RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

$a^{\ast}b = \frac{a + b}{2}\forall a, b\varepsilon N$

To find $\ast$ is commutative of or not.

Now, $a^{\ast}b \frac{a +b}{2} =\frac{b + a}{2} \therefore \text{(addition is commutative on N)}$

$= b^{\ast}a$

$\text{So a}^{\ast}b = b^{\ast}a$

$\therefore \ast \text{is commutative}$

To find $a^{\ast}(b^{\ast}c) = (a^{\ast}b)^\ast c $ or Not

Now $a^{\ast}(b^{\ast}c) = a^{\ast} = \frac{b + c}{2} =\frac{a+\bigg(\frac{b + c}{2}\bigg)}{2}= \frac{2a + b + c}{4}\dots\dots\text{(i)}$

$(a^{\ast}b)^{\ast}c = \bigg(\frac{a + b}{2}\bigg)^{\ast} c = \frac{\frac{a + b}{2} + c}{2}$

$= \frac{a + b + 2c}{4} \dots\dots\text{(ii)}$

From (i) and (ii)

$(a^{\ast}b)^{\ast} c \neq a^{\ast}(b^{\ast}c)$

Hence the operation is not associative.