For each binary operation * defined below, determine whether * is commutative or associative.
i. On Z⁺, define a * b = a - b
ii. On Q, define a * b = ab + 1
iii. On Q, define a * b = ab/2
iv. On Z⁺, define a * b = 2^ab
v. On Z⁺, define a * b = ab
vi. On R - {- 1}, define a * b = a/(b + 1)

Solution:

i. On Z⁺, define a * b = a - b

It can be observed that

1* 2 = 1- 2 = -1 and

2 *1 = 2 -1 = 1.

⇒ 1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also,

(1 * 2) * 3 = (1 - 2) * 3 = - 1 * 3 = - 1 - 3 = - 4

1 * (2 * 3) = 1 * (2 - 3) = 1 * - 1 = 1 - (- 1) = 2

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ Z

Hence, the operation * is not associative

ii. On Q, define a * b = ab + 1

ab = ba for all a, b ∈ Q

⇒ ab + 1 = ba + 1 for all a, b ∈ Q

⇒ a * b = b * a for all a, b ∈ Q

Hence, the operation * is commutative

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3)

Hence, the operation * is not associative

iii. On Q, define a * b = ab/2

ab = ba for all a, b ∈ Q

⇒ ab/2 = ab/2 for all a, b ∈ Q

⇒ a * b = b * a for all a, b ∈ Q

Hence, the operation * is commutative

(a * b) * c = (ab/2) * c = [(ab/2) * c]/2 = abc/4

And,

a * (b * c) = a * (bc/2) = [a * (bc/2)]/2 = abc/4

⇒ (a * b) * c = a * (b * c) where a, b, c ∈ Q

Hence, the operation * is associative

iv. On Z⁺, define a * b = 2^ab

ab = ba for all a, b ∈ Z

⇒ 2^ab = 2^ba for all a, b ∈ Z

⇒ a * b = b * a for all a, b ∈ Z

Hence, the operation * is commutative

(1 * 2) * 3 = 2^{1 x 2} * 3 = 4 * 3 = 2^{4 x 3} = 2¹²

1 * (2 * 3) = 1 * 2^{2 x 3} = 1 * 2⁶ = 1 * 64 = 2⁶⁴

⇒ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ Z⁺

Hence, the operation * is not associative

v. On Z⁺, define a * b = ab

1 * 2 = 1² = 1

2 * 1 = 2¹ = 2

⇒ 1* 2 ≠ 2 * 1 where 1, 2, ∈ Z⁺

Hence, the operation * is not commutative

(2 * 3)* 4 = 2³ * 4 = 8 * 4 = 8⁴ = 2¹²

2 * (3 * 4) = 2 * 3⁴ = 2 * 81 = 2⁸¹

⇒ (2 * 3) * 4 ≠ 2 * (3 * 4) where 2, 3, 4 ∈ Z⁺

Hence, the operation * is not associative

vi. On R - {- 1}, define a * b = a/(b + 1)

1 * 2 = 1/(2 + 1) = 1/3

2 * 1 = 2/(1 + 1) = 2/2 = 1

⇒ 1 * 2 ≠ 2 * 1 where 1, 2, ∈ R - {- 1}

Hence, the operation * is not commutative

(1 * 2) * 3 = 1/3 * 3 = (1/3)/(3 + 1) = 1/12

1 * (2 * 3) = 1 * 2/(3 + 1) = 1 * 2/4 = 1 * 1/2 = 1/(1/2 + 1) = 1/(3/2) = 2/3

⇒ (1* 2) * 3 ≠ 1* (2 * 3) where 1, 2, 3 ∈ R - {- 1}

Hence, the operation * is not associative

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.4 Question 2

For each binary operation * defined below, determine whether * is commutative or associative.
i. On Z⁺, define a * b = a - b ii. On Q, define a * b = ab + 1 iii. On Q, define a * b = ab/2 iv. On Z⁺, define a * b = 2^ab v. On Z⁺, define a * b = ab vi. On R - {- 1}, define a * b = a/(b + 1)

Summary:

For  each of the binary operation * defined below,i). On Z⁺, define a * b = a - b ii). On Q, define a * b = ab + 1 iii). On Q, define a * b = ab/2 iv). On Z⁺, define a * b = 2^ab v). On Z⁺, define a * b = ab vi. On R - {- 1}, define a * b = a/(b + 1), we have shown that whether the given operation is commutative or associative