Problem: Prove the following,

$\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad +bc}{bd}$ if b,d $\displaystyle \not=$ 0

I'm sorry to say I analyze this utterly elementary proof to the point of being unable to move.

Proof from book: $\displaystyle (ad + bc)/(bd) = (ad+bc)(bd)^{-1}$

$\displaystyle (ad + bc)(bd)^{-1}$ by (iii) = $\displaystyle ab^{-1} + cd^{-1} = \frac {a}{d} + \frac{c}{d}$

the book then cites the following proof as justification for the latter equivalence:

(iii) = $\displaystyle (ab)^{-1} = a^{-1}b^{-1}$ if a,b \not= 0

proof of (iii) from book, $\displaystyle ab(a^{-1}b^{-1}) = (a * a^{-1})(b * b^{-1}) = 1, so a^{-1} * b^{-1} = (ab)^{-1} $

I get that $\displaystyle (a)^{-1} = \frac{1}{a}$

I also understand that $\displaystyle bd(bd)^{-1} = 1$

There is a continuity gap in my comprehension, educate me please.