If $p=p^*+e_1$ and $q=q^*+e_2$, then the relative error in the quotient of $p/q$ is roughly equal to the sum of individual relative errors in p and q.
We want to show $$ \frac{\mid p/q-p^*/q^* \mid}{\mid p/q \mid} \approx \frac{\mid p-p^* \mid}{\mid p \mid}+\frac{\mid q-q^* \mid}{\mid q \mid}$$
I've tried to algebraically manipulating it many ways but can't seem to find the solution. Can anyone help?
We want to show the following is true $$ \frac{\mid p/q-p^*/q^* \mid}{\mid p/q \mid} \approx \frac{\mid p-p^* \mid}{\mid p \mid}+\frac{\mid q-q^* \mid}{\mid q \mid}$$ Proof: $$\frac{p^*}{q^*}=\frac{p-e_1}{q-e_2}$$ $$\frac{p^*}{q^*}=\frac{(p-e_1)}{(q-e_2)}\frac{(q+e_2)}{(q+e_2)}=\frac{pq+pe_2-qe_1-e_1e_2}{q^2-e_2^2}$$ Since $e_1$ and $e_2$ are already small, we can assume that $e_1e_2 \approx 0$ and $e_2^2 \approx 0$. $$\frac{p^*}{q^*}=\frac{pq+pe_2-qe_1}{q^2}$$ $$\frac{p}{q}-\frac{p^*}{q^*}=\frac{p}{q}-\frac{pq+pe_2-qe_1}{q^2}$$ $$\frac{p}{q}-\frac{p^*}{q^*}=\frac{pq-pq-pe_2+qe_1}{q^2}=\frac{qe_1-pe_2}{q_2}$$ $$\frac{\frac{p}{q}-\frac{p^*}{q^*}}{\frac{p}{q}}=\frac{(qe_1-pe_2)}{q_2}*\frac{q}{p}$$ $$\frac{\frac{p}{q}-\frac{p^*}{q^*}}{\frac{p}{q}}=\frac{qe_1-pe_2}{pq}$$ $$\frac{\frac{p}{q}-\frac{p^*}{q^*}}{\frac{p}{q}}=\frac{e_1}{p}+\frac{-e_2}{q}.$$