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Let $S \subset \mathbb{R}$ be the set of all real numbers $x$ for which there are infinitely many pairs of prime numbers $p$ and $q$ such that $$ \left|x-\frac{p}{q}\right| < \frac{1}{q^2}. $$ Do the following assertions hold?:

  • The set $S$ is dense in $\mathbb{R}$.

  • The set $S$ has Lebesgue measure $0$.

  • The set $S$ does not contain any algebraic numbers.

Remark: Without the requirement of primality of $p$ and $q$, the set $S$ would contain precisely the irrational numbers.

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    $\begingroup$ I think that $S$ has Lebesgue measure 0. The numbers in $(-m, m)$ well approximated with denominator a fixed prime $q$ and numerator some prime have measure $ 2m\frac{1+o(1)}{q\log(qm)}$ by PNT. Since $\sum_{q \in P} \frac{1}{q\log(q)} < \infty$ by PNT, Borel-Cantelli finishes. $\endgroup$
    – Matthew Bolan
    Commented Sep 1 at 22:14
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    $\begingroup$ About $x = \sqrt{2}$: one has $|\sqrt{2}-\frac{p}{q}| < \frac{1}{q^2}$ if and only if $|2q^2-p^2| < 2\sqrt{2}+(\frac{p}{q}-\sqrt{2})$, so for large $p,q$, the condition becomes $2q^2-p^2 \in \{-2,-1,0,1,2\}$, but $2q^2-p^2$ will be odd, so either $2q^2-p^2 = -1$ or $2q^2-p^2 = 1$. I think $2q^2-p^2 = -1$ can be ruled out: $2q^2 = (p-1)(p+1)$ implies $q \mid p-1$ or $q \mid p+1$, but $p \sim \sqrt{2}q$, so this is impossible. Maybe $2q^2-p^2 = 1$ can be ruled out by working over $\mathbb{Z}[i]$? $\endgroup$
    – mathworker21
    Commented Sep 2 at 13:50
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    $\begingroup$ @mathworker21 For $x = \sqrt{2}$, one finds the solutions $(p,q) = (3,2)$, $(p,q) = (7,5)$, $(p,q) = (41,29)$, $(p,q) = (63018038201,44560482149)$ and $(p,q) = (19175002942688032928599,13558774610046711780701)$, and there is a chance that these are all (if there are more, $p$ and $q$ have more than $1000$ decimal digits). $\endgroup$
    – Stefan Kohl ♦
    Commented Sep 2 at 20:38
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    $\begingroup$ Mathworker21's idea gives an easy proof for $\varphi = \frac{1+\sqrt 5}{2}$. We have $|\varphi - \frac{p}{q}| < \frac{1}{q^2}$ iff $|q^2+pq-p^2| < |\frac{p}{q}-\bar{\varphi}|$. Now when $q$ is large the fact the lefthand side is an odd integer implies $|q^2+pq-p^2| = 1$, so we must in fact satisfy the stronger $|\varphi - \frac{p}{q}| < \frac{1}{2q^2}$. By a theorem of Legendre's, $p/q$ is then a convergent of the continued fraction for $\varphi$, so $q,p$ are consecutive prime Fibonacci numbers. This only happens twice as $F_{n}|F_{2n}$ means $F_{2n}$ is not prime for $n>2$. $\endgroup$
    – Matthew Bolan
    Commented Sep 3 at 3:11
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    $\begingroup$ $S$ is dense in $\mathbb R^+$. The point is that $$S_n=\bigcup_{p,q \in P; q > n} \left(\frac{p}{q}-\frac{1}{q^2},\frac{p}{q}+\frac{1}{q^2}\right)\setminus\{\frac{p}{q}\}$$ is open and dense for each $n$, since for large enough $q$ there is always a prime in $(aq,bq)$ for any fixed $b>a\ge 0$. Thus $S = \bigcap_n S_n$ is dense by the Baire Category theorem. I think the third question is potentially hard, but heuristically I don't think any algebraic numbers should work. $\endgroup$
    – Matthew Bolan
    Commented Sep 3 at 13:09

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I will show $S$ is dense in $\mathbb R^+$ and $S$ is measure $0$.

Let $$I_{p,q} = \left(\frac{p}{q}-\frac{1}{q^2},\frac{p}{q}+\frac{1}{q^2}\right).$$

To see $S$ is dense, note that the open sets $$S_n = \bigcup_{\substack{q\in P \\ q > n}}\bigcup_{p \in P} I_{p,q} $$ are dense, since for fixed $b > a \ge 0$ and any large enough $q$ there is always a prime in $(aq,bq)$ by the prime number theorem. Thus $S = \bigcap_{n=1}^\infty S_n$ is dense by the Baire category theorem. Note that this would work with any exponent besides $2$, and even the analog of Liouville numbers will be dense.

To see $S$ is Lebesgue measure $0$, for $m>0$ write $$ S \cap [0,m) = \bigcap_{n=1}^\infty \bigcup_{\substack{q\in P \\ q > n}} \bigcup_{\substack{p\in P \\ p < mq}} I_{p,q}.$$ Now $$\sum_{q \in P}\lambda\left(\bigcup_{\substack{p\in P \\ p < mq}} I_{p,q}\right) = 2m\sum_{q \in P}\frac{1+o(1)}{q\log{(mq)}} < \infty$$ as $\sum_{q \in P}\frac{1}{q\log{q}} < \infty$ (see here), so by the Borel-Cantelli lemma $\lambda(S \cap [0,m)) = 0$ and thus $\lambda(S)=0$.

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