Proving the relative error of division.

Question

The problem says to show that the relative error for division on a computer is

$$Rel(\frac{x_{A}}{y_{A}})=\frac{Rel(x_{A})-Rel(y_{A})}{1-Rel(y_{A})}$$

$$\approx Rel(x_{A})-Rel(y_{A})$$

provided that the relative error of $y_{A}$ is small compared to one.

I know that $$Rel(x_{A})=\frac{x_{T}-x_{A}}{x_{T}}$$

and $x_{A}=x_{T}(1-e_{x})$ with $e_{x}$ being the error.

but I'm really not sure how to proceed from here.

Edit again: I emailed the professor and he sent out a class-wide email totally rearranging it so that's probably where confusion stems from. This is the new and actual problem.

Answer 1

Your notation is a complete mess (I made a correction, but it's still all wrong). You cannot start to trying proving something if you cannot make sense of that something. Try, for example, to work out a numeric example, to get some consistent notation.

I'll try. I define $e_X = (x_T- x_A)/x_T$, so $x_A = x_T (1-e_X)$ ($x_T$ is the true value, $x_A$ the approximate or actual value, $e_X$ the relative error).

Then $$z_A=\frac{x_A}{y_A}=\frac{x_T(1-e_X)}{y_T(1-e_Y)} = z_T \frac{1-e_X}{1-e_Y} \approx z_T (1-e_X)(1+e_Y) \approx z_T (1 - e_X + e_Y)$$

where the last approximations assumes $e_Y \ll 1$. But the sign of the relative error is immaterial, hence $ e_Z= e_X + e_Y$.

Update: following the revised question:

$$z_A = z_T \frac{1-e_X}{1-e_Y} = z_T (1 - e_Z) \implies e_Z = 1 - \frac{1-e_X}{1-e_Y} = \frac{e_X - e_Y}{1-e_Y}$$

Again, the denominator tends to 1 if $e_Y$ is small, and the numerator should be writen as $e_X + e_Y$ if we are computing propagation of errors.