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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.
Solution 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)`
Solution 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)`
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