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# Let a * b = 1 cm (a, b) for all values ofa, b ∈ N.

Question:

Let a * b = 1 cm (a, b) for all values ofa, b ∈ N.

(i) Find $(12 * 16)$.

(ii) Show that * is commutative on $\mathrm{N}$.

(iii) Find the identity element in $\mathrm{N}$.

(iv) Find all invertible elements in N.

Solution:

To find: (i)

LCM of 12 and 16

Prime factorizing 12 and 16 we get.

$20=2^{2} \times 3$

$16=2^{4}$

$\Rightarrow \mathrm{LCM}$ of 20 and $16=2^{4} \times 3=48$

(ii) To find LCM highest power of each prime factor has been taken from both the numbers and multiplied.

So it is irrelevant in which order the number are taken as their prime factors will remain the same.

So LCM(a,b) = LCM(b,a)

So * is commutative.

(iii)let $x \in N$ and $x^{*} 1=\operatorname{Icm}(x, 1)=x=\operatorname{Icm}(1, x)$

1 is the identity element.

(iv)let there exist $y$ in $n$ such that $x^{*} y=e=y^{*} x$

Here $e=1$

$\operatorname{Lcm}(x, y)=1$

This happens only when x = y = 1.

1 is the invertible element of n with respect to *.