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Binary Operations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsBinary Operations5 Marks
Question
On Q, the set of all rational numbers a binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}.$ Show that * is not associative on Q.
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Answer

Let $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{b}+\text{c}}{2}\Big)$
$=\frac{\text{a}+\big(\frac{\text{b}+\text{c}}{2}\big)}{2}$
$=\frac{2\text{a}+\text{b}+\text{c}}{4}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{a}+\text{b}}{2}\big)+\text{c}}{2}$
$=\frac{\text{a}+\text{b}+2\text{c}}{4}$
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
If a = 1, b = 2, c = 3
$1\ ^*\ (2\ ^*\ 3)=1\ ^*\ \Big(\frac{2+3}{2}\Big)$
$=1\ ^*\ \frac{5}{2}$
$=\frac{1+\frac{5}{2}}{2}$
$=\frac{7}{4}$
$(1\ ^*\ 2)\ ^*3=\Big(\frac{1+2}{2}\Big)\ ^*\ 3$
$=\frac{3}{2}\ ^*\ 3$
$=\frac{\frac{3}{2}+3}{2}$
$=\frac{9}{4}$
Therefore, $\exists\text{ a}=1,\text{b}=2,\text{c}=3\in\text{Q}$ such that $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.

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