Question

a*b = a+b+ab on Q-{-1} ,Find the identity element for the given binary operation and inverse of any element in case it exists (provided identity exists).

Answer 1

Answer:

1. Identitty element of  G is 0

2. Inverse of a is

$\frac{-a}{1+a}$

Step-by-step explanation:

Let G = Q-{-1}

1.

$\text{Let e be the identity element of G}$

Then, by definition of identity element, a*e=e*a=a $\text{for all}a\in\:G$

$\implies\:a*e=a$

$\implies\:a+e+ae=a$

$\implies\:e+ae=0$

$\implies\:e(1+a)=0$

$\text{But }a\neq\:-1$

$\implies\:e=0$

2.

$\text{Let }a\in\:G$

$\text{Let }a^{-1}\:\text{be the inverse of a}$

Then, by definition of inverse, $a*a^{-1}=a^{-1}*a=e$

$\implies\:a*a^{-1}=e$

$\implies\:a+a^{-1}+a\:a^{-1}=0$

$\implies\:a^{-1}+a\:a^{-1}=-a$

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

$\implies\:a^{-1}=\frac{-a}{1+a}\:\in\:G$ $\:(\because\:a\neq\:-1)$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

An operation * is defined as :

$\sf{ \: a*b = a+b+ab \: \: \: \: \: \: \: \forall \: a , b \in\mathbb{Q} - \{ - 1 \} \: }$

TO DETERMINE

• The identity element for the given binary operation

• The inverse of any element in case it exists (provided identity exists)

CALCULATION

CLOSURE PROPERTY UNDER OPERATION *

$\sf{For \: \: a, b \in \mathbb{Q} - \{ - 1 \}}$

$\sf{ a+b+ab \in \mathbb{Q}- \{ - 1 \} \: }$

$\implies \: \sf{ \: a*b \in \mathbb{Q} - \{ - 1 \} \: }$

So

$\sf{\mathbb{Q} - \{ - 1 \} \: \: is \: closed \: under \: the \: operation\: }$

EXISTENCE OF IDENTITY

Let e be the identity element

$\sf{let \: \: a \in \mathbb{Q} - \{ - 1 \}}$

$\sf{Then \: \: \: a*e = e*a = a}$

$\sf{Now \: \: \: a*e = a \: \: \: gives}$

$\sf {a + e + ae = a}$

$\implies \: \sf { e + ae = 0}$

$\implies \: \sf { e (1 + a) = 0}$

$\implies \: \sf { e = 0} \: \: \: \: ( \because \: \: a \in \mathbb{Q} - \{ - 1 \} \: )$

Hence 0 is the identity element

EXISTENCE OF INVERSE

$\sf{let \: \: a \in \mathbb{Q} - \{ - 1 \}}$

$\sf{let \: \: b \in \mathbb{Q} - \{ - 1 \}} \: be \: the \: inverse \: of \: a$

$\sf{Then \: \: \: a*b = b*a = e}$

$\implies \: \sf{ a*b = b*a = 0}$

Now

$\: \sf{ a*b = 0} \: \: gives$

$\implies \: \sf{a + b - ab = 0}$

$\implies \: \sf{ b - ab = - a}$

$\implies \: \sf{ b(1 - a) = - a}$

$\displaystyle \: \implies \: \sf{ b= \frac{a}{a - 1} }$

Hence

$\displaystyle \: \: \sf{ \frac{a}{a - 1} } \: \: is \: the \: inverse \: of \: \: a$

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