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Let * be a binary operation on the set Q of rational number as follows : (i) a*b = a − b (ii) a*b = a2 + b2 (iii) a*b = a + ab (iv) a*b = (a − b)2 (v) a*b = ab/4 (vi) a*b = ab2 Find which of the binary operation are commutative and which are associative? |
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Answer» (i) On Q, the operation * is defined as a*b = a − b. It can be observed that for 2,3,4 ∈ Q, we have 2*3 = 2 − 3 = −1 and 3*2 = 3 − 2 = 1 ⇒ 2*3 ≠ 3*2 Thus, the operation is not commutative. It can also be observed that (2*3)*4 = (−1)*4 = −1 − 4 = − 5 and 2*(3*4) = 2*(−1) = 2 − (−1) = 3 2*(3*4) ≠ 2* (3*4) Thus, the operation * is not associative. (ii) On Q, the operation * is defined as a*b = a2 + b2. For a,b ∈ Q, we have a*b = a2 + b2 = b2 + a2 = b*a Therefore, a*b = b*a Thus, the operation * is commutative. It can be observed that (1* 2)* 3 = (12 + 22) * 3 (1 + 4) * 4 = 5 * 4 = 52 + 42 = 41 and 1*(2* 3) = 1* (22 + 32) = 1 * (4 + 9) = 1*13 = 12 + 132 = 170 ⇒ (1*2)* 3 ≠ 1*(2*3) where 1,2, 3 ∈ Q Thus, the operation * is not associative. (iii) On Q, the operation is defined as a*b = a + ab It can be observed that 1*2 = 1 + 1 × 2 = 1 + 2 = 3, 2*1 = 2 + 2 × 1 = 2 + 2 = 4 ⇒ 1*2 ≠ 2*1 where 1, 2 ∈ Q Thus, the operation * is not commutative. It can also be observed that (1*2)*3 = (1 + 1 × 2)*3 = 3*3 = 3 + 3 × 3 = 3 + 9 = 12 and 1*(2* 3) = 1*(2 + 2 × 3) = 1*8 = 1 + 1 × 8 = 9 ⇒ (1*2)* 3 ≠ 1*(2*3) where 1,2, 3 ∈ Q Thus, the operation * is not associative. (iv) On Q, the operation * is defined by a*b = (a − b)2. For a,b ∈ Q, we have a*b = (a − b)2 and b*a = (b − a)2 = [−(a − b)]2 = (a − b)2 Therefore, a*b = b*a Thus, the operation * is commutative. It can be observed that (1* 2) * 3 = (1 – 2)2 * 3 = ( 1) * 3 = (1 – 3)2 = 4 and 1 * (2 *3) = 1 * (2 – 3)2 = 1 * (1) = (1 – 1)2 = 0 ⇒ (1*2)* 3 ≠ 1*(2* 3) where 1,2, 3 ∈ Q Thus, the operation * is not associative. (v) On Q, the operation * is defined as a*b = ab/4 For a,b ∈ Q, we have a * b = ab/4 = ba/4 =b*a Therefore, a*b = b*a Thus, the operation * is commutative. For a,b,c ∈ Q, we have a*(b*c) = ab/4 * c = ((ab/4)c)/4 = abc/16 and a*(b*c) = a* bc/4 = (a(bc/4))/4 = abc/16 Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative. (vi) On Q, the operation is defined as a*b = ab2 It can be observed that for 23 ∈ Q 2 * 3 = 2 x 32 = 18 and 3*2 = 3 x 22 = 12 Hence, 2*3 ≠ 3*2 Thus, the operation is not commutative. It can also be observed that for 1,2,3 ∈ Q (1*2)*3 = (1. 22)*3 = 4 *3 = 4.32 = 36 and 1*(2* 3) = 1*(2. 32) = 1 *18 = 1.182 = 324 ⇒ (1*2)*3 ≠ 1*(2*3) Thus, the operation * is not associative. Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative. |
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