Let A = {1, 2, 3, 4, 5}. Let R be a relation | vedclass

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(C) Given the set $A = \{1, 2, 3, 4, 5\}$ and the relation $R$ defined by $4x \leq 5y$. First,we list the elements $(x, y)$ such that $4x \leq 5y$: Fo

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$25$

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(C) Given the set $A = \{1, 2, 3, 4, 5\}$ and the relation $R$ defined by $4x \leq 5y$. First,we list the elements $(x, y)$ such that $4x \leq 5y$: For $x=1$: $4 \leq 5y \implies y \in \{1, 2, 3, 4, 5\}$ ($5$ elements) For $x=2$: $8 \leq 5y \implies y \in \{2, 3, 4, 5\}$ ($4$ elements) For $x=3$: $12 \leq 5y \implies y \in \{3, 4, 5\}$ ($3$ elements) For $x=4$: $16 \leq 5y \implies y \in \{4, 5\}$ ($2$ elements) For $x=5$: $20 \leq 5y \implies y \in \{4, 5\}$ ($2$ elements) Total elements $m = 5 + 4 + 3 + 2 + 2 = 16$. To make $R$ symmetric,for every $(x, y) \in R$ where $x 
eq y$,we must have $(y, x) \in R$. The elements in $R$ where $x 
eq y$ are: $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5), (5, 4)$. There are $11$ such pairs. We check which of their symmetric counterparts $(y, x)$ are $NOT$ already in $R$: $(2, 1) 
otin R$,$(3, 1) 
otin R$,$(4, 1) 
otin R$,$(5, 1) 
otin R$,$(3, 2) 
otin R$,$(4, 2) 
otin R$,$(5, 2) 
otin R$,$(4, 3) 
otin R$,$(5, 3) 
otin R$. Note that $(5, 4) \in R$ and $(4, 5) \in R$,so this pair is already symmetric. The elements to be added are $\{(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)\}$. Thus,$n = 9$. Therefore,$m + n = 16 + 9 = 25$.

Let $A = \{1, 2, 3, 4, 5\}$. Let $R$ be a relation on $A$ defined by $x R y$ if and only if $4x \leq 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m+n$ is equal to:

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