Let A = Q × Q, where Q is the set of all rational numbers, and * be a

1.	Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) ∈ A. Then find (i). The identity element of * in A.(ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and (\(\frac{1}{2}\),4)
Answer» (i). Let (x, y) be the identity element in A. Now, (a, b) * (x, y) = (a, b) = (x, y) * (a, b) ∀ (a, b) ∈ A ⇒ (ax, b + ay) = (a, b) = (xa, y + bx) Equating corresponding terms, we get ⇒ ax = a, b + ay = b or a = xa, b = y + bx, ⇒ x = 1 and y = 0 Hence, (1, 0) is the identity element in A. (ii). Let (a,b) be an invertible element in A and let (c,d) be its inverse in A. Now, (a,b) * (c,d) = (1,0) = (c,d) * (a,b) ⇒ (ac,b+ad) = (1,0) = (ca,d+c) ⇒ ac = 1, b + ad = 0 or 1 = ca, 0 = d + bc [By equating coefficients] ⇒ \(c = \frac{1}{a}\) and \(d=-\frac{b}{a}\) Where, a ≠ 0 Therefore, all (a,b) ∈ A is an invertible element of A if a ≠ 0, and inverse of (a,b) is (\( \frac{1}{a}\), \(-\frac{b}{a}\)) For inverse of (5,3) Inverse of (5,3) = (\(\frac{1}{5},\,-\frac{3}{5}\)) (∵ Inverse of (a,b) =\( \frac{1}{a}\),\(-\frac{b}{a}\)) For inverse of (\( \frac{1}{2}\),4) Inverse of (\( \frac{1}{2}\),4) = (2, -8)

Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) ∈ A. Then find (i). The identity element of * in A.(ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and (\(\frac{1}{2}\),4)

Answer»

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

Now,

(a, b) * (x, y) = (a, b)

= (x, y) * (a, b) ∀ (a, b) ∈ A

⇒ (ax, b + ay) = (a, b) = (xa, y + bx)

Equating corresponding terms, we get

⇒ ax = a, b + ay = b or a = xa, b = y + bx,

⇒ x = 1 and y = 0

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

(ii). Let (a,b) be an invertible element in A and let (c,d) be its inverse in A.

Now,

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

⇒ (ac,b+ad) = (1,0) = (ca,d+c)

⇒ ac = 1, b + ad = 0

or 1 = ca, 0 = d + bc [By equating coefficients]

⇒ \(c = \frac{1}{a}\) and \(d=-\frac{b}{a}\)

Where, a ≠ 0

Therefore, all (a,b) ∈ A is an invertible element of A if a ≠ 0, and inverse of (a,b) is (\( \frac{1}{a}\), \(-\frac{b}{a}\))

For inverse of (5,3)

Inverse of (5,3) = (\(\frac{1}{5},\,-\frac{3}{5}\)) (∵ Inverse of (a,b) =\( \frac{1}{a}\),\(-\frac{b}{a}\))