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I have this question out of Spivak's Calculus. From the field axioms, I have to prove that, for non-zero real numbers $a$ and $b$ we have that $(ab)^{-1} = a^{-1}b^{-1}$.

So this is what I have, where $e$ is the multiplicative identity: $$(ab)a^{-1}b^{-1} = (aa^{-1})(bb^{-1})$$ $$(ab)a^{-1}b^{-1} = (e)(e)$$

$$(ab)a^{-1}b^{-1} = e$$ $$(ab)^{-1}(ab)a^{-1}b^{-1} = (ab)^{-1}(e)$$ $$(e)a^{-1}b^{-1} = (ab)^{-1}$$ $$a^{-1}b^{-1} = (ab)^{-1}$$

Is this proof correct, and is there any way to write it in a more formal way.

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    @SimpleArt "$e$" is a standard notation for the identity in a group . . . – 2017-01-05
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    @NoahSchweber Ah, I ought to know such things. – 2017-01-05
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    Are these supposed to be numbers, elements of an arbitrary ring, a commutative ring? It is not true for groups in general that $(ab)^{-1}=a^{-1}b^{-1}$. – 2017-01-05
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    What kind of algebraic structure are we in? A field? – 2017-01-05
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    Only true in a structure with a commutativity property. – 2017-01-05
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    Sorry, I should have been more specific. This is over the field of real numbers – 2017-01-05

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I am not saying your method is completely right or wrong.

But first thing $(ab)^{-1} = b^{-1}a^{-1}$

One method to prove.

\begin{align*} (ab)^{-1}(ab) &= e \newline (ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline (ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline (ab)^{-1}(a)(e) &= (b^{-1}) \newline (ab)^{-1}(a) &= (b^{-1}) \newline (ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline (ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline (ab)^{-1} &= (b^{-1})(a^{-1}) \newline \end{align*}