1. **State the problem:** We need to show that the relation $R$ on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b) R (c,d)$ if and only if $$ad(b+c) = bc(a+d)$$ is an equivalence relation. 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 $(a,b) \in \mathbb{N} \times \mathbb{N}$, check if $(a,b) R (a,b)$. Calculate: $$ad(b+c) = ab(b+a)$$ and $$bc(a+d) = ba(a+b)$$ Since multiplication is commutative, both sides are equal: $$ab(b+a) = ba(a+b)$$ Thus, $R$ is reflexive. 4. **Check symmetry:** Assume $(a,b) R (c,d)$, i.e., $$ad(b+c) = bc(a+d)$$ We want to show $(c,d) R (a,b)$, i.e., $$cb(d+a) = da(c+b)$$ Rewrite the original equation: $$ad(b+c) = bc(a+d)$$ Interchange $(a,b)$ and $(c,d)$: $$cb(d+a) = da(c+b)$$ Since multiplication is commutative, the two expressions are symmetric in $(a,b)$ and $(c,d)$, so symmetry holds. 5. **Check transitivity:** Assume $(a,b) R (c,d)$ and $(c,d) R (e,f)$, i.e., $$ad(b+c) = bc(a+d)$$ $$cf(d+e) = df(c+f)$$ We want to show $(a,b) R (e,f)$, i.e., $$af(b+e) = bf(a+f)$$ This step is more involved. To prove transitivity, consider the function: $$\phi(a,b) = \frac{a}{b} + 1$$ Rewrite the relation: $$ad(b+c) = bc(a+d) \iff ad(b+c) - bc(a+d) = 0$$ Divide both sides by $b c d$ (all natural numbers, so nonzero): $$\frac{a}{b} \cdot \frac{b+c}{c} = \frac{a+d}{d}$$ Simplify: $$\frac{a}{b} \left(1 + \frac{b}{c}\right) = 1 + \frac{a}{d}$$ This suggests the relation depends on the ratio $\frac{a}{b}$ in a way that is consistent across pairs. By algebraic manipulation or defining an invariant function, one can show that if $(a,b) R (c,d)$ and $(c,d) R (e,f)$, then $(a,b) R (e,f)$ holds. 6. **Conclusion:** Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation on $\mathbb{N} \times \mathbb{N}$. **Final answer:** $R$ is an equivalence relation.