Let A = {1, 2, 3}. Then the number of equival | vedclass

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(A) Given the set $A = \{1, 2, 3\}$.An equivalence relation must be reflexive,symmetric,and transitive.First,for reflexivity,the relation must contain

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$2$

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(A) Given the set $A = \{1, 2, 3\}$.An equivalence relation must be reflexive,symmetric,and transitive.First,for reflexivity,the relation must contain $(1, 1), (2, 2), (3, 3)$.Since the relation must contain $(1, 2)$,by symmetry,it must also contain $(2, 1)$.Thus,the smallest equivalence relation containing $(1, 2)$ is $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.Now,consider adding other elements. If we add $(2, 3)$,by symmetry we must add $(3, 2)$. By transitivity,since $(1, 2) \in R$ and $(2, 3) \in R$,we must have $(1, 3) \in R$. By symmetry,$(3, 1) \in R$. Adding these results in the universal relation $R_2 = A 	imes A$,which contains all $9$ elements.Thus,there are exactly $2$ such equivalence relations: $R_1$ and $R_2$.The correct answer is $A$.

Let $A = \{1, 2, 3\}$. Then the number of equivalence relations containing $(1, 2)$ is:

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