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Referos · #3 July 3rd, 2009, 07:17 PM

2) The inverse of $a$ is $-a$ . Assume that there exists another inverse element denoted as $a^{-1}$ such that $a^{-1}\neq-a$

Now, by the definition of inverse element, we have:

$a+a^{-1}=0$

But from the third axiom, we have:

$a+(-a)=0$

Ergo:

$a+a^{-1}=a+(-a)$

$a^{-1}=-a$

Which contradicts our original assumption $a^{-1}\neq-a$ , so the inverse element must be unique.