Let A = {1, 2, 3}. Show that the number of re | vedclass

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(C) Let $A = \{1, 2, 3\}$. $A$ relation $R$ on $A$ is reflexive if $(1, 1), (2, 2), (3, 3) \in R$. Given $R$ contains $(1, 2)$ and $(2, 3)$,by transit

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$3$

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(C) Let $A = \{1, 2, 3\}$. $A$ relation $R$ on $A$ is reflexive if $(1, 1), (2, 2), (3, 3) \in R$. Given $R$ contains $(1, 2)$ and $(2, 3)$,by transitivity,it must contain $(1, 3)$. So,the smallest reflexive and transitive relation containing $(1, 2)$ and $(2, 3)$ is $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. This relation $R_0$ is not symmetric because $(2, 1) 
otin R_0$. We can add other elements to $R_0$ while maintaining reflexivity and transitivity: $1$. $R_1 = R_0 \cup \{(2, 1)\}$: This is reflexive and transitive,but not symmetric since $(3, 2) 
otin R_1$. $2$. $R_2 = R_0 \cup \{(3, 2)\}$: This is reflexive and transitive,but not symmetric since $(2, 1) 
otin R_2$. $3$. $R_3 = R_0 \cup \{(2, 1), (3, 2), (3, 1)\}$: This is reflexive and transitive,but not symmetric since $(1, 3) \in R_3$ but $(3, 1) \in R_3$,however $(1, 2) \in R_3$ and $(2, 1) \in R_3$,but $(2, 3) \in R_3$ and $(3, 2) \in R_3$. Wait,checking symmetry: $R_3$ is actually symmetric if we add $(2, 1), (3, 2), (3, 1)$. Actually,the relations are $R_0, R_0 \cup \{(2, 1)\}, R_0 \cup \{(3, 2)\}$. Thus,there are $3$ such relations.

Let $A = \{1, 2, 3\}$. Show that the number of relations on $A$ containing $(1, 2)$ and $(2, 3)$ which are reflexive and transitive but not symmetric is $3$. (Note: The original prompt claimed $4$,but the correct count for this specific set is $3$).

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