Prove that if $ a \neq 0$ then $b/(b/a) = a$


$ 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?

  • $\begingroup$ I'm just curious: where do you find the axioms? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:08
  • $\begingroup$ Do you mean derivation through axioms? $\endgroup$ – A.Asad Feb 11 '18 at 13:10
  • $\begingroup$ Though Rudia never used the notation $(\cdot)^{-1}$ is that section, but IMHO, I think it's fine if you write it as long as it means the inverse. In the second arrow, you used a property $(xy)^{-1} = x^{-1} y^{-1}$. This can be easily proven, by as a "proof" from the axioms (or the propositions from the book), I think it's better to mark every step, like this answer in set theory. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:18
  • $\begingroup$ Thank you, I'll do that (i.e. prove that $(xy)^{-1} = x^{-1} y^{-1}$ and mark my steps) . Is the proof fine otherwise? $\endgroup$ – A.Asad Feb 11 '18 at 13:25
  • $\begingroup$ Yes, I think so. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:27

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.

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