Question

Let $A=Q\times Q$ and let ${}^{\ast }$ be a binary operation on A defined by $(a,b{)}^{\ast }(c,d)=(ac,b+ad\left)\forall \right(a,b),(c,d)ϵA$ then, with respect to ${}^{\ast }$ on A. Find the identify and inverse element of A.

Answer 1

Here $\ast$ is a binary operation on $A\times A$

here $(a,b)\ast (c,d)=(ac,b+ad)\forall (a,b),(c,d)\in A$

let's assume $(e,e^{\prime })$ is the identity element on $A$ by ${}^{\prime }{\ast }^{\prime }$ operation.

then from the formula and rule of identity element

$(a,b)\ast (e,e^{\prime })=(ae,b+a{e}^{\prime })=(a,b)$

$ae=a$ $b+a{e}^{\prime }=b$

$e=1$ $a{e}^{\prime }=0$ [by law of equity]

$e^{\prime }=0$

so identity element is $(1,0)$

Let's assume inverse of $(a,b)$ is $(a^{\prime },b^{\prime })$

where $(a,b),(a^{\prime },b^{\prime })\in A$

So, $(a,b)\ast (a^{\prime },b^{\prime })=(1,0)$

$(a{a}^{\prime },b+a{b}^{\prime })=(1,0)$

$a{a}^{\prime }=1$ $b+a{b}^{\prime }=0$

$a^{\prime }=\frac{1}{a}$ $b^{\prime }=\frac{-b}{a}$

So inverse of $(a,b)$ is $(\frac{1}{a},\frac{-b}{a})$ on $A$ by the operation ${}^{\prime }{\ast }^{\prime }$