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

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  • $\begingroup$ What have you tried? Where are you stuck? Also, note that this is only true if $Rel(y)$ is small (or if $y$ has been rounded up). If you try out a couple of examples with $Rel(y)=0.5$, for instance comparing $\frac11$ to $\frac{1}{0.5}$, you will find that the actual bound is quite a bit larger. $\endgroup$
    – Arthur
    Commented Nov 16, 2018 at 0:00
  • $\begingroup$ I completly have no idea about this,I proved something like this for addition and multiplication but I can't do anything about division,and also all errors are small $\endgroup$ Commented Nov 16, 2018 at 1:15
  • $\begingroup$ How did you do multiplication, then? Why doesn't that work for division? $\endgroup$
    – Arthur
    Commented Nov 16, 2018 at 2:34
  • $\begingroup$ $Rel(xy)= \frac{| xy-(x-a)(y-b) | }{ | xy |} = \frac{ |ay+bx-ab | }{ | xy | } \leq \frac{ | ay |+ | bx | + | ab |}{ | xy | } = \frac{|a|}{|x|} + \frac{|b|}{|y|}+\frac{|a|}{|x|}\frac{|b|}{|y|}=Rel(x)+Rel(y)+Rel(x)Rel(y)$ $\endgroup$ Commented Nov 16, 2018 at 10:42
  • $\begingroup$ Cool. And if you try that for division, what happens? $\endgroup$
    – Arthur
    Commented Nov 16, 2018 at 10:55

1 Answer 1

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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.

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