How to prove " multiplicative inverse of inverse of $a$ is $a$ itself" ( with $a$, say, a real number) in basic arithmetics? ( $\frac{1}{1/a}$ = $a$.) Suppose I want to treat basic arithmetics on real numbers as a little deductive system ( without using abstract algebra). 
In order to prove the " divide by a fraction " rule ( $\frac{a}{b/c}$ = $\frac{ac}{b}$), I need the " inverse of inverse rule" (  How to deduce the "divide by a fraction" formula from the definition of division), namely : 
$\frac{1} {1/a}$ = $a$  ( provided a is not equal to 0). 
How can this rule be proved without using the " divide by a fraction" rule? 

I have done this : 
Assuming 


*

*for all $a$, $\frac aa$$=$$1$ ( provided $a$ is not null) 

*for all $a$, $b$, $\frac ab$$=$$a\times\frac 1b$ ( provided $b$ is not null) 

*number $1$ is the identity element for multiplication and for division. 

*for all $a,b,c,d$, $\frac {ac}{bd}$ $=$ $\frac ab\times\frac cd$ ( with $c,d$ not equal to $0$). 
$\frac{1}{\frac 1a}$= $\frac{\frac aa}{\frac 1a}$= $\frac{\frac a1\times\frac 1a} {1\times\frac 1a}$= $\frac {\frac a1}{1}\times\frac{\frac 1a}{\frac 1a}$= $\frac a1\times1$= $a\times1$= $a$ 
provided $a$ is not null. 

Multiply both sides by the inverse of $a$. You know from the definition of inverse that $$a\cdot\frac 1a=\frac1a \cdot a=1$$
Then the right hand side is $1$
On the left hand side use that the inverse of $a$ is $b$. then you have $$\frac 1{1/a}\cdot \frac 1a=\frac 1b\cdot b=1$$
Subtract first term and last term from the two lines and you get 
$$\left(\frac1{1/a}-a\right)\frac 1a=0$$
We multiply both sides by $a$, then move $a$ to the right hand side. 
The only thing that I used are associativity, commutativity, definition of inverse, $a\ne 0$ and $1/a\ne 0$. 
