Let * be a binary operation defined on Q+ by the rule a*b= ab 3 a… - Vidyadip

Question

Let * be a binary operation defined on Q⁺ by the rule $\text{a}*\text{b}=\frac{\text{ab}}3\forall\text{ a, b}\in \text{Q}^+$. The inverse of 4 * 6 is:

$\frac{9}{8}$
$\frac{2}3$
$\frac{3}2$
None of these.

✓

Answer

$\frac{9}8$

Solution:

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,

$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$

e = 3, $\forall\text{ a}\in\text{Q}^+$

Thus, 3 is the identity element in Q⁺ with respect to *.

Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$

$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$

Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.

Given: $\text{a}*\text{b}=\frac{\text{ab}}3$

$4*6=\frac{4\times6}3=8$

Now,

$\text{a}^{-1}=\frac{9}{\text{a}}$

$(4*6)^{-1}=8^{-1}$

$=\frac{9}8$

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