Explain the effect of multiplication or division of errors on the final result.

(N/A) Suppose two physical quantities $A$ and $B$ have measured values $A \pm \Delta A$ and $B \pm \Delta B$ respectively,where $\Delta A$ and $\Delta B$ are their absolute errors.
For a product: Let $Z = AB$. Then the measured value is $Z \pm \Delta Z = (A \pm \Delta A)(B \pm \Delta B) = AB \pm A \Delta B \pm B \Delta A \pm \Delta A \Delta B$.
Dividing by $Z = AB$,we get $1 \pm \frac{\Delta Z}{Z} = 1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}$.
Since $\frac{\Delta A}{A}$ and $\frac{\Delta B}{B}$ are very small,their product is neglected. For maximum relative error,$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.
For a quotient: Let $Z = \frac{A}{B}$. Then $Z \pm \Delta Z = \frac{A \pm \Delta A}{B \pm \Delta B} = \frac{A(1 \pm \Delta A/A)}{B(1 \pm \Delta B/B)} = Z(1 \pm \Delta A/A)(1 \pm \Delta B/B)^{-1}$.
Using binomial expansion $(1 \pm x)^n \approx 1 \pm nx$ for small $x$,we get $1 \pm \frac{\Delta Z}{Z} \approx (1 \pm \frac{\Delta A}{A})(1 \mp \frac{\Delta B}{B}) \approx 1 \pm \frac{\Delta A}{A} \mp \frac{\Delta B}{B}$.
Thus,the maximum relative error is $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.