Error in ratio of two numbers

Question

How tow find error associated with a ratio $R$, when both the numerator and denominator contains error. For example

$$ R=\frac{(A \pm \Delta A)}{(B\pm\Delta B)}.$$

Answer 1

Assume that $\displaystyle \left|\frac{\Delta B}{B}\right|<1$, then you may write $$ R=\frac{(A \pm \Delta A)}{(B\pm\Delta B)}=\frac{(A \pm \Delta A)}{B}\times \frac{1}{1\pm\frac{\Delta B}{B}}=\frac{(A \pm \Delta A)}{B}\times \left( 1\mp\frac{\Delta B}{B}+\mathcal{O}\left(\frac{(\Delta B)^2}{B^2}\right)\right) $$ giving, to the first order: $$ R=\frac{(A \pm \Delta A)}{(B\pm\Delta B)}\approx \frac{A}{B}\left(1-\frac{\Delta B}{B}\right) +\frac{\Delta A}{B}.$$

Answer 2

I'll assume $\Delta A$ and $\Delta B$ are non-negative.

If $\Delta B = |B|$ then there is no upper bound on $R$. If $\Delta B > |B|$ then there is neither an upper bound nor lower bound on $R$. So let's suppose that $0 \leq \Delta B < |B|$. Then the informal notation,

$$ R=\frac{(A \pm \Delta A)}{(B\pm\Delta B)},$$

interpreted as $R = \frac ab$ where $A - \Delta A \leq a \leq A + \Delta A$ and $B - \Delta B \leq b \leq B + \Delta B$, implies that

$$ \frac{(A - \Delta A)}{(B + \Delta B)} \leq R \leq \frac{(A + \Delta A)}{(B - \Delta B)}. $$

Let $$ R_0 = \frac{AB + (\Delta A)(\Delta B)}{B^2 - (\Delta B)^2} $$

and let $$ \Delta R = \frac{A\Delta B + B \Delta A}{B^2 - (\Delta B)^2}. $$

Then $$ R_0 - \Delta R \leq R \leq R_0 + \Delta R.$$

Note that $R_0 \geq \frac AB,$ with equality if and only if $\Delta B = 0.$