The set of all rational numbers other than $1$ forms a commutative group with respect to the operation $*$ defined by $$a * b = a + b - ab$$ for all $a, b \in \mathbb{Q} \setminus \{1\}$.
How do I verify that
$$( a * b )^{-1} = a^{-1} * b^{-1}?$$
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1Follows immediately by transport of structure as in the linked dupe. – Bill Dubuque Jul 23 '22 at 14:00
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1You can just use associativity and commutativity to prove it directly (i.e. consider $(ab)(a^{-1}*b^{-1})$) – Lorago Jul 23 '22 at 14:02