Let R_{1} and R_{2} be relations on the set { | vedclass

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(D) The set is $S = \{1, 2, \ldots, 50\}$. The primes $p \in S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ (total $15$ primes).F

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$8$

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(D) The set is $S = \{1, 2, \ldots, 50\}$. The primes $p \in S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ (total $15$ primes).For $R_{1}$,we need $p^{n} \leq 50$ where $n \geq 0$:- For $p=2$: $2^{0}, 2^{1}, 2^{2}, 2^{3}, 2^{4}, 2^{5} \leq 50$ ($6$ elements).- For $p=3$: $3^{0}, 3^{1}, 3^{2}, 3^{3} \leq 50$ ($4$ elements).- For $p=5$: $5^{0}, 5^{1}, 5^{2} \leq 50$ ($3$ elements).- For $p=7$: $7^{0}, 7^{1}, 7^{2} \leq 50$ ($3$ elements).- For $p \in \{11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ ($11$ primes): $p^{0}, p^{1} \leq 50$ ($2$ elements each,total $11 	imes 2 = 22$ elements).Total elements in $R_{1} = 6 + 4 + 3 + 3 + 22 = 38$.For $R_{2}$,we need $n=0$ or $n=1$:- For each of the $15$ primes,we have $(p, p^{0})$ and $(p, p^{1})$.Total elements in $R_{2} = 15 	imes 2 = 30$.$R_{1} - R_{2}$ contains elements where $n \geq 2$:- $p=2$: $(2, 2^{2}), (2, 2^{3}), (2, 2^{4}), (2, 2^{5})$ ($4$ elements).- $p=3$: $(3, 3^{2}), (3, 3^{3})$ ($2$ elements).- $p=5$: $(5, 5^{2})$ ($1$ element).- $p=7$: $(7, 7^{2})$ ($1$ element).Total elements in $R_{1} - R_{2} = 4 + 2 + 1 + 1 = 8$.

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

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