Explain effect of multiplication or division of error on final result.

Suppose two physical quantities $\mathrm{A}$ and $\mathrm{B}$ have measured values $\mathrm{A} \pm \Delta \mathrm{A}, \mathrm{B} \pm \Delta \mathrm{B}$ respectively. Where $\triangle \mathrm{A}$ and $\triangle \mathrm{B}$ are their absolute errors.

For a product: Suppose the product of $\mathrm{A}$ and $\mathrm{B}$ is $\mathrm{Z}$ and the absolute error in $\mathrm{Z}$ is $\Delta Z$.

$\therefore \mathrm{Z}=\mathrm{AB}$

$\therefore \mathrm{Z} \pm \Delta \mathrm{Z}=(\mathrm{A} \pm \Delta \mathrm{A})(\mathrm{B} \pm \Delta \mathrm{B})$

$\therefore \mathrm{Z} \pm \Delta \mathrm{Z}=\mathrm{AB} \pm \mathrm{A} \Delta \mathrm{B} \pm \mathrm{B} \Delta \mathrm{A} \pm(\Delta \mathrm{A} \Delta \mathrm{B})$

Dividing the left side by $Z$ and right side by $A B$,

$1 \pm \frac{\Delta Z}{Z}=1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm\left(\frac{\Delta A}{A}\right)\left(\frac{\Delta B}{B}\right)$

$\frac{\Delta \mathrm{A}}{\mathrm{A}}$ and $\frac{\Delta \mathrm{B}}{\mathrm{B}}$ are very small hence their product will be very small and can be neglected. Neglecting negative sign for maximum error,

$\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}$

For a Quotient : Suppose the quotient of $A$ and $B$ is $Z$ and the absolute error of $Z$ is $\Delta Z$.

$\therefore \mathrm{Z}=\frac{\mathrm{A}}{\mathrm{B}}$

$\therefore \mathrm{Z} \pm \Delta \mathrm{Z}=\frac{\mathrm{A} \pm \Delta \mathrm{A}}{\mathrm{B} \pm \Delta \mathrm{B}}$

$Z\left(1 \pm \frac{\Delta Z}{Z}\right)=\frac{A\left(1 \pm \frac{\Delta A}{A}\right)}{B\left(1 \pm \frac{\Delta B}{B}\right)}$

Now, dividing the left side by $Z$ and right side by $\frac{A}{B}$,

$1 \pm \frac{\Delta Z}{Z}=\frac{1 \pm \frac{\Delta A}{A}}{1 \pm \frac{\Delta B}{B}}$

$\therefore 1 \pm \frac{\Delta Z}{Z}=\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)^{-1}$