Let A = {1, 2, 3, 4} and R = {(1, 2), (2, 3), | vedclass

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(A) An equivalence relation $S$ must be reflexive,symmetric,and transitive.Given $R = \{(1, 2), (2, 3), (1, 4)\}$.$1$. Reflexivity: $S$ must contain $

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$16$

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(A) An equivalence relation $S$ must be reflexive,symmetric,and transitive.Given $R = \{(1, 2), (2, 3), (1, 4)\}$.$1$. Reflexivity: $S$ must contain $(1, 1), (2, 2), (3, 3), (4, 4)$.$2$. Symmetry: Since $(1, 2) \in S$,then $(2, 1) \in S$. Since $(2, 3) \in S$,then $(3, 2) \in S$. Since $(1, 4) \in S$,then $(4, 1) \in S$.$3$. Transitivity: Since $(1, 2) \in S$ and $(2, 3) \in S$,then $(1, 3) \in S$. By symmetry,$(3, 1) \in S$.Since $(1, 3) \in S$ and $(3, 2) \in S$,we already have $(1, 2) \in S$.Since $(1, 4) \in S$ and $(4, 1) \in S$,we have $(1, 1) \in S$ (already included).Since $(2, 3) \in S$ and $(3, 2) \in S$,we have $(2, 2) \in S$ (already included).Since $(1, 2) \in S$ and $(2, 1) \in S$,we have $(1, 1) \in S$.Checking the equivalence classes: $1$ is related to $2, 3, 4$. Thus,$1, 2, 3, 4$ are all in the same equivalence class.If all elements are in one equivalence class,$S = A 	imes A$.The number of elements $n = 4 	imes 4 = 16$.

Let $A = \{1, 2, 3, 4\}$ and $R = \{(1, 2), (2, 3), (1, 4)\}$ be a relation on $A$. Let $S$ be the smallest equivalence relation on $A$ such that $R \subset S$. If the number of elements in $S$ is $n$,then the value of $n$ is:

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