# Let Q0 be the set of all nonzero rational numbers

Question:

Let $Q_{0}$ be the set of all nonzero rational numbers. Let $*$ be a binary operation on $Q_{0}$, defined by

$a^{*} b=\frac{a b}{4}$ for $a l l a, b \in Q_{0}$

(i) Show that $*$ is commutative and associative.

(ii) Find the identity element in $\mathrm{Q}_{0}$.

(iii) Find the inverse of an element $\mathrm{a}$ in $\mathrm{Q}_{0}$.

Solution:

(i) For commutative binary operation, $a^{*} b=b^{*} a$.

$a^{*} b=\frac{a b}{4}$ and $b^{*} a=\frac{b a}{4}$

as multiplication is commutative ab = ba so a*b = b*a. Hence * is commutative binary operation.

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

$a^{*}\left(b^{*} c\right)=a^{*} \frac{b c}{4} \Rightarrow \frac{a \cdot \frac{b c}{4}}{4}=\frac{a b c}{16}$

$(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\frac{\mathrm{ab}}{4} * \mathrm{c} \Rightarrow \frac{\frac{\mathrm{ab}}{4} \cdot \mathrm{c}}{4}=\frac{\mathrm{abc}}{16}$

Since a*(b*c) = (a*b) *c, hence * is an associative binary operation.

(ii) For a binary operation *, e identity element exists if a*e = e*a = a. As a*b = a+ b- ab

$a^{*} e=\frac{a e}{4}(1)$

$e^{*} a=\frac{e a}{4}(2)$

using a*e = a

$\frac{\mathrm{ae}}{4}=\mathrm{a} \Rightarrow \frac{\mathrm{ae}}{4}-\mathrm{a}=0 \Rightarrow \frac{\mathrm{a}}{4}(\mathrm{e}-4)=0$

Either a = 0 or e = 4 as given a≠0, so e = 4.

Identity element e = 4.

(iii) 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 = 4

$\frac{\mathrm{ab}}{4}=4 \Rightarrow \mathrm{b}=\frac{16}{\mathrm{a}}$

$a^{-1}=\frac{16}{a}$