Question

Let $A=Q\times Q$ and let $\ast$ be a binary operation on $A$ defined by $(a,b)\ast (c,d)=(ac,b+ad)$ for $(a,b)(c,d)ϵA$. Determine, whether $\ast$ is commutative and associative. Then, with respect to $\ast$ on $A$

$\left(i\right)$ Find the identity element in $A$

$\left(ii\right)$ Find the invertible elements of $A$.

Answer 1

Given binary operation is :
$(a,b)\ast (c,d)=(ac,b+ad)\dots \left(i\right)$
$(c,d)\ast (a,b)=(ca,d+cb)\dots \left(ii\right)$
Since $\left(i\right)\ne \left(ii\right)$
Thus, $\ast$ is not commutative.
Now, $(a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)$
$=(a,b)\ast (ce,d+cf)$
$=(ace,b+ad+acf)\dots \left(iii\right)$
And, $\left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)$
$=(ac,b+ad)\ast (e,f)$
$=(ace,b+ad+acf)\dots \left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right)$, we have
$(a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)=\left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)$.
Thus, $\ast$ is associative.
$\left(i\right)$ Let $(x,y)$ be the identity element in $A$.
Now, $(a,b)\ast (x,y)=(a,b)=(x,y)\ast (a,b)\forall (a,b)\in A$
$\Rightarrow (ax,b+ay)=(a,b)=(xa,y+bx)$
Equating corresponding terms, we have
$\Rightarrow ax=a,b+ay=bora=xa,b=y+bx$
$\Rightarrow x=1$ and $y=0$
Hence, $(1,0)$ is the identity element in $A$.
$\left(ii\right)$ Let $(a,b)$ be an invertible element in $A$ and let $(c,d)$ be its inverse in $A$.
Now, $(a,b)\ast (c,d)=(1,0)=(c,d)\ast (a,b)$
$\Rightarrow (ac,b+ad)=(1,0)=(ca,d+bc)$
[by equating corresponding elemensts]
$\Rightarrow ac=1,b+ad=0$ or $1=ca,0=d+bc$
$\Rightarrow c=\frac{1}{a}$ and $d=-\frac{b}{a}$, where, $a\ne 0$
Therefore, $(a,b)\in A$ is an invertible element of $A$ if $a\ne 0$, and inverse of $(a,b)$ is$(\frac{1}{a},-\frac{b}{a})$.