Question

Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.

(i) Show that ‘O’ is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A.

Answer 1

(i) Let X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

S0,

X O Y = Y O X, ∀ X, Y ∈ A

Thus, O commutative on A.

Let us check the associativity of O

Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R₀ and b, d, f ∈ R

X O (Y O Z) = (a, b) O (ce, de + f)

= (ace, bce + de + f)

(X O Y) O Z = (ac, bc + d) O (e, f)

= (ace, (bc + d) e + f)

= (ace, bce + de + f)

So, X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A

(ii) Let E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R

Such that,

XOE = X = EOX, ∀ X ∈ A

XOE = X and EOX = X

(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b)

Consider (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 [since x = 1]

Consider (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 [since x = 1]

So, (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Consider (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a [since m = 1/a]

Consider (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

So, the inverse of (a, b) ∈ A with respect to O is (1/a, -1/a)