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If $*$ be binary operation defined on $\mathrm{R}$ by $\mathrm{a} * \mathrm{~b}=1+\mathrm{ab}, \forall$
$\mathrm{a}, \mathrm{b} \in \mathrm{R}$. Then, the operation $*$ is
(i) commutative but not associative.
(ii) associative but not commutative.
(iii) neither commutative nor associative.
(iv) both commutative and associative.
$\mathrm{a}, \mathrm{b} \in \mathrm{R}$. Then, the operation $*$ is
(i) commutative but not associative.
(ii) associative but not commutative.
(iii) neither commutative nor associative.
(iv) both commutative and associative.
Solution:
2496 Upvotes
Verified Answer
Since, $\mathrm{a} * \mathrm{~b}=1+\mathrm{ab}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}$
$$
a * b=a b+1=b * a
$$
So, $*$ is a commutative binary operation.
Also, $a *(b * c)=a *(1+b c)=1+a(1+b c)$
$\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1+\mathrm{a}+\mathrm{abc}$
$(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1+\mathrm{ab}) * \mathrm{c}$
$=1+(1+a b) c$
$=1+c+a b c$
From Eqs. (i) and (ii),
$$
\mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}
$$
So, $*$ is not associative
Hence, $*$ is commutative but not associative.
$$
a * b=a b+1=b * a
$$
So, $*$ is a commutative binary operation.
Also, $a *(b * c)=a *(1+b c)=1+a(1+b c)$
$\mathrm{a} *(\mathrm{~b} * \mathrm{c})=1+\mathrm{a}+\mathrm{abc}$
$(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(1+\mathrm{ab}) * \mathrm{c}$
$=1+(1+a b) c$
$=1+c+a b c$
From Eqs. (i) and (ii),
$$
\mathrm{a} *(\mathrm{~b} * \mathrm{c}) \neq(\mathrm{a} * \mathrm{~b}) * \mathrm{c}
$$
So, $*$ is not associative
Hence, $*$ is commutative but not associative.
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