Let $R_{1}$ and $R_{2}$ be relations on the set $\{1,2, \ldots, 50\}$ such that $R_{1}=\left\{\left(

Question

Let $R_{1}$ and $R_{2}$ be relations on the set $\{1,2, \ldots, 50\}$ such that $R_{1}=\left\{\left(p, p^{n}\right): p\right.$ is a prime and $n \geq 0$ is an integer $\}$ and $R_{2}=\left\{\left(p, p^{n}\right): p\right.$ is a prime and $n=0$ or 1$\}$. Then, the number of elements in $R_{1}-R_{2}$ is

Professor PC Thomas Classes · Answer

(8): We have, $R_{1}=\left\{\left(2,2^{0}ight),\left(2,2^{1}ight), \ldots,(2ight.$, $\left.2^{5}ight),\left(3,3^{0}ight),\left(3,3^{1}ight),\left(3,3^{2}ight),\left(3,3^{3}ight),\left(5,5^{0}ight),\left(5,5^{1}ight)$, $\left(5,5^{2}ight),\left(7,7^{0}ight),\left(7,7^{1}ight),\left(7,7^{2}ight),\left(11,11^{0}ight),(11,11)$, $\left(13,13^{0}ight),(13,13),\left(17,17^{0}ight),(17,17),\left(19,19^{0}ight)$, $(19,19),\left(23,23^{0}ight),(23,23),\left(29,29^{0}ight),(29,29)$, $\left(31,31^{0}ight),(31,31),\left(37,37^{0}ight),(37,37),\left(41,41^{0}ight)$, $\left.(41,41),\left(43,43^{0}ight),(43,43),\left(47,47^{0}ight),(47,47)ight\}$\ Similarly, $R_{2}=\left\{\left(2,2^{0}ight)\left(2,2^{1}ight),\left(3,3^{0}ight)ight.$, $\left.\left(3,3^{1}ight), \ldots,\left(47,47^{0}ight),(47,47)ight\}$\ $	herefore \quad$ Number of elements in $R_{1}-R_{2}=8$\