Define $*$ on $\mathrm{N}$ by $\mathrm{m} * \mathrm{n}=1 \mathrm{~cm}(\mathrm{~m}, \mathrm{n})$. Show that $*$ is a binary operation which is commutative as well as associative.
$*$ is an operation as $m * n=\operatorname{LCM}(m, n)$ where $m, n \in N$. Let $m=2$ and $b=3$ two natural numbers.
m*n = 2*3
$=\operatorname{LCM}(2,3)$
$=6 \in \mathrm{N}$
So, $*$ is a binary operation from $\mathrm{N} \times \mathrm{N} \rightarrow \mathrm{N}$.
For commutative,
$\mathrm{n} * \mathrm{~m}=3 * 2$
$=\operatorname{LCM}(3,2)$
$=6 \in \mathrm{N}$
Since $m^{*} n=n^{*} m$, hence $*$ is commutative operation.
Again, for associative, let p = 4
$\mathrm{m} *(\mathrm{n} * \mathrm{p})=2 * \operatorname{LCM}(3,4)$
$=2 * 12$
$=\operatorname{LCM}(2,12)$
$=12 \in \mathrm{N}$
$(m * n) * p=\operatorname{LCM}(2,3) * 4$
$=6 * 4$
$=\operatorname{LCM}(6,4)$
$=12 \in \mathrm{N}$
As $m *(n * p)=(m * n) * p$, hence $*$ an associative operation.