On the set Q+ of all positive rational numbers, define an operation

Solution:

(i) $*$ is an operation as $\mathrm{a}^{*} \mathrm{~b}=\frac{\mathrm{ab}}{2}$ where $\mathrm{a}, \mathrm{b} \in \mathrm{Q}^{+} .$Let $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=2$ two integers.

$\mathrm{a}^{*} \mathrm{~b}=\frac{1}{2} * 2 \Rightarrow 1 \in \mathrm{Q}^{+}$

So, $*$ is a binary operation from $\mathrm{Q}^{+} \times \mathrm{Q}^{+} \rightarrow \mathrm{Q}^{+}$.

(ii) For commutative binary operation, $a * b=b * a$.

$\mathrm{b}^{*} \mathrm{a}=2 \cdot \frac{1}{2} \Rightarrow 1 \in \mathrm{Q}^{+}$

Since $a * b=b * a$, hence $*$ is a commutative binary operation.

(iii) For associative binary operation, $a *(b * c)=(a * b) * c$.

$a^{*}\left(b^{*} c\right)=a^{*} \frac{b c}{2} \Rightarrow \frac{a \cdot \frac{b c}{2}}{2}=\frac{a b c}{4}$

$(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\frac{\mathrm{ab}}{2} * \mathrm{c} \Rightarrow \frac{\frac{\mathrm{ab}}{2} \cdot \mathrm{c}}{2}=\frac{\mathrm{abc}}{4}$

As $a^{*}\left(b^{*} c\right)=(a * b) * c$, hence $*$ is an associative binary operation.

For a binary operation $*$, e identity element exists if $\mathrm{a} * \mathrm{e}=\mathrm{e} * \mathrm{a}=\mathrm{a}$.

$\mathrm{a}^{*} \mathrm{e}=\frac{\mathrm{ae}}{2}(1)$

$\mathrm{e}^{*} \mathrm{a}=\frac{\mathrm{ea}}{2}(2)$

using $a^{*} e=a$

$\frac{\mathrm{ae}}{2}=\mathrm{a} \Rightarrow \frac{\mathrm{ae}}{2}-\mathrm{a}=0 \Rightarrow \frac{\mathrm{a}}{2}(\mathrm{e}-2)=0$

Either $a=0$ or $e=2$ as given $a \neq 0$, so $e=2$

For a binary operation $*$ if $\mathrm{e}$ is identity element then it is invertible with respect to $*$ if for an element b, $a * b=e=b * a$ where $b$ is called inverse of $*$ and denoted by $a^{-1}$.

$a * b=2$

$\frac{\mathrm{ab}}{2}=2 \Rightarrow \mathrm{b}=\frac{4}{\mathrm{a}}$

$a^{-1}=\frac{4}{a}$