How to show that show that $\frac{v+u}{1+ uv/c^2}=c$ when $u=c$?

Question

I am trying to show that $\dfrac{v+u}{1+\dfrac{uv}{c^2}}=c$ when $u=c$.

Context

It's needed for a physics proof that I'm working on. This is the formula for relative velocity, $u$ represents the velocity of the observer, $v$ represents the velocity of a projectile and $c$ represents the speed of light.

Einstein said that this is right, so I'm sure that it is I'm just not sure how it simplifies.

Answer 1

If you mean $$\frac{u+v}{1+\frac{uv}{c^2}} $$ as noted in the comments. Then let $u = c$, and we get \begin{align} \require{cancel} \frac{u+v}{1+\frac{uv}{c^2}} &= \frac{c+v}{1+\frac{\cancel{c}v}{c^\cancel{2}}}\\ &=\frac{c+v}{1+\frac{v}{c}}\\ &=\frac{c+v}{1+\frac{v}{c}}\frac{c}{c}\\ &= \frac{c\cancel{(c+v)}}{\cancel{c+v}}\\ &= c. \end{align}

Answer 2

Lets plut in $u=c$:

$u+\frac{v}{1+\frac{uv}{u^2}}=u$

$\iff 0=\frac{v}{1+\frac{v}{u}}= \frac{v}{\frac{u+v}{u}}$

$\iff 0 = \frac{uv}{u+v}$

$\iff u=0$ or $v=0$, so that is what you can do algebraically, but from the point of view of physics you should tell us what all those variables stand for.

Answer 3

It is not correct. If you let $v=c/2$ for example, $$u+\dfrac{v}{1+\dfrac{uv}{c^2}}=c+\frac {\frac c2}{1+\frac{c^2}{2c^2}}=c+\frac{c^3}{3c^2}=\frac43c$$ If you mean $$\dfrac{u+v}{1+\dfrac{uv}{c^2}}=\dfrac{c+v}{1+\dfrac{cv}{c^2}}=\frac{c^3+c^2v}{c^2+cv}=c$$