1. **Problem Statement:** Prove that the relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)R(c,d)$ if and only if $a + d = c + b$ is an equivalence relation. Also, find the equivalence class of $(4,6)$. 2. **Recall the definition of an equivalence relation:** A relation is an equivalence relation if it is reflexive, symmetric, and transitive. 3. **Check reflexivity:** For any $(a,b) \in \mathbb{N} \times \mathbb{N}$, we need to show $(a,b)R(a,b)$. $$a + b = a + b$$ This is true, so $R$ is reflexive. 4. **Check symmetry:** Suppose $(a,b)R(c,d)$, i.e., $$a + d = c + b$$ We need to show $(c,d)R(a,b)$, i.e., $$c + b = a + d$$ This is the same equation, so $R$ is symmetric. 5. **Check transitivity:** Suppose $(a,b)R(c,d)$ and $(c,d)R(e,f)$, i.e., $$a + d = c + b \quad \text{and} \quad c + f = e + d$$ We want to show $(a,b)R(e,f)$, i.e., $$a + f = e + b$$ From the first equation, rearranged: $$a - b = c - d$$ From the second: $$c - d = e - f$$ By equality, $$a - b = e - f$$ Rearranged: $$a + f = e + b$$ Thus, $R$ is transitive. 6. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. 7. **Find the equivalence class of $(4,6)$:** The equivalence class $[(4,6)]$ is the set of all $(x,y) \in \mathbb{N} \times \mathbb{N}$ such that $$(4,6)R(x,y) \implies 4 + y = x + 6$$ Rearranged: $$x - y = -2$$ So, $$[(4,6)] = \{(x,y) \in \mathbb{N} \times \mathbb{N} : x - y = -2\}$$ This means all pairs where the first component is exactly 2 less than the second. **Final answer:** $R$ is an equivalence relation, and the equivalence class of $(4,6)$ is $\{(x,y) \in \mathbb{N} \times \mathbb{N} : x - y = -2\}$.