Answer :

(i) Let (e, f) be the identity element for *

∴ for (a, b) ∈ Q × Q

We have-

(a, b) * (e, f) = (a, b) = (e, f) * (a, b)

⇒ (ae, af+b) = (a, b) = (ea, eb+f)

From 1st & 2nd part-

ae = a ⇒ e = 1 ...(1)

and, af+b = b ⇒ af = 0 ...(2)

From 2nd & 3rd part-

a = ea ⇒ e = 1 ...(3)

and, b = ef+b

⇒ b = (1)f + b [using (3)]

⇒ b = f + b

⇒ f = 0 ...(4)

From (2) & (4), we find that

a need not to be '0'.

⇒ e = 1, f = 0

∴ (e, f) = (1, 0) ∈ Q × Q

∴ (1, 0) is the identity element of A.

(ii) Let (a, b) ∈ Q × Q

Let (c, d) ∈ Q × Q

such that

(a, b) * (c, d) = (1, 0) = (c, d) * (a, b)

⇒ (ac, ad + b) = (1, 0) = (ca, cb + d)

⇒ ac = 1, ad + b = 0, ca = 1, cb + d = 0

⇒ c = (1/a), d = (-b/a), (1/a)b + d = 0 (a ≠ 0)

∴ (c, d) = (1/a, -b/a) (a ≠ 0)

∴ for a ≠ 0, (a, b)^{-1} = (1/a, -b/a)

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