# Proving the relative error of division.

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

\begin{align}\textrm{Rel}\left(\frac{x_{A}}{y_{A}}\right)&=\frac{\textrm{Rel}(x_{A})-\textrm{Rel}(y_{A})}{1-\textrm{Rel}(y_{A})}\\ &\approx \textrm{Rel}(x_{A})-\textrm{Rel}(y_{A})\end{align}

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

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

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.

• What is $x_B$...? Commented Feb 13, 2013 at 14:41
• And what is $x_T$ in the definition of relative error? Commented Feb 13, 2013 at 14:45
• @Ross: I think it is the denominator in his division operation. But perhaps the OP can verify this. Commented Feb 13, 2013 at 14:48
• @rlgordonma: I see it there, but then the definition of relative error in $x_A$ doesn't make sense as it shouldn't refer to what you will divide it by. Commented Feb 13, 2013 at 14:50
• @RossMillikan: I see what you mean. So the OP needs to figure out this stuff for himself. Perhaps he could have stated things better by saying that $x_T = x_A (1 + \epsilon)$. But then again, this doesn't make a heck of a lot of sense either. Commented Feb 13, 2013 at 15:24

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.

• Since $e_X$ and $e_Z$ are $<<1,$ shouldn't $e_Z=e_X+e_Y$ as $e_Xe_Y$ is smaller? Commented Feb 13, 2013 at 19:33
• @RossMillikan: Of course, fixed, thanks. Commented Feb 13, 2013 at 20:03
• I think I follow this actually, thank you. I've edited the problem to show the revision my professor made, is the answer you gave still applicable? Commented Feb 13, 2013 at 21:01
• Deleted a couple comments because it was beginning to feel like spamming you. After the arrow you have ez=(error) and I understand that step, but when you go from that to the very last (ex-ey)/(1-ey) I am missing how that transition is made. Something tells me it is very easy algebra but my brain is fried right now. Care to explain quick? Commented Feb 13, 2013 at 22:05
• Basic algebra, common denominator $1 - (1-A)/(1-B) = [(1-B) - (1-A)]/(1-B) = (A-B)/(1-B)$ Commented Feb 14, 2013 at 0:02