Why are those equivalent transformations of inverse functions not the same thing?

Why are $\frac{1}{f}=\frac{1}{g}+\frac{1}{b}$ and $f=g+b$ not the same thing?

• Try $1/2 + 1/2$. Now try $2 + 2$. – aes Nov 22 '14 at 20:24

we have $\frac{1}{f}=\frac{g+b}{bg}$ and then we have $f=\frac{bg}{b+g}$