Solutions

(i)

$$\begin{aligned} \frac{a}{b} &= a b^{-1} \\ &= ab^{-1} \cdot 1 \\ &= ab^{-1} \cdot c c^{-1} \\ &= ac \cdot b^{-1} c^{-1} \\ &= ac {(bc)}^{-1} \\ &= \frac{ac}{bc} \end{aligned}$$

(ii)

$$\begin{aligned} \frac{a}{b} + \frac{c}{d} & = a b^{-1} + c d^{-1} \\ &= ab^{-1} \cdot 1 + cd^{-1} \cdot 1 \\ &= ab^{-1} \cdot d d^{-1} + cd^{-1} \cdot bb^{-1} \\ &= ad b^{-1} d^{-1} + bc b^{-1}d^{-1} && \text{Commutativity} \\ &= ad {(bd)}^{-1} + bc {(bd)}^{-1} && (i) \\ &= (ad + bc){(bd)}^{-1} && \text{Distributivity} \\ &= \frac{ad + bc}{bd} \end{aligned}$$

(iii)

$$\begin{aligned} a, b \neq 0 & \implies ab \neq 0 \\ & \implies \exists \ {(ab)}^{-1} \\ & \implies {(ab)}^{-1} = {(ab)}^{-1} \cdot 1 \cdot 1 = {(ab)}^{-1} \cdot aa^{-1} \cdot bb^{-1} = {(ab)}^{-1} (ab) a^{-1} b^{-1} = a^{-1} b^{-1} \end{aligned}$$

(iv)

$$\begin{aligned} b, d \neq 0 & \implies \exists \ b^{-1}, d^{-1} \\ & \implies \frac{a}{b} \cdot \frac{c}{d} = a b^{-1} \cdot c d^{-1} = ac \cdot b^{-1} d^{-1} = ac \cdot (bd)^{-1} = \frac{ac}{bd} && (iii) \end{aligned}$$

(v) We first need to prove that ${(a^{-1})}^{-1} = a$

$$\begin{aligned} {(a^{-1})}^{-1} &= {(a^{-1})}^{-1} \cdot 1 \\ &= {(a^{-1})}^{-1} \cdot (a a^{-1}) \\ &= {(a^{-1})}^{-1} \cdot a^{-1} \cdot a && \text{Commutativity} \\ &= a \end{aligned}$$

We can then go on and prove the actual exercise:

$$\begin{aligned} b, c, d \neq 0 & \implies \exists \ b^{-1}, c^{-1}, d^{-1} \\ & \implies \frac{a}{b} / \frac{c}{d} = a b^{-1} {(c d^{-1})}^{-1} = a b^{-1} c^{-1} {(d^{-1})}^{-1} = a b^{-1} c^{-1} d = ad {(bc)}^{-1} = \frac{ad}{bc} \end{aligned}$$

(vi)

$$\begin{aligned} \frac{a}{b} = \frac{c}{d} & \iff a b^{-1} = c d^{-1} \\ & \iff a b^{-1} (b d) = c d^{-1} (b d) \\ & \iff a (b^{-1} b) d = c(d^{-1}d)b && \text{Associativity and Commutativity} \\ & \iff ad = bc \end{aligned}$$