Question

Discuss the commutativity and associativity of binary operation $\ast$. defined on $A=Q-\left\{1\right\}$ by the rule $a\ast b=a-b+ab$ fo all $a,b\in A$. Also find the identity element of $\ast$ in $A$ and hence find the invertible elements of $A$.

Answer 1

Given # is a binary operation on $Q-\left\{1\right\}$

defined by $a\ast b=a-b+ab$

Commutativity:

for any $a,b\in A$

We have $a\ast b=a-b+ab$ and $b\ast a=b-a+ba$

Since, $a-b+ab\ne b-a+ab$

$\therefore a\ast b\ne b\ast a$

So, $\ast$ is not commutative on $A$

Associativity:

Let $a,b,c\in A(a\ast b)\ast c=(a-b+ab)\ast c$

$\Rightarrow (a\ast b)\ast c=(a-b+ab)-c+(a-b+ab)c$

$\Rightarrow (a\ast b)\ast c=a-b+ab-c+ac-bc+abc$

$a\ast (b\ast c)=a\ast (b-c+bc)$

$\Rightarrow a\ast (b\ast c)=a-(b-c+bc)+a(b-c+bc)$

$\Rightarrow a\ast (b\ast c)=a-b+c-bc+ab-ac+abc$

$\Rightarrow (a\ast b)\ast c\ne a\ast (b\ast c)$

so $\ast$ is not associative on $A$

Identity Element:

Let $e$ be the identity elements in $A$

the $a\ast e=a=e\ast aa\in Q-\left\{1\right\}$

$a-e+ae=a$

$(a-1)e=0$

$e=0$ $($ as $a\ne 1)$[so ${}^{\prime }{0}^{\prime }$ si identity element in $A]$

Inverse of an Element:

Let $a$ be an arbitary element of $A$ as $b$ be the inverse of $a$.

$\because$ $a\ast b=e=b\ast a$

$\Rightarrow a\ast b=e$

$\Rightarrow a-b+ab=0[\because e=0]$

$a=b(1-a)$

$b=a/1-a$

since $b\in Q-1$

So, every elements of $A$ is invertible.