1. **Problem Statement:** 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. **Definition of Equivalence Relation:** A relation is an equivalence relation if it is reflexive, symmetric, and transitive. 3. **Reflexivity:** For any $(a,b) \in \mathbb{N} \times \mathbb{N}$, check if $(a,b) R (a,b)$. $$a \cdot b = b \cdot a$$ This is true since multiplication is commutative. 4. **Symmetry:** Assume $(a,b) R (c,d)$, i.e., $ad = bc$. We need to show $(c,d) R (a,b)$. Given $ad = bc$, by commutativity, $cb = da$, so $(c,d) R (a,b)$. 5. **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., $af = be$. Multiply the first equation by $f$ and the second by $b$: $$adf = bcf$$ $$bcf = bde$$ Since $bcf$ appears in both, equate: $$adf = bde$$ Cancel $d$ from both sides (since $d \neq 0$ in $\mathbb{N}$): $$\cancel{d}af = b\cancel{d}e$$ So, $$af = be$$ Thus, $(a,b) R (e,f)$. 6. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation on $\mathbb{N} \times \mathbb{N}$. 7. **Equivalence Class of $(2,6)$:** The equivalence class $[(2,6)]$ is the set of all $(a,b)$ such that $(a,b) R (2,6)$. By definition, $$2b = 6a \implies \frac{a}{b} = \frac{1}{3}$$ So, $$[(2,6)] = \{(a,b) \in \mathbb{N} \times \mathbb{N} : \frac{a}{b} = \frac{1}{3} \}$$ This means all pairs $(a,b)$ where $a$ is one-third of $b$, or equivalently, $b = 3a$. **Final answer:** $$[(2,6)] = \{(a,3a) : a \in \mathbb{N} \}$$