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Let * be a binary operation on the set $Q$ of rational numbers as follows:
(i) $a{ }^* \mathbf{b}=\mathbf{a}-\mathbf{b}$
(ii) $a^* b=a^2+b^2$
(iii) $a^* \mathbf{b}=\mathbf{a}+\mathbf{a b}$
(iv) $a^* b=(a-b)^2$
(v) $a * b=\frac{a b}{4}$
(vi) $a * b=a b^2$
Find which of the binary operations are commutative and which are associative.
(i) $a{ }^* \mathbf{b}=\mathbf{a}-\mathbf{b}$
(ii) $a^* b=a^2+b^2$
(iii) $a^* \mathbf{b}=\mathbf{a}+\mathbf{a b}$
(iv) $a^* b=(a-b)^2$
(v) $a * b=\frac{a b}{4}$
(vi) $a * b=a b^2$
Find which of the binary operations are commutative and which are associative.
Solution:
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Verified Answer
Operation is on the set $\mathrm{Q}$.
(i) defined as $a * b=a-b$
(a) Now b* $a=b-a$ But $a-b \neq b-a$ $\therefore \mathrm{a} * \mathrm{~b} \neq \mathrm{b}^* \mathrm{a}$
$\therefore$ Operation * is not commutative.
(b) $a *(b * c)=a *(b-c)=a-(b-c)$
$=a-b+c(a * b) * c=(a-b) * c=a-b-c$
Thus $a *(b * c)^1(a * b)^* c=\left(a^2+b^2\right)^2+c^2$
$\Rightarrow \mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\therefore$ The operation * as defined is not associative.
(ii) (a) $\mathrm{a} * \mathrm{~b}=\mathrm{a}^2+$
$b^* a=b^2+a^2=a^2+b^2 \therefore a * b=b * a$
$\therefore$ This binary operation is commutative.
(b) $a *(b * c)=a *\left(b^2+c^2\right)$
$\left.=\mathrm{a}^2+\left(\mathrm{b}^2\right)^2+\mathrm{c}^2\right)^2$
$\Rightarrow(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\left(\mathrm{a}^2+\mathrm{b}^2\right) * \mathrm{c}=\left(\mathrm{a}^2+\mathrm{b}^2\right)+\mathrm{c}^2$
Thus $\mathrm{a}^*(\mathrm{~b} * \mathrm{c})(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\therefore$ The operation * given is not associative.
(iii) Operation * is defined as $a * b=a+a b$
(a) $\mathrm{b}^* \mathrm{a}=\mathrm{b}+\mathrm{ba} \quad \therefore \mathrm{a}^* \mathrm{~b} \neq \mathrm{b}^* \mathrm{a}$
$\therefore$ The operation is not commutative.
(b) $\mathrm{a} *(\mathrm{~b} * \mathrm{c})=\mathrm{a} *(\mathrm{~b}+\mathrm{bc})$
$=a+a(b+b c)=a+a b+a b c$
$(a * b) * c=(a+a b) * c=(a+a b)+(a+a b) \cdot c$
$=a+a b+a c+a b c$
$\Rightarrow \mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\Rightarrow$ The binary operation is not associative.
(iv) The binary operation is defined as $\mathrm{a} * \mathrm{~b}=(\mathrm{a}-\mathrm{b})^2$
(a) $b * a=(b-a)^2=(a-b)^2 \Rightarrow a * b=b * a$
$\therefore$ This binary operation * is commutative.
(b) $a *(b * c)=a *(b-c)^2=\left[a-(b-c)^2\right]^2(a * b) * c$
$=(a-b)^2 * c$
$=\left[(a-b)^2-c\right]^2$
$\Rightarrow a *(b * c) \neq(a * b) * c$
$\therefore$ the operation * is not associative.
(v) Commutative and associative.
(vi) Neither commutative nor associative.
(i) defined as $a * b=a-b$
(a) Now b* $a=b-a$ But $a-b \neq b-a$ $\therefore \mathrm{a} * \mathrm{~b} \neq \mathrm{b}^* \mathrm{a}$
$\therefore$ Operation * is not commutative.
(b) $a *(b * c)=a *(b-c)=a-(b-c)$
$=a-b+c(a * b) * c=(a-b) * c=a-b-c$
Thus $a *(b * c)^1(a * b)^* c=\left(a^2+b^2\right)^2+c^2$
$\Rightarrow \mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\therefore$ The operation * as defined is not associative.
(ii) (a) $\mathrm{a} * \mathrm{~b}=\mathrm{a}^2+$
$b^* a=b^2+a^2=a^2+b^2 \therefore a * b=b * a$
$\therefore$ This binary operation is commutative.
(b) $a *(b * c)=a *\left(b^2+c^2\right)$
$\left.=\mathrm{a}^2+\left(\mathrm{b}^2\right)^2+\mathrm{c}^2\right)^2$
$\Rightarrow(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\left(\mathrm{a}^2+\mathrm{b}^2\right) * \mathrm{c}=\left(\mathrm{a}^2+\mathrm{b}^2\right)+\mathrm{c}^2$
Thus $\mathrm{a}^*(\mathrm{~b} * \mathrm{c})(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\therefore$ The operation * given is not associative.
(iii) Operation * is defined as $a * b=a+a b$
(a) $\mathrm{b}^* \mathrm{a}=\mathrm{b}+\mathrm{ba} \quad \therefore \mathrm{a}^* \mathrm{~b} \neq \mathrm{b}^* \mathrm{a}$
$\therefore$ The operation is not commutative.
(b) $\mathrm{a} *(\mathrm{~b} * \mathrm{c})=\mathrm{a} *(\mathrm{~b}+\mathrm{bc})$
$=a+a(b+b c)=a+a b+a b c$
$(a * b) * c=(a+a b) * c=(a+a b)+(a+a b) \cdot c$
$=a+a b+a c+a b c$
$\Rightarrow \mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}$
$\Rightarrow$ The binary operation is not associative.
(iv) The binary operation is defined as $\mathrm{a} * \mathrm{~b}=(\mathrm{a}-\mathrm{b})^2$
(a) $b * a=(b-a)^2=(a-b)^2 \Rightarrow a * b=b * a$
$\therefore$ This binary operation * is commutative.
(b) $a *(b * c)=a *(b-c)^2=\left[a-(b-c)^2\right]^2(a * b) * c$
$=(a-b)^2 * c$
$=\left[(a-b)^2-c\right]^2$
$\Rightarrow a *(b * c) \neq(a * b) * c$
$\therefore$ the operation * is not associative.
(v) Commutative and associative.
(vi) Neither commutative nor associative.
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