Question about the proof of simple algebra rule $\frac{1}{\frac{1}{a}} = a$ I have a question about a proof I saw in a book about basic algeba rules. The  rule to prove is: 
\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}
And the proof: 
\begin{eqnarray*}
1 = a \frac{1}{a} \Longrightarrow 1 =  \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}
Why is it allowed to just replace $a$ with $1/a$? What's the explanation behind it? 
 A: Let $x=\frac{1}{a}$. Then:
\begin{eqnarray*}
1 = x \frac{1}{x} \Longrightarrow 1 =  \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}
Better? You're right, replacing $a\to\frac{1}{a}$ is a minor abuse of notation, because they're implicitly changing the variable without telling you. But you can fix that by just letting $x=\frac{1}{a}$ at the start.
A: $1/a$ is the multiplicative inverse $a^{-1}$ of 
$a (\not =0)$,  i .e. $a^{-1}a=1$.
Need to show:
$(a^{-1})^{-1} =a;$
Since
$(a^{-1})^{-1}(a^{-1})=1$;
$(a^{-1})^{-1}(a^{-1})a=1a=a$;
$(a^{-1})^{-1}(a^{-1}a)=a;$
$(a^{-1})^{-1}1=(a^{-1})^{-1}= a$.
A: I'm not terribly fond of this proof.
I would rather go on defining the inverse:


*

*$y$ is the inverse of $x\iff xy=1$ then we write it $y=\frac 1x$.

*since everything is symmetrical $x$ is also the inverse of $y=\frac 1x$.


From there on, we have $\frac 1{\frac 1a}$ is the inverse of $\frac 1a$ which is $a$.
A: I don't think they are replacing $a \to \frac1a$. I think the logic in the first implication is they are taking 1 over both sides. So $1/1\to 1$ on the LHS, and on the RHS, 
$$a \to \frac1a \text{ and } \frac1a \to \frac{1}{\frac1a}.$$
Similarly, in the next implication, they multiply both sides by $a$.
A: I proceed in steps:
The first implication follows because the reciprocal of $1$ is $1.$ Thus, they got $$1=\frac 1a\frac{1}{\frac 1a}$$ by taking reciprocals of both sides of $$1=a\frac 1a.$$
The second implication follows since they only multiplied both sides by $a.$ This gives the last equation, which is what was to be proved.

A shorter way is to first define $1/a=a^{-1}.$ Then it is almost trivial to see that $$a\frac 1a=1\implies a\frac 1a\left(\frac 1a\right)^{-1}=\left(\frac 1a\right)^{-1}\implies a=\frac{1}{\frac 1a}.$$
