Question

Let $R_{1}$ and $R_{2}$ be relations on the set $\{1,2, \ldots, 50\}$ such that
$R _{1}=\left\{\left( p , p ^{ n }\right)\right.$ : $p$ 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 ______

Answer 1

Here, $p , p ^{ n } \in\{1,2, \ldots 50\}$
Now $p$ can take values
$2,3,5,7,11,13,17,23,29,31,37,41,43$ and 47 .
we can calculate no. of elements in R, as
$\left(2,2^{\circ}\right),\left(2,2^{1}\right) . .\left(2,2^{5}\right)$
$\left(3,3^{\circ}\right), \ldots\left(3,3^{3}\right)$
$\left(5,5^{\circ}\right), \ldots\left(5,5^{2}\right)$
$\left(7,7^{\circ}\right), \ldots\left(7,7^{2}\right)$
$\left(11,11^{\circ}\right), \ldots\left(11,11^{1}\right)$
And rest for all other two elements each
$\therefore n \left( R _{1}\right)=6+4+3+3+(2 \times 10)=36 $
Similarly for $R _{2} $
$ \left(2,2^{\circ}\right),\left(2,2^{1}\right) $
$\left(47,47^{\circ}\right),\left(47,47^{1}\right)$
$\therefore n \left( R _{2}\right)=2 \times 14=28$
$\therefore n \left( R _{1}\right)- n \left( R _{2}\right)=36-28=8$