Prove that if $ a \neq 0$ then $b/(b/a) = a$
Proof:
$ b/(b/a) = b. 1/(b/a)$ (by Multiplication Axiom 4 (M4))
$\implies b/(b/a) $ = $b.(b/a)^{-1}$ (by M5)
$\implies $ $b/(b/a)$ = $b.(b.(1/a))^{-1}$
$\implies $ $b/(b/a)$ = $b.(b.(a)^{-1})^{-1}$
$\implies $ $b/(b/a)$ = $b.[1/(b.(a)^{-1})]$
$\implies $ $b/(b/a)$ = $b . (1/b) . (1/a^{-1})$
$\implies $ $b/(b/a)$ = $1 .(1/a^{-1})$
$\implies $ $b/(b/a)$ = $(a^{-1})^{-1}$
$\implies $ $b/(b/a)$ = a
Is this correct? Can anyone please verify?
As OP's comments suggest, to derive the proof from the five multiplication axioms in Baby Rudin, break $1$ into the product of $a$ and $1/a$ in the second step. $\require{action}$ $$ \begin{aligned} b / (b/a) &= \texttip{b \cdot 1 \cdot \left( \frac{1}{b/a} \right)} {(M4): multiplication by identity} \\ &= \texttip{b \cdot \left( \frac{1}{a} \right) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M5): existence of inverse} \\ &= \texttip{(b/a) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M3): regroup the leftmost two factors} \\ &= \texttip{a \cdot (b/a) \cdot \left( \frac{1}{b/a} \right)} {(M2): multiplication is commutative} \\ &= \texttip{a \cdot \left( (b/a) \cdot \left( \frac{1}{b/a} \right) \right)} {(M3): multiplication is associative} \\ &= \texttip{a \cdot 1}{(M5): multiplication by inverse} \\ &= \texttip{a}{(M4): multiplication by identity} \end{aligned} \bbox[4pt,border: 1px solid red]{ \begin{array}{l} \text{If you cannot figure out why a line}\\ \text{is true, move your mouse over}\\ \text{RHS of that line for hint.} \end{array}} $$ Remarks: The above proof is one line shorter than OP's one.