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On the set $Q$ of all rational numbers the operation * which is both associative and commutative is given by $a * b=$
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Verified Answer
The correct answer is:
$a+b+a b$
If $a * b=b * a$, then the operation is commutative.
If $(a * b) * c=a *(b * c)$, then the operation is associative.
By option (d),
$\begin{aligned}
&a * b=a+b+a b \\
&b * a=b+a+b a=a+b+a b
\end{aligned}$
Here, $a * b=b * a$, so the operation is commutative.
$\begin{aligned}
(a * b) * c &=(a+b+a b) * c \\
&=(a+b+a b)+c+(a+b+a b) c \\
&=a+b+a b+c+a c+b c+a b c \\
&=a+b+c+a b+a c+b c+a b c \\
a *(b * c) &=a *(b+c+b c) \\
&=a+b+c+b c+a(b+c+b c) \\
&=a+b+c+b c+a b+a c+a b c \\
\text { Here, }(a * b) & * c=a *(b * c) .
\end{aligned}$
So, the operation is associative.
If $(a * b) * c=a *(b * c)$, then the operation is associative.
By option (d),
$\begin{aligned}
&a * b=a+b+a b \\
&b * a=b+a+b a=a+b+a b
\end{aligned}$
Here, $a * b=b * a$, so the operation is commutative.
$\begin{aligned}
(a * b) * c &=(a+b+a b) * c \\
&=(a+b+a b)+c+(a+b+a b) c \\
&=a+b+a b+c+a c+b c+a b c \\
&=a+b+c+a b+a c+b c+a b c \\
a *(b * c) &=a *(b+c+b c) \\
&=a+b+c+b c+a(b+c+b c) \\
&=a+b+c+b c+a b+a c+a b c \\
\text { Here, }(a * b) & * c=a *(b * c) .
\end{aligned}$
So, the operation is associative.
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