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Joe Silverman
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Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$,$$A = \left\{\frac{a}{q}: 1\le a < q\le X,\ (a, q) = 1\right\},$$ and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$ (where here $p$ is prime). In addition we have$$B = \left\{\frac{a}{p}: 1\le a < p\le X,\ (a, p) = 1,\text{ and $p$ is prime}\right\}.$$ Note that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\gg\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$.$$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ge\frac{1}{qq'}\ge\frac{1}{X^2},\quad\text{where}\quad\lVert\beta\rVert := \min_{n\in\mathbb{Z}} |\beta - n|.$$ It therefore follows follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as , since $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$$$|B|^{-1}\ll X^{-2}\log X?$$ Note that the inclusion ($O(\log X)$ is$B\subseteq A$ gives the trivial bound one gets from the fact that $B\subseteq A$)?$O(\log X)$.

Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$, and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$ (where here $p$ is prime). In addition we have that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\gg\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$. It therefore follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$ ($O(\log X)$ is the trivial bound one gets from the fact that $B\subseteq A$)?

Let $X\ge 2$ be large. Let $$A = \left\{\frac{a}{q}: 1\le a < q\le X,\ (a, q) = 1\right\},$$ and let $$B = \left\{\frac{a}{p}: 1\le a < p\le X,\ (a, p) = 1,\text{ and $p$ is prime}\right\}.$$ Note that for distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ge\frac{1}{qq'}\ge\frac{1}{X^2},\quad\text{where}\quad\lVert\beta\rVert := \min_{n\in\mathbb{Z}} |\beta - n|.$$ It follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible, since $|A|\gg X^2$. Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $$|B|^{-1}\ll X^{-2}\log X?$$ Note that the inclusion $B\subseteq A$ gives the trivial bound $O(\log X)$.

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Mayank Pandey
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Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$, and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$ (where here $p$ is prime). In addition we have that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ll\frac{1}{X^2}$$$$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\gg\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$. It therefore follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$ ($O(\log X)$ is the trivial bound one gets from the fact that $B\subseteq A$)?

Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$, and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$. In addition we have that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ll\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$. It therefore follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$ ($O(\log X)$ is the trivial bound one gets from the fact that $B\subseteq A$)?

Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$, and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$ (where here $p$ is prime). In addition we have that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\gg\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$. It therefore follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$ ($O(\log X)$ is the trivial bound one gets from the fact that $B\subseteq A$)?

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Mayank Pandey
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Spacing of fractions with prime denominator

Let $X\ge 2$ be large. Let $A = \{\frac{a}{q}: 1\le a < q\le X, (a, q) = 1\}$, and let $B = \{\{\frac{a}{p}: 1\le a < p\le X, (a, p) = 1\}\}$. In addition we have that for any two distinct $\frac{a}{q}, \frac{a'}{q'}\in A$, we have $$\left\lVert \frac{a}{q} - \frac{a'}{q'}\right\rVert\ll\frac{1}{X^2}$$ where here $\lVert\beta\rVert = \min_{n\in\mathbb{Z}} |\beta - n|$. It therefore follows that in any interval of length $X^{-2}$ in $[0, 1]$, there are at most $O(1)$ elements of $A$, and this is clearly the best result possible (as $|A|\gg X^2$). Is it possible to get better results for $B$, since we have that $|B|\asymp X^2 / \log X$ by the prime number theorem. In particular, is it possible to get a bound of $o(\log X)$ for the number of elements of $B$ in an interval of length $|B|^{-1}\ll X^{-2}\log X$ ($O(\log X)$ is the trivial bound one gets from the fact that $B\subseteq A$)?