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**beebe** $\displaystyle Let G=\mathbb{Q}-\{1\}$

$\displaystyle a*b=a+b-ab \forall a,b \in G$

I need to show my group is closed under *, and I'm happy with my proof that all of the possible outcomes are rational numbers. So now I need to show that 1 is not a possible outcome. Is my proof of this acceptable?

Assume $\displaystyle a+b-ab=1$

$\displaystyle a-ab=1-b$

$\displaystyle a(1-b)=1-b$

$\displaystyle a=\frac{1-b}{1-b}$

$\displaystyle a=1$

$\displaystyle 1 \notin G$

Therefore, $\displaystyle a+b-ab \neq 1$