Let A = {2, 3, 5, 7, 9}. Let R be the relatio | vedclass

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(B) Given $A = \{2, 3, 5, 7, 9\}$. The relation $R$ is defined by $2x \le 3y$, which implies $y \ge \frac{2x}{3}$.For each $x \in A$, we find the corr

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$25$

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(B) Given $A = \{2, 3, 5, 7, 9\}$. The relation $R$ is defined by $2x \le 3y$, which implies $y \ge \frac{2x}{3}$.For each $x \in A$, we find the corresponding $y \in A$:If $x = 2$, $y \ge 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).If $x = 3$, $y \ge 2 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).If $x = 5$, $y \ge 3.33 \implies y \in \{5, 7, 9\}$ ($3$ elements).If $x = 7$, $y \ge 4.66 \implies y \in \{5, 7, 9\}$ ($3$ elements).If $x = 9$, $y \ge 6 \implies y \in \{7, 9\}$ ($2$ elements).Total elements $l = 5 + 5 + 3 + 3 + 2 = 18$.For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x)$ must be in $R$.The elements currently in $R$ are: $(2,2), (2,3), (2,5), (2,7), (2,9), (3,2), (3,3), (3,5), (3,7), (3,9), (5,5), (5,7), (5,9), (7,5), (7,7), (7,9), (9,7), (9,9)$.The pairs $(x, y)$ such that $(x, y) \in R$ but $(y, x) 
otin R$ are: $(2,5), (2,7), (2,9), (3,5), (3,7), (3,9), (5,9)$.To make $R$ symmetric, we must add their reverses: $(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)$.Thus, $m = 7$.Therefore, $l + m = 18 + 7 = 25$.

Let $A = \{2, 3, 5, 7, 9\}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x \le 3y$. Let $l$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $l + m$ is equal to:

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