MCQ
On the set $Q^+$ of all positive rational numbers a binary operation $*$ is defined by $\text{a}^*\text{b}=\frac{\text{ab}}2\forall\text{ a, b}\in \text{Q}^+$. The inverse of 8 is:
- A$\frac{1}{8}$
- ✓$\frac{1}2$
- C$2$
- D$4$
✓
Answer
Correct option: B.
$\frac{1}2$
Let e be the identity element in $Q^+$ with respect to $*$ such that
$a^ * e = a = e^ * a, \forall\text{ a}\in\text{Q}^+$
$a^ * e = a$ and $e^ * a = a, \forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{2}=\text{a}$ and $\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 2, \forall\text{ a}\in\text{Q}^+$
Thus$, 2$ is the identity element in $Q^+$ with respect to $*.$
Let $\text{b}\in\text{Q}^+$ be the inverse of $8.$ Then,
$8^ * b = e = b^ * 8$
$8^ * b = e$ and $b^ * 8 = e$
$\frac{(8)\text{b}}2=2$ and $\frac{\text{b}(8)}2=2$ $[\because\ \text{e}=2]$
$b = 12$
Thus, $\frac{1}2$ is the inverse of $8.$
$a^ * e = a = e^ * a, \forall\text{ a}\in\text{Q}^+$
$a^ * e = a$ and $e^ * a = a, \forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{2}=\text{a}$ and $\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 2, \forall\text{ a}\in\text{Q}^+$
Thus$, 2$ is the identity element in $Q^+$ with respect to $*.$
Let $\text{b}\in\text{Q}^+$ be the inverse of $8.$ Then,
$8^ * b = e = b^ * 8$
$8^ * b = e$ and $b^ * 8 = e$
$\frac{(8)\text{b}}2=2$ and $\frac{\text{b}(8)}2=2$ $[\because\ \text{e}=2]$
$b = 12$
Thus, $\frac{1}2$ is the inverse of $8.$
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