Home » Blog » Prove that the inverse of a product is the product of the inverses

Prove that the inverse of a product is the product of the inverses

by RoRi

Prove that (ab)^{-1} = a^{-1} b^{-1} if a, b \neq 0.

Proof. Since a, b \neq 0 we know there exist elements a^{-1}, b^{-1} \in \mathbb{R} such that aa^{-1} = 1 and bb^{-1} = 1. So,

    \begin{align*} (ab)^{-1} &= (ab)^{-1} \cdot (aa^{-1}) \cdot (bb^{-1}) \\ &= (ab)^{-1} \cdot (ab) \cdot (a^{-1} b^{-1}) & (\text{Using assoc. and comm.}) \\ &= a^{-1} b^{-1}. \qquad \blacksquare \end{align*}

One comment

  1. Anonymous says:

    My solution:

    Since a ≠ 0 and b ≠ 0, by Theorem I.11, it follows that ab ≠ 0. Hence, the multiplicative inverse of ab exists. To show that this multiplicative inverse is nothing more than the product of the multiplicative inverses of a and b, we can notice the following:

    (ab)(a^{-1}b^{-1}) = (ba)(a^{-1}b^{-1}) = b(aa^{-1})b^{-1} = b(1)b^{-1} = bb^{-1} = 1.

    Hence, the multiplicative inverse of ab is a^{-1}b^{-1}; that is, (ab)^{-1} = a^{-1}b^{-1}.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):