Question

Let A = Q x Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) ∈ A. Determine, whether * is commutative and associative. Then, with respect to * on A

1) Find the identity element in A

2) Find the invertible elements of A.

Answer 1

(a, b) * (c, d) = (ac, b + ad)

(c, d) * (a, b) = (ca, d + cf)

Not commutative

(a, b) * [(c, d) * (e, f)]

= (a, b) * [ce, d + cf]

= [ace, b + ad + acf]

Now,
[(a, b) * (c, d)] * (e, f)

= [ac, b + ad] * (e, f)

= [ace, b + ad + acf]

∴ Associative

∵ (a, b) * [(c, d) * (e, f)]= [(a, b)] * [(c, d) * (e, f)]

1) (a, b) * e = (a, b)

⇒ a = ac

⇒ c = 1 and b = b + ad ⇒ ad = 0

⇒ d = 0

∴ (a, b) * (1, 0) = (a, b + a × 0) = (a, b)

⇒ (1,0) is identify

2) (a, b) * (c, d) = e = (1, 0)

⇒ ac = 1 and b + ad = 0

⇒ `d = (-b)/a`

∴ Inverse of element

∴ Inverse of element of a, b is `(1/a, (-b)/a)`

Answer 2

Let A=Q×Q and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d)∈A.

Commutativity:

Let X = (a, b) and Y = (c, d) ∈ A,∀ a, c ∈ Q and b, d ∈ Q. Then,

X * Y =(ac, b + ad)

Y * X=(ca, d + cb)

Therefore, X * Y ≠ Y * X ∀ X,Y∈A

Thus, * is not commutative on A.

Associativity:

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

X*(Y*Z)=(a, b)*(ce, d+cf)

=(ace, b + ad + acf)

(X * Y)* Z=(ac, b + ad)*(e,f)

= (ace, b + ad + acf)

∴ X*(Y * Z) = (X * Y)*Z, ∀ X, Y, Z ∈ A

Thus,* is associative on A.

1)  Let E = (x, y) be the identity element in A with respect to *,∀ x ∈ Q and y ∈ Q such that

X * E = X = E * X,∀ X ∈ A

⇒ X * E = X and E * X = X

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

Considering (ax, b+ay)=(a, b)

⇒ ax = a     

⇒ x = 1   and  b + ay = b

⇒ y = 0                 

Considering (xa, y + xb) = (a, b)

⇒ xa = a

⇒ x = 1 and y + xb = b

⇒y = 0               [∵ x=1]

∴ (1, 0) is the identity element in A with respect to *.

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

X * F = E and F * X = E

⇒(am, b + an) = (1, 0) and (ma, n + mb) = (1, 0)

Considering (am, b + an)=(1, 0)

⇒ am = 1

⇒m = 1/a and b + an = 0

`=> n = (-b)/a`

Considering (ma, n+mb)=(1, 0)

⇒ ma = 1

`=> m = 1/a and n + mb = 0`

`=> n = (-b)/a`   [∵ m = 1/a]

∴ The inverse of (a, b) ∈ A with respect to * is `(1/a, (-b)/a)`