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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.

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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 $z\neq0$, $z\cdot\frac1z=1$.

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Suggestion:

Let $x\not =0.$

$x^{-1}= 1/x.$

Starting with the second line:

$ (xx^{-1})(x^{-1})^{-1}= $

$x(x^{-1}(x^{-1})^{-1})=x \cdot 1=x$.

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$$\frac1{\dfrac1x}=x\iff \dfrac1x\frac1{\dfrac1x}=\dfrac1xx\iff1=1.$$

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