Question

Let $A=Q\times Q$, where Q is the set of all rational numbers, and * be a binary operation on $A$ defined by $(a,b)\ast (c,d)=(ac,b+ad)$ for $(a,b),(c,d)$ $ϵ$ $A.$ Then find
(i) The identify element of $\ast$ in $A.$
(ii) Invertible elements of $A$, and hence write the inverse of elements $(5,3)$ and $(\frac{1}{4},4)$.

Answer 1

$(a,b)\ast (c,d)=(ac,ad+b)$

$\left(1\right)$ identity element

let (e,f) be the identity element .

Then $(a,b)\ast (e,f)=(ae,af+b)=(a,b)$

$[a\ast e=a,$ where e is the identity element $]$

$(ae,af+b)=(a,b)$

$⟹ae=a$ and $af+b=b$

$⟹e=1$ and $f=0$

$(1,0)$ is the identity element .

$\left(2\right)$ invertible element

Let $(c,d)$ be the inverse of $(a,b)$

then

$(a,b)\ast (c,d)=(1,0)[a\ast b$, e is the identity $]$

$⟹(ac,ad+b)=(1,0)$

$ac=1;ad+b=0$

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

So, $(\frac{1}{a},\frac{-b}{a})$ is the inverse.

The inverse of $(5,3)$ is $(\frac{1}{5},\frac{-3}{5})$

The inverse of $(\frac{1}{4},4)$ is $(4,-16)$