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 $(a,b),(c,d)\in A$. Determine, whether $\ast$ is commutative and associative. Then, with respect to $\ast$ on $A$.
(i) Find the identify element in A.
(ii) Find the invertible element of A.

Answer 1

$A=Q\times Q$ ....... [Given]

For any $(a,b),(c,d)\in A$, $\ast$ is defined by

$(a,b)\ast (c,d)=(ac,b+ad)$ ..... [Given]

To check $\ast$ is commutative i.e. to check $(a,b)\ast (c,d)=(c,d)\ast (a,b)$ for any $(a,b),(c,d)\in A$

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

$(c,d)\ast (a,b)=(ca,d+cb)=(ac,d+bc)\ne (ac,b+ad)$

$\therefore (a,b)\ast (c,d)\ne (c,d)\ast (a,b)$

Thus, $\ast$ is not commutative ....... $\left(1\right)$

To check associativity

Let $(a,b),(c,d),(e,f)\in A$

Now, $(a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)=(a,b)\ast (ce,d+cf)$

$=(ace,b+a(d+cf\left)\right)$

$=(ace,b+ad+acf)$ ...... $\left(2\right)$

$\therefore (a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)=(ace,b+ad+acf)$

$\left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)=(ac,b+ad)\ast (e,f)$

$=(ace,b+ad+acf)$

$=(a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)$ ..... From $\left(2\right)$

$\therefore (a,b)\ast \left(\right(c,d)\ast (e,f\left)\right)=\left(\right(a,b)\ast (c,d\left)\right)\ast (e,f)$

Thus, $\ast$ is associative ....... $\left(3\right)$

(i) To find identity element

Let $e=(a^{\prime },b^{\prime })$ be identity element of $A$

$⟹(a,b)\ast (a^{\prime },b^{\prime })=(a,b)=(a^{\prime },b^{\prime })\ast (a,b)$

As $(a,b)\ast (a^{\prime },b^{\prime })=(a,b)$

$⟹(a{a}^{\prime },b+a{b}^{\prime })=(a,b)$ ........ Using definition of $\ast$

$⟹a{a}^{\prime }=a$ and $b+a{b}^{\prime }=b$

$⟹a^{\prime }=1$ and $b^{\prime }=0$

We can verify it as follows

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

Similarly, $(a,b)\ast (a^{\prime },b^{\prime })=(a,b)$

Hence, $e=(1,0)$ is the identity element in $A$.

(ii) To find inverse element

Let $f=(c^{\prime },d^{\prime })$ be inverse element of $(a,b)\in A$

$⟹(a,b)\ast (c^{\prime },d^{\prime })=(1,0)=(c^{\prime },d^{\prime })\ast (a,b)$ ...... Using definition of inverse element

Now, $(a,b)\ast (c^{\prime },d^{\prime })=(1,0)$

$⟹(a{c}^{\prime },b+a{d}^{\prime })=(1,0)$ ........ [Using definition of $\ast$]

$⟹a{c}^{\prime }=1$ and $b+a{d}^{\prime }=0$

$⟹c^{\prime }=\frac{1}{a}$ and $d^{\prime }=-\frac{b}{a}$

We can verify it as follows

$(c^{\prime },d^{\prime })\ast (a,b)=(c^{\prime }a,d^{\prime }+c^{\prime }b)=(\frac{1}{a}\times a,-\frac{b}{a}+\frac{1}{a}\times b)=(1,0)$

Hence, $f=(\frac{1}{a},-\frac{b}{a})$ is the inverse element of $A$