# Let Q be the set of all rational numbers. Define an operation on

Question:

Let $\mathrm{Q}$ be the set of all rational numbers. Define an operation on $\mathrm{Q}-\{-1\}$ by $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}$. Show that

(i) $*$ is a binary operation on $\mathrm{Q}-\{-1\}$,

(ii) * is Commutative,

(iii) * is associative,

(iv) zero is the identity element in $\mathrm{Q}-\{-1\}$ for $*$,

Solution:

(i) $*$ is an operation as $\mathrm{a}^{*} \mathrm{~b}=\mathrm{a}+\mathrm{b}+\mathrm{ab}$ where $\mathrm{a}, \mathrm{b} \in \mathrm{Q}-\{-1\}$. Let $\mathrm{a}=1$ and $\mathrm{b}=\frac{-3}{2}$ two rational numbers.

$\mathrm{a} * \mathrm{~b}=1+\frac{-3}{2}+1 \cdot \frac{-3}{2} \Rightarrow \frac{2-3}{2}-\frac{3}{2}=\frac{-1-3}{2} \Rightarrow \frac{-4}{2}=-2 \in \mathrm{Q}-\{-1\}$

So, $*$ is a binary operation from $Q-\{-1\} \times Q-\{-1\} \rightarrow Q-\{-1\}$

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

$\mathrm{b}^{*} \mathrm{a}=\frac{-3}{2}+1+\frac{-3}{2} \cdot 1 \Rightarrow \frac{-3+2}{2}-\frac{3}{2}=\frac{-1-3}{2} \Rightarrow \frac{-4}{2}=-2 \in \mathrm{Q}-\{-1\}$

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

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

a+(b*c) = a*(b+ c+ bc) = a+ (b+ c+ bc) +a(b+ c+ bc)

= a+ b+ c+ bc+ ab+ ac+ abc

(a*b) *c = (a+ b+ ab) *c = a+ b+ ab+ c+ (a+ b+ ab) c

= a+ b+ c+ ab+ ac+ bc+ abc

Now as a*(b*c) = (a*b) *c, hence an associative binary operation.

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

a*e = a+ e+ ae (1)

e*a = e+ a +e a (2)

using a*e = a

a+ e+ ae = a

e+ ae = 0

e(1+a) = 0

either e = 0 or a = -1 as operation is on Q excluding -1 so a≠-1, hence e = 0

So identity element e = 0.

(v) for a binary operation $*$ if 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 $a^{-1}$.

$a^{*} b=0$

$a+b+a b=0$

$\mathrm{b}(1+\mathrm{a})=-\mathrm{a}$

$b=\frac{-a}{(1+a)}$

$a^{-1}=\frac{-a}{(a+1)}$