51. Let * be a binary operation on the set Q of rational numbers 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 operations are commutative and which are associative.

(i) On Q the operation* is defined as a*b=a-b. It can be observed that: 12.13=12−13=3−26=16"and"13.12=13−12=2−36=−16∴12.13≠13.12where,12,13∈Q Thus, the operation* is not commutative. It can also be observed that: (12.13).14=(12−13).14=16.14=16−14=2−312=−11212.(13.14)=12.(13−14)=12.112=12−112=6−112=512∴(12.13).14≠12.(13.14)where,12,13,14∈Q Thus, the operation* is not associative. (ii) On Q the operation* is defined as a*b=a2+b. For a,b∈Q , we have: a.b=a2+b2=b2+a2=b.a∴a*b=b*a Thus, the operation* is commutative. It can be observed that: (1*2)*3=(12+22)*3=(1+4)*3=5*3=52+32=25+9=341*(2*3)=1*(22+32)=1*(4+9)=1*13=12+132=1+169=170∴(1*2)*3≠1*(2*3),where,1,2,3∈Q Thus, the operation* is not commutative. (iii) On Q the operation* is defined as a*b=a+ab. It can be observed that: 1.2=1+1*2=1+2=32.1=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=121.(2.3)=1.(2+2*3)=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 as a*b=(a-b)2 For a,b∈Q , we have: a*b=(a−b)2b*a=(b−a)2=[−(a−b)]2=(a−b)2∴a*b=b*a Thus, the operation* is commutative. It can be observed that: (1.2).3=(1.2)2.3=(−1)2.3=1.3=(1.3)2=(−2)2=41.(2.3)=1.(2.3)2=1.(−1)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=ab4=ba4=b.a∴a*b=b*a Thus, the operation* is commutative. For a,b,c∈Q , we have: (a.b).c=ab4.c=ab4.c4=abc16a.(b.c)=a.bc4=a.bc44=abc16=∴(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: 12.13=12.(13)2=12.19=11813.12=13.(12)2=13.14=112∴12.13≠13.12,where,12,13∈Q Thus, the operation* is not commutative. It can be observed that: (12.13).14=[12.(13)2].14=118.14=118.(14)2=118*1612.(13.14)=12.[13.(14)2]=12.148=12.(148)2=12*(48)2∴(12.13).14≠12.(13.14),where,12,13,1.4∈Q Thus, the operation* is not associative. Hence, the operations defined (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

**51. Let * be a binary operation on the set Q of rational numbers 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 operations are commutative and which are associative.

11 Views|Posted 9 months ago

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Vishal Baghel | Contributor-Level 10

9 months ago

(i) On Q the operation* is defined as a*b=a-b.

It can be observed that:

$\begin{array}{l} \frac{1}{2} . \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} " a n d " \frac{1}{3} . \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = \frac{- 1}{6} \\ ∴ \frac{1}{2} . \frac{1}{3} \neq \frac{1}{3} . \frac{1}{2} w h e r e, \frac{1}{2}, \frac{1}{3} \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} = (\frac{1}{2} - \frac{1}{3}) . \frac{1}{4} = \frac{1}{6} . \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = \frac{- 1}{12} \\ \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) = \frac{1}{2} . (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} . \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12} \\ ∴ (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} \neq \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) w h e r e, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in Q \end{array}$

Thus, the operation* is not associative.

(ii) On Q the operation* is defined as a*b=a²+b.

For $a, b \in Q$ , we have:

$\begin{array}{l} a . b = a^{2} + b^{2} = b^{2} + a^{2} = b . a \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative.

It can be observed tha

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