1. **Problem statement:** Prove that if $R$ is reflexive, then $R$ is symmetric only if for every $a \in A$, whenever $(a,a) \in R$, it implies $(a,a) \in R$ always (which is trivial). Explain why symmetry places structural constraints on the relation. 2. **Definitions:** - Reflexive: $\forall a \in A, (a,a) \in R$ - Symmetric: $\forall a,b \in A, (a,b) \in R \implies (b,a) \in R$ - Transitive: $\forall a,b,c \in A, (a,b) \in R \wedge (b,c) \in R \implies (a,c) \in R$ 3. **Step 1: Reflexivity implies $(a,a) \in R$ for all $a$** By definition, reflexivity means every element relates to itself. 4. **Step 2: Symmetry condition on $(a,a)$** Symmetry requires that if $(a,a) \in R$, then $(a,a) \in R$ must also hold. This is trivially true because $(a,a)$ is its own symmetric pair. 5. **Explanation of structural constraints by symmetry:** Symmetry forces that for any related pair $(a,b)$, the reverse $(b,a)$ must also be in $R$. This restricts $R$ from being one-directional and ensures pairs come in "mirror" pairs, shaping the structure of $R$. --- 6. **Problem statement:** Prove that if $R$ is symmetric and transitive, then the set of all equivalence classes of $R$ forms a partition of $A$. 7. **Recall equivalence classes:** For $a \in A$, the equivalence class $[a] = \{ x \in A \mid (a,x) \in R \}$. 8. **Step 1: Show equivalence classes cover $A$** Since $R$ is reflexive (implied for equivalence relation), $a \in [a]$ for all $a$, so union of all $[a]$ is $A$. 9. **Step 2: Show equivalence classes are disjoint or equal** If $[a] \cap [b] \neq \emptyset$, then there exists $x$ with $(a,x) \in R$ and $(b,x) \in R$. 10. **Step 3: Use symmetry and transitivity to prove $[a] = [b]$** Since $R$ is symmetric, $(x,b) \in R$. By transitivity, $(a,b) \in R$ because $(a,x) \in R$ and $(x,b) \in R$. Similarly, for any $y \in [a]$, $(a,y) \in R$, so by transitivity and symmetry, $(b,y) \in R$, thus $y \in [b]$. By symmetry, $[a] \subseteq [b]$ and $[b] \subseteq [a]$, so $[a] = [b]$. 11. **Conclusion:** The equivalence classes form a partition of $A$ because they cover $A$ and are pairwise disjoint or equal. **Final answer:** The proofs show the trivial symmetry condition on reflexive pairs and that symmetric and transitive relations induce partitions via equivalence classes.