Question 845451
If *[tex \large a > 0], then *[tex \large a^{-1} = \frac{1}{a}] since (1/a)*a = a*(1/a) = 1. If *[tex \large a < 0], then *[tex \large a^{-1} = a]. In each of these cases, the inverse is unique.


To show *[tex \large a*(b*c) = (a*b)*c] for all *[tex \large a,b,c \in \mathbb{R} \backslash \{0\}], we have four cases:


*[tex \large a,b > 0 \Rightarrow a*(b*c) = (a*b)*c = abc]
*[tex \large a > 0, b < 0 \Rightarrow a*(b*c) = (a*b)*c = \frac{ab}{c}]
*[tex \large a < 0, b > 0 \Rightarrow a*(b*c) = (a*b)*c = \frac{a}{bc}]
*[tex \large a, b < 0 \Rightarrow a*(b*c) = (a*b)*c = \frac{ac}{b}]


Associativity holds in all four cases, so the group operation is associative. Hence (R, *) is a group.