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Let $*$ be a binary operation on the set $\mathrm{Q}$ of rational numbers as follows:
(i) $a * b=a-b$
(ii) $a^* b=a^2+b^2$
(iii) $a^* b=a+a b$
(iv) $a * b=(a-b)^2$
Find which of the binary operations are commutative.
(i) $a * b=a-b$
(ii) $a^* b=a^2+b^2$
(iii) $a^* b=a+a b$
(iv) $a * b=(a-b)^2$
Find which of the binary operations are commutative.
Solution:
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Verified Answer
Operation is on the set Q.
(i) Defined as $a * b=a-b$
Now $b^* a=b-a$ But $a-b \neq b-a$ $\therefore \quad \mathrm{a} * \mathrm{~b} \neq \mathrm{b} * \mathrm{a}$
$\therefore$ Operation $*$ is not commutative.
(ii) $\mathrm{a} * \mathrm{~b}=\mathrm{a}^2+\mathrm{b}^2$
$b * a=b^2+a^2=a^2+b^2$
$\therefore \mathrm{a} * \mathrm{~b}=\mathrm{b} * \mathrm{a}$
$\therefore$ This binary operation is commutative.
(iii) Operation $*$ is defined as $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{ab}$ $\mathrm{b} * \mathrm{a}=\mathrm{b}+\mathrm{ba} \quad \therefore \mathrm{a} * \mathrm{~b} \neq \mathrm{b} * \mathrm{a}$
$\therefore$ The operation is not commutative.
(iv) The binary operation is defined as
$$
\begin{aligned}
&a * b=(a-b)^2 \\
&b * a=(b-a)^2=(a-b)^2 \\
&\Rightarrow a * b=b * a
\end{aligned}
$$
$\therefore$ This binary operation $*$ is commutative.
(i) Defined as $a * b=a-b$
Now $b^* a=b-a$ But $a-b \neq b-a$ $\therefore \quad \mathrm{a} * \mathrm{~b} \neq \mathrm{b} * \mathrm{a}$
$\therefore$ Operation $*$ is not commutative.
(ii) $\mathrm{a} * \mathrm{~b}=\mathrm{a}^2+\mathrm{b}^2$
$b * a=b^2+a^2=a^2+b^2$
$\therefore \mathrm{a} * \mathrm{~b}=\mathrm{b} * \mathrm{a}$
$\therefore$ This binary operation is commutative.
(iii) Operation $*$ is defined as $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{ab}$ $\mathrm{b} * \mathrm{a}=\mathrm{b}+\mathrm{ba} \quad \therefore \mathrm{a} * \mathrm{~b} \neq \mathrm{b} * \mathrm{a}$
$\therefore$ The operation is not commutative.
(iv) The binary operation is defined as
$$
\begin{aligned}
&a * b=(a-b)^2 \\
&b * a=(b-a)^2=(a-b)^2 \\
&\Rightarrow a * b=b * a
\end{aligned}
$$
$\therefore$ This binary operation $*$ is commutative.
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