Let R be a relation from A = {2, 3, 4, 5} to | vedclass

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(D) The relation $R$ is defined as $R = \{(a, b) \mid a 	ext{ divides } b, a \in A, b \in B\}$.We check each element $a \in A$ to see which $b \in B$

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$5$

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(D) The relation $R$ is defined as $R = \{(a, b) \mid a 	ext{ divides } b, a \in A, b \in B\}$.We check each element $a \in A$ to see which $b \in B$ it divides:For $a = 2$: $2$ divides $6$ and $10$,so $(2, 6) \in R$ and $(2, 10) \in R$.For $a = 3$: $3$ divides $3$ and $6$,so $(3, 3) \in R$ and $(3, 6) \in R$.For $a = 4$: $4$ does not divide any element in $B$.For $a = 5$: $5$ divides $10$,so $(5, 10) \in R$.Thus,$R = \{(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)\}$.The inverse relation $R^{-1}$ is defined as $R^{-1} = \{(b, a) \mid (a, b) \in R\}$.Therefore,$R^{-1} = \{(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)\}$.The number of elements in $R^{-1}$ is the same as the number of elements in $R$,which is $5$.

Let $R$ be a relation from $A = \{2, 3, 4, 5\}$ to $B = \{3, 6, 7, 10\}$ defined by $R = \{(a, b) \mid a \text{ divides } b, a \in A, b \in B\}$. Then,the number of elements in $R^{-1}$ will be:

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