Prologue - Problem 3

We have to prove the following assertions.

(i) $\frac{a}{b} = \frac{ac}{bc}$, if $b,c \neq 0$. \[ \begin{align*} \frac{a}{b} &= ab^{-1} \\ &= ab^{-1}cc^{-1} \\ &= acb^{-1}c^{-1} \\ &= ac(bc)^{-1} && \text{| see (iii)} \\ &= \frac{ac}{bc} \end{align*} \]

(ii) $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$. \[ \begin{align*} \frac{a}{b} + \frac{c}{d} &= ab^{-1} + cd^{-1} \\ &= ab^{-1}dd^{-1} + cd^{-1}bb^{-1} \\ &= adb^{-1}d^{-1} + cbd^{-1}b^{-1} \\ &= ad(bd)^{-1} + bc(bd)^{-1} && \text{| see (iii)} \\ &= (ad + bc)(bd)^{-1} \\ &= \frac{ad + bc}{bd} \end{align*} \]

(iii) $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b \neq 0$. \[ \begin{align*} aa^{-1}bb^{-1} &= 1 \\ aba^{-1}b^{-1} &= 1 \\ (ab)(a^{-1}b^{-1}) &= 1 \\ (ab)(a^{-1}b^{-1}) &= (ab)(ab)^{-1} && \text{| inverse } (ab) \\ a^{-1}b^{-1} &= (ab)^{-1} \end{align*} \]

(iv) $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{db}$, if $b,d \neq 0$. \[ \begin{align*} \frac{a}{b} \cdot \frac{c}{d} &= ab^{-1}cd^{-1} \\ &= ac(bd)^{-1} \\ &= \frac{ac}{bd} = \frac{ac}{db} \end{align*} \]

(v) $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$, if $b,c \neq 0$. \[ \begin{align*} ab^{-1}(cd^{-1})^{-1} &= ab^{-1}c^{-1}d && \text{| } (a^{-1})^{-1} = a \\ &= ad(bc)^{-1} \\ &= \frac{ad}{bc} \end{align*} \]

(vi) If $b,d \neq 0$, then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$. Also determine when $\frac{a}{b} = \frac{b}{a}$. \[ \begin{align*} \frac{a}{b} &= \frac{c}{d} \\ ab^{-1} &= cd^{-1} \\ ab^{-1}bd &= cd^{-1}bd \\ ad &= cb \\ ad &= bc \end{align*} \] And: \[ \begin{align*} ad &= bc \\ aa &= bb \\ a &= b \end{align*} \]