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 $(a,b)$, check if $(a,b) R (a,b)$. $$a \cdot b = b \cdot a$$ This is true since multiplication is commutative. 4. **Symmetry:** If $(a,b) R (c,d)$, then $ad = bc$. We need to show $(c,d) R (a,b)$. Since $ad = bc$, it follows that $cb = da$, so symmetry holds. 5. **Transitivity:** If $(a,b) R (c,d)$ and $(c,d) R (e,f)$, then $ad = bc$ and $cf = de$. We need 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 $d$ (common factor) carefully: $$\cancel{d}af = b\cancel{d}e$$ So, $$af = be$$ Thus, transitivity holds. 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 $a$ and $b$ are natural numbers and $a$ is one-third of $b$. **Final answer:** $R$ is an equivalence relation, and $$[(2,6)] = \{(a,b) \in \mathbb{N} \times \mathbb{N} : 3a = b \}$$