1. **Problem Statement:** Given the set $A = \{1,2,3,4\}$ and relation $R = \{(1,2), (2,3), (3,4), (1,4), (2,2), (3,3)\}$, determine if $R$ is reflexive, symmetric, antisymmetric, and/or transitive. 2. **Reflexivity:** A relation $R$ on set $A$ is reflexive if for every element $a \in A$, $(a,a) \in R$. - Check elements: $(1,1)$ is missing, $(2,2)$ and $(3,3)$ are present, $(4,4)$ is missing. - Since $(1,1)$ and $(4,4)$ are not in $R$, $R$ is **not reflexive**. 3. **Symmetry:** $R$ is symmetric if whenever $(a,b) \in R$, then $(b,a) \in R$. - For $(1,2) \in R$, check if $(2,1) \in R$; it is not. - Therefore, $R$ is **not symmetric**. 4. **Antisymmetry:** $R$ is antisymmetric if whenever $(a,b) \in R$ and $(b,a) \in R$ with $a \neq b$, then this never happens. - There are no pairs $(a,b)$ and $(b,a)$ with $a \neq b$ in $R$. - Hence, $R$ is **antisymmetric**. 5. **Transitivity:** $R$ is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$. - Check pairs: - $(1,2)$ and $(2,3)$ imply $(1,3)$ should be in $R$, but it is not. - So, $R$ is **not transitive**. 6. **Find the transitive closure $R^+$:** The transitive closure adds the minimal pairs to make $R$ transitive. - From $(1,2)$ and $(2,3)$ add $(1,3)$. - From $(2,3)$ and $(3,4)$ add $(2,4)$. - From $(1,3)$ (new) and $(3,4)$ add $(1,4)$ (already in $R$). - So, $R^+ = R \cup \{(1,3), (2,4)\} = \{(1,2), (2,3), (3,4), (1,4), (2,2), (3,3), (1,3), (2,4)\}$. 7. **Check if $R$ can be a partial order:** A partial order requires reflexivity, antisymmetry, and transitivity. - $R$ is not reflexive and not transitive. - Therefore, $R$ **cannot be a partial order**. - Because it is not transitive and reflexive, a Hasse diagram cannot be constructed. **Final answers:** - Reflexive: No - Symmetric: No - Antisymmetric: Yes - Transitive: No - Transitive closure $R^+$: $\{(1,2), (2,3), (3,4), (1,4), (2,2), (3,3), (1,3), (2,4)\}$ - Partial order: No, fails reflexivity and transitivity