Prove that .

*Proof.*On the one hand since 1 is the multiplicative identity we have,

On the other hand, from Theorem I.10 (Exercise I.3.3, #1 part (d), we have . Hence,

Therefore, since ,

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Stumbling Robot

A Fraction of a Dot
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Multiplicative identity is its own multiplicative inverse

* Proof. * On the one hand since 1 is the multiplicative identity we have,
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### Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):

Prove that .

On the other hand, from Theorem I.10 (Exercise I.3.3, #1 part (d), we have . Hence,

Therefore, since ,

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)