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Let ${ }^*$ be the binary operation on N defined by $a^* b=$ H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on $\mathbf{N}$ ?
MathematicsRelations and Functions (Class 12)
Solution:
2842 Upvotes Verified Answer
Binary operation on set $\mathrm{N}$ is defined as $\mathrm{a} * \mathrm{~b}=\mathrm{HCF}$ of $\mathrm{a}$ and $\mathrm{b}$
(a) We know HCF of a, b= HCF of b, a
$\therefore \mathrm{a}^* \mathrm{~b}=\mathrm{b} * \mathrm{a} \quad \therefore$ Binary operation $*$ is commutative.
(b) $\mathrm{a}^*(\mathrm{~b} * \mathrm{c})=\mathrm{a}^*(\mathrm{HCF}$ of b $\mathrm{c})=\mathrm{HCF}$ of a and (HCF of $\left.\mathrm{b}, \mathrm{c}\right)$ $=\mathrm{HCF}$ of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$
Similarly $(a * b) *$ c $=$ HCF of $a, b$, and $c$ $\Rightarrow(a * b) * c=a *(b * c)$
Binary operation * as defined above is associative.
(c) $1 * \mathrm{a}=\mathrm{a}^* 1=1 \neq \mathrm{a}$
$\therefore$ There does not exists any identity element.

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