Ok, check this out. I have the following information;

$\displaystyle R'=\frac{V}{I}$

$\displaystyle I=\frac{V}{R}+\frac{V}{r_v}$

Taking $\displaystyle I$ and plugging it into $\displaystyle R'$, the following happens;

$\displaystyle R'=\frac{V}{\frac{V}{R}+\frac{V}{r_v}}=\frac{V}{\f rac{V}{R+r_v}}=\frac{V(R+r_v)}{V}=R+r_v$

So far so good? Well, $\displaystyle R+r_v$ is the same as $\displaystyle \frac{Rr_v}{R+r_v}$ right? I mean, $\displaystyle \frac{Rr_v}{R}+\frac{Rr_v}{r_v}$ once simplified, gives $\displaystyle R+r_v$ right?

The problem I have asks to show what $\displaystyle \frac{|R'-R|}{R}$ is in $\displaystyle R$ and $\displaystyle r_v$.

Solving looks like this; $\displaystyle \frac{|R'-R|}{R}=\frac{|\frac{Rr_v}{R+r_v}-R|}{R}=\frac{|\frac{Rr_v-R(R+r_v)}{R+r_v}|}{R}=\frac{|Rr_v-R(R+r_v)|}{R(R+r_v)}=\frac{|r_v-(R+r_v)|}{R+r_v}$

$\displaystyle =\frac{|-R|}{R+r_v}=\color{red}\frac{R}{R+r_v}$

However, if I plug in $\displaystyle R'=R+r_v$ instead of $\displaystyle R'=\frac{Rr_v}{R+r_v}$ the following ensues;

$\displaystyle \frac{|R'-R|}{R}=\frac{|(R+r_v)-R|}{R}=\color{red}\frac{r_v}{R}\neq{\frac{R}{R+r_v }}$ right?

Or can someone show me that $\displaystyle \frac{r_v}{R}=\frac{R}{R+r_v}$ because that would be AWESOME.