1. **State the problem:** We have a relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $$ (a,b) R (c,d) \iff ad = bc $$ for all $(a,b), (c,d) \in \mathbb{N} \times \mathbb{N}$. We need to show that $R$ is an equivalence relation and find the equivalence class of $(2,6)$. 2. **Recall the definition of an equivalence relation:** A relation $R$ on a set is an equivalence relation if it is: - Reflexive: $x R x$ for all $x$. - Symmetric: If $x R y$, then $y R x$. - Transitive: If $x R y$ and $y R z$, then $x R z$. 3. **Show reflexivity:** For any $(a,b) \in \mathbb{N} \times \mathbb{N}$, $$ (a,b) R (a,b) \iff a \cdot b = b \cdot a $$ which is true since multiplication is commutative. 4. **Show symmetry:** Assume $(a,b) R (c,d)$, i.e., $$ ad = bc $$ Then, $$ cb = da $$ which means $$ (c,d) R (a,b) $$ so $R$ is symmetric. 5. **Show transitivity:** Assume $(a,b) R (c,d)$ and $(c,d) R (e,f)$, i.e., $$ ad = bc \quad \text{and} \quad cf = de $$ We want to show $(a,b) R (e,f)$, i.e., $$ af = be $$ From $ad = bc$, multiply both sides by $f$: $$ adf = bcf $$ From $cf = de$, multiply both sides by $b$: $$ bcf = bde $$ Equate the right sides: $$ adf = bde $$ Cancel $d$ (since $d \in \mathbb{N}$ and $d \neq 0$): $$ \cancel{d}af = b\cancel{d}e $$ which simplifies to $$ af = be $$ Thus, $(a,b) R (e,f)$ and $R$ is transitive. 6. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation on $\mathbb{N} \times \mathbb{N}$. 7. **Find the equivalence class of $(2,6)$:** The equivalence class $[(2,6)]$ is the set of all $(a,b) \in \mathbb{N} \times \mathbb{N}$ such that $$ (a,b) R (2,6) \iff 2b = 6a $$ Simplify: $$ 2b = 6a \implies b = 3a $$ So, $$ [(2,6)] = \{(a,b) \in \mathbb{N} \times \mathbb{N} : b = 3a \} $$ This means all pairs where the second component is three times the first component belong to the equivalence class of $(2,6)$. **Final answer:** $R$ is an equivalence relation on $\mathbb{N} \times \mathbb{N}$. The equivalence class of $(2,6)$ is $$ [(2,6)] = \{(a,b) \in \mathbb{N} \times \mathbb{N} : b = 3a \}.$$