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In the group $Q-\{1\}$ under the binary operation * defined by $a * b=a+b+a b$ the inverse of 10 is
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Verified Answer
The correct answer is:
$\frac{-10}{11}$
Let the inverse element of 10 be $e$, then
$$
a * e=0
$$
Therefore, $10 \div e=10+e+10 e=0$
$$
\Rightarrow \quad 10+11 e=0
$$
$\begin{array}{rr}\Rightarrow & 11 e=-10 \\ \Rightarrow & e=\frac{-10}{11}\end{array}$
$$
a * e=0
$$
Therefore, $10 \div e=10+e+10 e=0$
$$
\Rightarrow \quad 10+11 e=0
$$
$\begin{array}{rr}\Rightarrow & 11 e=-10 \\ \Rightarrow & e=\frac{-10}{11}\end{array}$
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