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Question
Let ' * ' be a binary operation on set Q − {1} defined bya * b = a + b − ab for all a, b ∈ Q − {1}.
Then, which of the following statement(s) is/are true?
Then, which of the following statement(s) is/are true?
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Solution
The correct option is D * is associative on Q− {1}
Let e be the identity element.⇒a*e=e*a=a ∀a∈Q−(1}⇒a+e−ae=a⇒e−ae=0⇒e(1−a)=0⇒e=0 (Since a≠1]∴ The identity element with respect to '*' on Q−(1} is 0.If 'b' is the inverse element of 'a' wrt * on Q−(1}, thena*b=b*a=e, a∈Q−(1}⇒a+b−ab=0⇒ a+b(1−a)=0⇒b=aa−1Here, b is defined for all a∈Q−(1}.⇒Every element of Q−(1} is invertible and the inverse is given by aa−1.
Let a, b, c∈Q−(1}Consider (a*b)*c =(a+b−ab)*c =a+b−ab+c−(a+b−ab)c =a+b+c−ab−bc−ca−abcNow, a*(b*c) =a*(b+c−bc) =a+b+c−bc−a(b+c−bc) =a+b+c−ab−bc−ca−abcThus, (a*b)*c= a*(b*c)for all a, b, c∈Q−(1}∴ * is associative on Q−(1}.
Let e be the identity element.⇒a*e=e*a=a ∀a∈Q−(1}⇒a+e−ae=a⇒e−ae=0⇒e(1−a)=0⇒e=0 (Since a≠1]∴ The identity element with respect to '*' on Q−(1} is 0.If 'b' is the inverse element of 'a' wrt * on Q−(1}, thena*b=b*a=e, a∈Q−(1}⇒a+b−ab=0⇒ a+b(1−a)=0⇒b=aa−1Here, b is defined for all a∈Q−(1}.⇒Every element of Q−(1} is invertible and the inverse is given by aa−1.
Let a, b, c∈Q−(1}Consider (a*b)*c =(a+b−ab)*c =a+b−ab+c−(a+b−ab)c =a+b+c−ab−bc−ca−abcNow, a*(b*c) =a*(b+c−bc) =a+b+c−bc−a(b+c−bc) =a+b+c−ab−bc−ca−abcThus, (a*b)*c= a*(b*c)for all a, b, c∈Q−(1}∴ * is associative on Q−(1}.
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