# Can you please check this proof?

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.

• @SimpleArt "$e$" is a standard notation for the identity in a group . . . – Noah Schweber Jan 5 '17 at 1:25
• @NoahSchweber Ah, I ought to know such things. – Simply Beautiful Art Jan 5 '17 at 1:26
• Are these supposed to be numbers, elements of an arbitrary ring, a commutative ring? It is not true for groups in general that $(ab)^{-1}=a^{-1}b^{-1}$. – juan arroyo Jan 5 '17 at 1:26
• What kind of algebraic structure are we in? A field? – Milo Brandt Jan 5 '17 at 1:27
• Only true in a structure with a commutativity property. – ÍgjøgnumMeg Jan 5 '17 at 1:47

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