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RELATIONS AND FUNCTIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSRELATIONS AND FUNCTIONS3 Marks
Question
  1. Is the binary operation $\ast,$ defined on set N, given by a$a^{\ast}b = \frac{a + b}{2}$ for all $a, b\varepsilon N,$ commutative?
  2. Is the above binary operation $\ast$ associative?
✓

Answer

$(i) \text{Given N be the set}$$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.

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