Let $\mathrm{A}=\mathrm{\mathbb{Q}}\times \mathrm{\mathbb{Q}}$ and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) 𝜖 A. Determine, whether * is commutative and associative. Then, with respect to * on A
(i) Find the identity element in A.
(ii) Find the invertible elements of A.

A =ℚ×ℚ....... [Given]
For any (a, b), (c, d) ∈A, ∗ is defined by
(a, b) ∗ (c, d) = (ac, b + ad) ..... [Given]
To check ∗ is commutative i.e. to check (a, b) ∗ (c, d) = (c, d) ∗ (a, b) for any (a, b), (c, d) ∈ A
Now, (a, b) ∗ (c, d) = (ac, b + ad)
(c, d) ∗ (a, b) = (ca, d + cb) = (ac, d + bc) = (ac, b + ad)
∴(a, b) ∗ (c, d) = (c, d) ∗ (a, b)
Thus, ∗ is not commutative ....... (1)
To check associativity
Let (a, b), (c, d), (e, f) ∈ A
Now, (a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (ce, d + cf)
= (ace, b + a(d + cf))
= (ace, b + ad + acf) ...... (2)
∴ (a, b) ∗ ((c, d) ∗ (e, f)) = (ace, b + ad + acf)
((a, b) ∗ (c, d)) ∗ (e, f) = (ac, b + ad) ∗ (e, f)
= (ace, b + ad + acf)
= (a, b) ∗ ((c, d) ∗ (e, f)) ..... From (2)
∴ (a, b) ∗ ((c, d) ∗ (e, f)) = ((a, b) ∗ (c, d)) ∗ (e, f)
Thus, ∗ is associative ....... (3)
(i) To find identity element
Let e = (a′, b′) be identity element of A
⟹ (a, b) ∗ (a′, b′) = (a, b) = (a′, b′) ∗ (a, b)
As (a, b) ∗ (a′, b′) = (a, b)
⟹ (aa′, b + ab′) = (a, b) ........ Using definition of ∗
⟹ aa′ = a and b + ab′=b
⟹ a′ = 1 and b′ = 0
We can verify it as follows
(a′, b′) ∗ (a, b) = (a′a, b′ + a′b) = (1⋅a, 0+1⋅b) = (a, b)
Similarly, (a, b) ∗ (a′, b′) = (a, b)
Hence, e= (1, 0) is the identity element in A.
(ii) To find inverse element
Let f = (c′, d′) be inverse element of (a, b) ∈ A
⟹ (a, b) ∗ (c′, d′) = (1, 0) = (c′, d′) ∗ (a, b) ...... Using definition of inverse element
Now, (a, b) ∗ (c′, d′) = (1,0)
⟹ (ac′, b + ad′) = (1,0) ........ [Using definition of ∗]
⟹ ac′ = 1 and b + ad′ =0
$\Rightarrow {\mathrm{c}}^{\mathrm{\prime }}=\frac{1}{\mathrm{a}}\text{ and }{\mathrm{d}}^{\mathrm{\prime }}=-\frac{\mathrm{b}}{\mathrm{a}}$
We can verify it as follows
$\left({\mathrm{c}}^{\mathrm{\prime }},{\mathrm{d}}^{\mathrm{\prime }}\right)\ast (\mathrm{a},\mathrm{b})=\left({\mathrm{c}}^{\mathrm{\prime }}\mathrm{a},{\mathrm{d}}^{\mathrm{\prime }}+{\mathrm{c}}^{\mathrm{\prime }}\mathrm{b}\right)=\left(\frac{1}{\mathrm{a}}\times \mathrm{a},-\frac{\mathrm{b}}{\mathrm{a}}+\frac{1}{\mathrm{a}}\times \mathrm{b}\right)=(1,0)$
Hence, $\mathrm{f}=\left(\frac{1}{\mathrm{a}},-\frac{\mathrm{b}}{\mathrm{a}}\right)$ is the inverse element of A