Question

Let $A=Q\times Q$, where $Q$ is the set of all rational numbers, and $\ast$ is a binary operation on $A$ defined by $(a,b)\ast (c,d)=(ac,b+ad)$ for $(a,b),(c,d)\in A$.

Then find (i) The identity element of $\ast$ in $A.$

(ii) Invertible elements of $A,$ and hence write the inverse of elements $(5,3)$ and $(\frac{1}{2},4)$.

Answer 1

(i) Let $(x,y)$ be the identity element in $A.$

$\therefore (a,b)\ast (x,y)=(a,b)=(x,y)\ast (a,b)$

$(ax,b+ay)=(a,b)$

$\Rightarrow ax=a,b+ay=b$

$\Rightarrow x=1,y=0$

$\Rightarrow (a,0)\in A$ (set of rational numbers)

$\therefore (1,0)$ is an identity element.

(ii) Here, $(1,0)$ is the identity element

$\therefore (a,b)\ast (c,d)=(1,0)$

$\Rightarrow (ac,b+ad)=(1,0)$

$\Rightarrow ac=1,b+ad=0$

$\Rightarrow c=\frac{1}{a},d=\frac{-b}{a}$

$\therefore A^{-1}=(\frac{1}{a},\frac{-b}{a})$

Also, inverse of $(5,3)=(\frac{1}{5},\frac{-3}{5})$ and inverse of $(\frac{1}{2},4)=(2,-8)$