I'm having trouble proving this theorem from Rudin: that if $x \neq 0$ then $\frac{1}{\frac{1}{x}} = x$.
Rudin seems to solve this by referring to an earlier result that $xy = xz$ for $x \neq 0$ implies that $y = z$. I haven't quite been able to grasp this approach, as we don't seem to have access to this assumption. Another, perhaps more intuitive approach, is to deduce it from the field axioms:
\begin{align*}
\frac{1}{\frac{1}{x}} & = 1 \cdot \frac{1}{\frac{1}{x}} & & \text{Mult Identity} \\
& = \left(x \cdot \frac{1}{x}\right) \cdot \frac{1}{\frac{1}{x}} & & \text{Mult Inverse} \\
& = x \left(\frac{1}{x \cdot \frac{1}{x}} \right) & & \text{Associativity, simplification} \\
& = x \cdot 1 & & \text{Mult Inverse} \\
& = x & & \text{Mult Identity}
\end{align*}
I'm particularly unsure on whether we can perform the third line of this proof, wherein we write that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = \frac{1}{x \cdot \frac{1}{x}}$. This is surely the multiplication law for rational numbers, but our only assumption is that $x$ is an element of some field. Considering all of the possible fields — reals, rationals, complex, finite fields, etc. — it seems that this law would work, but I can't think of an axiom by which it would other than the fact that multiplication is defined and behaves as we would expect in fields, so this seems like a standard result. I'm unsure on whether I ought to prove such a result prior to using it, or if it simply follows from the definition.
Any helpful insights or hints would be greatly appreciated.
Best Answer
This is correct, except that the justification for the third equality is not appropriate. There is no “simplification” axiom. But, by definition of multiplicative inverse,$$\frac1x\cdot\frac1{\frac1x}=1,$$since, for each $z\neq0$, $z\cdot\frac1z=1$.