Ingram's work, so it's good tp acknowledge its sufficiency. $\endgroup$ Commented 18 hours ago
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    $\begingroup$ In science, it is a good thing to strive for the simplest explanation (cf. en.wikipedia.org/wiki/Occam%27s_razor and en.wikipedia.org/wiki/Proofs_from_THE_BOOK). So deducing a result from a simpler theorem (or fewer axioms or fewer hypotheses) is a desirable thing, unless the proof becomes significantly more complicated. $\endgroup$ Commented 14 hours ago
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It is stated in the following paper

ISTVÁN MEZŐ, A class of dense subsets on the real line, The Mathematical Gazette, Vol. 98, No. 542 (July 2014), pp. 332-334 (3 pages). https://www.jstor.org/stable/24496675

that:

Let $(a_n)$ and $(b_n)$ be two positive integer sequences which tend to infinity, and then let $$D=\{\pm \frac{a_n}{b_m} \;|\; n,m=1, 2, 3, \dots\}.$$ Then $D$ is dense in the real line regardless of the actual forms of the sequences $(a_n)$ and $(b_n)$.

This will positively answer your first question:

Is the set of all pure rational numbers a dense subset of $\mathbb{R}$?

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    $\begingroup$ That can't be right. Let $a_n=b_n=2^n$, then $a_n/b_m$ is a power of two, and those are definitely not dense in the real line. $\endgroup$ Commented 21 hours ago
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    $\begingroup$ This sounds wong. How can we get close to $2/3$ if $a_n=b_n=2^n$?Ah, the same comment as @GerryMyerson $\endgroup$ Commented 21 hours ago
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    $\begingroup$ @Giorgio, beat you by twenty seconds. $\endgroup$ Commented 20 hours ago
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    $\begingroup$ This works if, e.g., $a_n$ and $b_n$ are two unbounded real sequences such that $a_{n+1}/a_n$ or $b_{n+1}/b_n$ tends to $1$ (which is what the prime number theorem gives, as noted elsewhere by Wojowu). $\endgroup$ Commented 18 hours ago
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    $\begingroup$ I looked at the proof in the quoted article. The mistake is around the end: "$[e,f]$ is wide enough to contain an element from the sequence $(b_n)$". $\endgroup$ Commented 6 hours ago

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