# Verify Equivalence Relations

Find an example of three relations $R_{1}$, $R_{2}$, and $R_{3}$ on the set $S=\{1,2,3,4,5\}$ such that

• $R_{1}$ is reflexive but not transitive,
• $R_{2}$ is transitive but neither symmetric nor reflexive, and
• $R_{3}$ is symmetric but not reflexive.

Solution

Let $R_{1} = \{(1,1),(2,2),(3,3),(4,4),(4,5),(5,5)\}$. It is symmetric but not transitive because $(1,2),(2,3),(3,4),(1,3),(2,4),(3,5) \notin R_{1}$.

Let $R_{2} = \{(1,2),(2,3),(1,3),(3,4),(4,5),(3,5)\}$. It is transitive but not symmetric because ${(2,1),(3,2),(4,3),(5,4)} \notin R_{2}$.

Let $R_{3} = \{(1,2),(2,1),(2,3),(3,2),(4,5),(5,4)\}$. It is symmetric but not reflexive because $(1,1),(2,2),(3,3),(4,4),(5,5) \notin R_{3}$.

Can anyone please check my solution.

• Your $R_1$ is actually transitive. To prevent this, try (for example) adding $(1,2)$ and $(2,3)$, but not $(1,3)$.
• Your $R_2$ is not transitive, since $(2,3),(3,4) \in R_2$ but $(2,4) \notin R_2$. To fix this, take the transitive closure by adding $(2,4), (2,5), (1,4), (1,5)$.
• Your $R_3$ works.
• You correctly made $R_1$ reflexive and $R_2,R_3$ not reflexive. My answer gave some suggestions to make $R_1$ not transitive and to make $R_2$ transitive. – Adriano Sep 11 '14 at 0:29