Relative error of division

Question

How can I proof that $Rel(\frac{x}{y})$ $\leq$ $Rel(x)+Rel(y)$ where $Rel(x)$ is relative error of $x$

Answer 1

If the rounded value of $x$ is $x(1+e_x)$ and the rounded value of $y$ is $y(1+e_y)$, then this means $|e_x|=Rel(x)$ and $|e_y|=Re(y)$. We get $$ \frac{x(1+e_x)}{y(1+e_y)} =\frac xy\cdot(1+e_x)\cdot\frac1{1+e_y} $$ Since $e_y$ is small (specifically, between $-1$ and $1$), we have $$ \frac1{1+e_y}=1-e_y+e_y^2-e_y^3+\cdots $$ (Usually you go the other way, from the infinite geometric series to the fraction, but the equality is just as valid here.)

This gives us $$ \frac xy\cdot(1+e_x)\cdot\frac1{1+e_y}=\frac xy\cdot(1+e_x)(1-e_y+e_y^2-\cdots)\\ =\frac xy(1+e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-\cdots) $$ This means that $$ Rel\left(\frac xy\right)=|e_x-e_y-e_xe_y+e_y^2+e_xe_y^2-\cdots| $$ which, after using the triangle inequality, and also removing all higher degree terms, becomes the expression you were after.