Prove that 
.
Proof. On the one hand since 1 is the multiplicative identity we have,
![\[ 1 \cdot 1^{-1} = 1^{-1}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-72dfbc7b2b53cd920b48d38de7791fd4_l3.png "Rendered by QuickLaTeX.com")
On the other hand, from Theorem I.10 (Exercise I.3.3, #1 part (d), we have 
. Hence,
![\[ 1^{-1} \cdot (1^{-1})^{-1} = 1 \qquad \implies \qquad 1^{-1} \cdot 1 = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8811a1ee3008641f98bfe214903bb0e1_l3.png "Rendered by QuickLaTeX.com")
Therefore, since 
,
![\[ 1 = 1^{-1}. \qquad \blacksqaure\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5f340e9a7caf432089182b559b923576_l3.png "Rendered by QuickLaTeX.com")
Multiplicative identity is its own multiplicative inverse
Proof. On the one hand since 1 is the multiplicative identity we have,
Hi, RoRi. Your proof is of course correct; but I think there’s no need of using the reciprocal of the reciprocal. So, I will offer an alternative.
You could simply say that:
On the one hand, by Axiom 4, we know that 1 is the multiplicative identity, so 1 * (1^(-1)) = 1^(-1).
On the other hand, by Theorem I.8. (and Axiom 6), we know that 1 * (1^(-1)) = 1 (definition of reciprocal).
Therefore, you have 1^(-1) = 1 * (1^(-1)) = 1, i.e., 1^(-1) = 1.
(P.S: Sorry, for not using LaTeX, I don’t know how to use it yet. I hope the answer is clear though)