1. **Problem 1:** Show that the relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)R(c,d) \iff \frac{a}{c} = \frac{b}{d}$ is an equivalence relation. 2. **Recall the definition of an equivalence relation:** It must be reflexive, symmetric, and transitive. 3. **Reflexivity:** For any $(a,b)$, check if $(a,b)R(a,b)$ holds. $$\frac{a}{a} = \frac{b}{b} \implies 1 = 1$$ True, so $R$ is reflexive. 4. **Symmetry:** Assume $(a,b)R(c,d)$, i.e., $\frac{a}{c} = \frac{b}{d}$. Then, $$\frac{c}{a} = \frac{d}{b}$$ which means $(c,d)R(a,b)$. So $R$ is symmetric. 5. **Transitivity:** Assume $(a,b)R(c,d)$ and $(c,d)R(e,f)$, i.e., $$\frac{a}{c} = \frac{b}{d} \quad \text{and} \quad \frac{c}{e} = \frac{d}{f}$$ From the first, $ad = bc$; from the second, $cf = de$. Multiply the two equalities: $$ad \times cf = bc \times de$$ Simplify: $$a d c f = b c d e$$ Cancel common factors $c$ and $d$: $$a \cancel{d} \cancel{c} f = b \cancel{c} \cancel{d} e \implies a f = b e$$ Thus, $$\frac{a}{e} = \frac{b}{f}$$ which means $(a,b)R(e,f)$. So $R$ is transitive. 6. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. --- 7. **Problem 2:** Show that the relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)R(c,d) \iff a - c = b - d$ is an equivalence relation. 8. **Reflexivity:** For any $(a,b)$, $$a - a = b - b \implies 0 = 0$$ True, so $R$ is reflexive. 9. **Symmetry:** Assume $(a,b)R(c,d)$, i.e., $$a - c = b - d$$ Then, $$c - a = d - b$$ which means $(c,d)R(a,b)$. So $R$ is symmetric. 10. **Transitivity:** Assume $(a,b)R(c,d)$ and $(c,d)R(e,f)$, i.e., $$a - c = b - d \quad \text{and} \quad c - e = d - f$$ Add the two equations: $$(a - c) + (c - e) = (b - d) + (d - f)$$ Simplify: $$a - e = b - f$$ which means $(a,b)R(e,f)$. So $R$ is transitive. 11. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.