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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
Consider the binary operations *: R × R → R and o: R × R → R defined as $\text{a}∗\text{b} = |\text{a } – \text{ b}|\text{ and}\text{ a} o \text{b} = \text{a},\forall\text{a},\text{b}\in\text{R}.$ Show that * is commutative but not associative, o is associative but not commutative. Further, show that $\forall\text{a},\text{b},\text{c}\in\text{R},\text{a} *(\text{b}o\text{c}) = (\text{a} *\text{b})o(\text{a}*\text{b}).$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
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Answer

It is given that *: R × R → and o: R × R → R is defined as
a * b = |a - b| and aob = a, $\Box\text{a},\text{b}\in\text{R}.$
For $\text{a},\text{b}\in\text{R},$ we have:
a * b = |a - b|
b * a = |b - a| = |-(a - b)| = |a - b|
$\therefore$ The operation * is commutative.
It can be observed that,
(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2
1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0
$\therefore(1*2) *3\neq1*(2*3)(\text{where }1, 2,3\in\text{R})$
$\therefore$ The operation * is not associative.
Now, consider the operation o:
It can be observed that 1o2 = 1 and 2o1 = 2.
$\therefore1o2\neq2o1\ (\text{where }1,2\in\text{R})$
$\therefore$ The operation o is not commutative.
Let $\text{a},\text{b},\text{c}\in\text{R}.$ Then, we have:
(aob)oc = aoc = a
ao(boc) = aob = a
⇒ (aob)oc = ao(boc)
$\therefore$ The operation o is associative.
Now, let $\text{a},\text{b},\text{c}\in\text{R},$ then we have:
a * (boc) = a * b = |a - b|
(a * b)o(a * c) = (|a - b|)o(|a - c|) = |a - b|
Hence, a * (boc) = (a * b)o(a * c).
Now,
1o(2 * 3) = 1o(|2 - 3|) = 1o1 = 1
(1o2) * (1o3) = 1 * 1 = |1 - 1| = 0
$\therefore1o(2 * 3)\neq(1o2)*(1o3)\ (\text{where }1,2,3\in\text{R})$
$\therefore$ The operation o does not distribute over *.

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