If $p=p^*+e_1$ and $q=q^*+e_2$, then the relative error in the product of $p*q$ is roughly equal to the sum of individual relative errors in p and q.
Here is my attempt: We want to show $$ \frac{\mid pq-p^*q^* \mid}{\mid pq \mid}= \frac{\mid p-p^* \mid}{\mid p \mid}+\frac{\mid q-q^* \mid}{\mid q \mid}$$ $$RHS= \frac{\mid p^*+e_1-p^* \mid}{\mid p \mid}+\frac{\mid q^*+e_2-q^* \mid}{\mid q \mid}$$ $$= \frac{\mid e_1\mid}{\mid p \mid}+\frac{\mid e_2 \mid}{\mid q \mid}$$ $$= \frac{\mid q*e_1\mid}{\mid p \mid}+\frac{\mid p*e_2 \mid}{\mid q \mid}$$ $$= \frac{\mid q*e_1+p*e_2\mid}{\mid pq \mid}$$ $$= \frac{\mid q(p-p^*)+p(q-q^*)\mid}{\mid pq \mid}$$ $$= \frac{\mid qp-qp^*+pq-pq^*\mid}{\mid pq \mid}$$ $$= \frac{\mid 2pq-pq^*-qp^*\mid}{\mid pq \mid}$$
Here is where I get stuck and I'm not sure how to proceed. Any help would be greatly appreciated.
Let $$pq=(p^*+e_1)(q^*+e_2)=p^* q^*\left(1+\frac{e_1}{p^*}\right)\left(1+\frac{e_2}{q^*}\right)$$ Take logarithms $$\log(pq)=\log(p^* q^*)+\log\left(1+\frac{e_1}{p^*}\right)+\log\left(1+\frac{e_2}{q^*}\right)$$ Using Taylor $$\log(pq)\sim\log(p^* q^*)+\frac{e_1}{p^*}+\frac{e_2}{q^*}\implies \frac{pq}{p^* q^*}\sim \exp\left(\frac{e_1}{p^*}+\frac{e_2}{q^*}\right)$$ Use Taylor again $$\frac{pq}{p^* q^*}\sim 1+\frac{e_1}{p^*}+\frac{e_2}{q^*}\implies\frac{pq-p^* q^*}{p^* q^*}\sim \frac{e_1}{p^*}+\frac{e_2}{q^*}$$