# Proving the relative error of division.

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.

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What is $x_B$...? –  Ron Gordon Feb 13 '13 at 14:41
And what is $x_T$ in the definition of relative error? –  Ross Millikan Feb 13 '13 at 14:45
@Ross: I think it is the denominator in his division operation. But perhaps the OP can verify this. –  Ron Gordon Feb 13 '13 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. –  Ross Millikan Feb 13 '13 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. –  Ron Gordon Feb 13 '13 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.

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Since $e_X$ and $e_Z$ are $<<1,$ shouldn't $e_Z=e_X+e_Y$ as $e_Xe_Y$ is smaller? –  Ross Millikan Feb 13 '13 at 19:33
@RossMillikan: Of course, fixed, thanks. –  leonbloy Feb 13 '13 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? –  BMEdwards37 Feb 13 '13 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? –  BMEdwards37 Feb 13 '13 at 22:05
Basic algebra, common denominator $1 - (1-A)/(1-B) = [(1-B) - (1-A)]/(1-B) = (A-B)/(1-B)$ –  leonbloy Feb 14 '13 at 0:02