# RD Sharma Solutions For Class 12 Maths Exercise 3.2 Chapter 3 Binary Operations

RD Sharma Solutions Class 12 Maths Exercise 3.2 Chapter 3 Binary OperationsÂ  is provided here for students to study and score good marks in their board exams. This exercise deals with types of binary operations. Exercise 3.2 of Chapter 3 consists of problems based on commutativity and associativity of binary operations. Students can refer to the solutions as a major study material to improve their speed in solving problems accurately.

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1. Let ‘*’ be a binary operation onÂ NÂ defined by aÂ *Â bÂ = l.c.m. (a,Â b) for allÂ a,Â bÂ âˆˆÂ N
(i) Find 2 * 4, 3 * 5, 1 * 6.

(ii) Check the commutativity and associativity of ‘*’ on N.

Solution:

(i) Given aÂ *Â bÂ = 1.c.m. (a,Â b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5)

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) We have to prove commutativity of *

Let a, b âˆˆ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

Therefore

a * b = b * a âˆ€ a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ N

a * (b * c ) = a * l.c.m. (b, c)

= l.c.m. (a, (b, c))

= l.c.m (a, b, c)

(a * b) * c = l.c.m. (a, b) * c

= l.c.m. ((a, b), c)

= l.c.m. (a, b, c)

Therefore

(a * (b * c) = (a * b) * c, âˆ€ a, b , c âˆˆ N

Thus, * is associative on N.

2. Determine which of the following binary operation is associative and which is commutative:

(i) * onÂ NÂ defined byÂ aÂ *Â bÂ = 1 for allÂ a,Â bÂ âˆˆÂ N

(ii) * on Q defined by a * b = (a + b)/2 for all a, b âˆˆ Q

Solution:

(i) We have to prove commutativity of *

Let a, b âˆˆ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ N

Then a * (b * c) = a * (1)

= 1

(a * b) *c = (1) * c

= 1

Therefore a * (b * c) = (a * b) *c for all a, b, c âˆˆ N

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

Let a, b âˆˆ N

a * b = (a + b)/2

= (b + a)/2

= b * a

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ N

a * (b * c) = a * (b + c)/2

= [a + (b + c)]/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= [(a + b)/2 + c] /2

= (a + b + 2c)/4

Thus, a * (b * c) â‰  (a * b) * c

If a = 1, b= 2, c = 3

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

= 1 * (5/2)

= [1 + (5/2)]/2

= 7/4

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

= 3/2 * 3

= [(3/2) + 3]/2

= 4/9

Therefore, there exist a = 1, b = 2, c = 3 âˆˆ N such that a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N.

3. LetÂ AÂ be any set containing more than one element. Let ‘*’ be a binary operation onÂ A defined byÂ aÂ *Â bÂ =Â bÂ for allÂ a,Â bÂ âˆˆÂ AÂ Is ‘*’ commutative or associative onÂ A?

Solution:

Let a, b âˆˆ A

Then, a * b = b

b * a = a

Therefore a * b â‰  b * a

Thus, * is not commutative on A

Now we have to check associativity:

Let a, b, c âˆˆ A

a * (b * c) = a * c

= c

Therefore

a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ A

Thus, * is associative on A

4. Check the commutativity and associativity of each of the following binary operations:

(i) ‘*’ onÂ ZÂ defined byÂ aÂ *Â bÂ =Â aÂ +Â bÂ +Â a bÂ for allÂ a,Â bÂ âˆˆÂ ZÂ

(ii) â€˜*â€™ on N defined by a * b = 2ab for all a,Â bÂ âˆˆ N

(iii) â€˜*â€™ on Q defined by a * b = a â€“ b for all a, b âˆˆ Q

(iv) â€˜âŠ™â€™ on Q defined by a âŠ™ b = a2 + b2 for all a, b âˆˆ Q

(v) â€˜oâ€™ on Q defined by a o b = (ab/2) for all a, b âˆˆ Q

(vi) â€˜*â€™ on Q defined by a * b = ab2 for all a, b âˆˆ Q

(vii) â€˜*â€™ on Q defined by a * b = a + a b for all a, b âˆˆ Q

(viii) â€˜*â€™ on R defined by a * b = a + b -7 for all a, b âˆˆ R

(ix) â€˜*â€™ on Q defined by a * b = (a â€“ b)2 for all a, b âˆˆ Q

(x) â€˜*â€™ on Q defined by a * b = a b + 1 for all a, b âˆˆ Q

(xi) â€˜*â€™ on N defined by a * b = ab for all a, b âˆˆ N

(xii) â€˜*â€™ on Z defined by a * b = a â€“ b for all a, b âˆˆ Z

(xiii) â€˜*â€™ on Q defined by a * b = (ab/4) for all a, b âˆˆ Q

(xiv) â€˜*â€™ on Z defined by a * b = a + b â€“ ab for all a, b âˆˆ Z

(xv) â€˜*â€™ on Q defined by a * b = gcd (a, b) for all a, b âˆˆ Q

Solution:

(i) First we have to check commutativity of *

Let a, b âˆˆ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Z

Now we have to prove associativity of *

Let a, b, c âˆˆ Z, Then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

Therefore,

a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ Z

Thus, * is associative on Z.

(ii) First we have to check commutativity of *

Let a, b âˆˆ N

a * b = 2ab

= 2ba

= b * a

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus, * is commutative on N

Now we have to check associativity of *

Let a, b, c âˆˆ N

Then, a * (b * c) = a * (2bc)

=

$$\begin{array}{l}2^{a*2^{bc}}\end{array}$$

(a * b) * c = (2ab) * c

=

$$\begin{array}{l}2^{ab*2^{c}}\end{array}$$

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N

(iii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = a â€“ b

b * a = b â€“ a

Therefore, a * b â‰  b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c)

= a â€“ (b â€“ c)

= a â€“ b + c

(a * b) * c = (a â€“ b) * c

= a â€“ b â€“ c

Therefore,

a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q

(iv) First we have to check commutativity of âŠ™

Let a, b âˆˆ Q, then

a âŠ™ b = a2 + b2

= b2 + a2

= b âŠ™ a

Therefore, a âŠ™ b = b âŠ™ a, âˆ€ a, b âˆˆ Q

Thus, âŠ™ on Q

Now we have to check associativity of âŠ™

Let a, b, c âˆˆ Q, then

a âŠ™ (b âŠ™ c) = a âŠ™ (b2 + c2)

= a2 + (b2 + c2)2

= a2 + b4 + c4 + 2b2c2

(a âŠ™ b) âŠ™ c = (a2 + b2) âŠ™ c

= (a2 + b2)2 + c2

= a4 + b4 + 2a2b2 + c2

Therefore,

(a âŠ™ b) âŠ™ c â‰  a âŠ™ (b âŠ™ c)

Thus, âŠ™ is not associative on Q.

(v) First we have to check commutativity of o

Let a, b âˆˆ Q, then

a o b = (ab/2)

= (b a/2)

= b o a

Therefore, a o b = b o a, âˆ€ a, b âˆˆ Q

Thus, o is commutative on Q

Now we have to check associativity of o

Let a, b, c âˆˆ Q, then

a o (b o c) = a o (b c/2)

= [a (b c/2)]/2

= [a (b c/2)]/2

= (a b c)/4

(a o b) o c = (ab/2) o c

= [(ab/2) c] /2

= (a b c)/4

Therefore a o (b o c) = (a o b) o c, âˆ€ a, b, c âˆˆ Q

Thus, o is associative on Q.

(vi) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = ab2

b * a = ba2

Therefore,

a * b â‰  b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (bc2)

= a (bc2)2

= ab2 c4

(a * b) * c = (ab2) * c

= ab2c2

Therefore a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

(vii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

Therefore, a * b â‰  b * a

Thus, * is not commutative on Q.

Now we have to prove associativity on Q.

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b + b c)

= a + a (b + b c)

= a + ab + a b c

(a * b) * c = (a + a b) * c

= (a + a b) + (a + a b) c

= a + a b + a c + a b c

Therefore a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

(viii) First we have to check commutativity of *

Let a, b âˆˆ R, then

a * b = a + b â€“ 7

= b + a â€“ 7

= b * a

Therefore,

a * b = b * a, for all a, b âˆˆ R

Thus, * is commutative on R

Now we have to prove associativity of * on R.

Let a, b, c âˆˆ R, then

a * (b * c) = a * (b + c â€“ 7)

= a + b + c -7 -7

= a + b + c â€“ 14

(a * b) * c = (a + b â€“ 7) * c

= a + b â€“ 7 + c – 7

= a + b + c â€“ 14

Therefore,

a * (b * c ) = (a * b) * c, for all a, b, c âˆˆ R

Thus, * is associative on R.

(ix) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = (a â€“ b)2

= (b â€“ a)2

= b * a

Therefore,

a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c)2

= a * (b2 + c2 â€“ 2 b c)

= (a â€“ b2 â€“ c2 + 2bc)2

(a * b) * c = (a â€“ b)2 * c

= (a2 + b2 â€“ 2ab) * c

= (a2 + b2 â€“ 2ab â€“ c)2

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

(x) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (bc + 1)

= a (b c + 1) + 1

= a b c + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

(xi) First we have to check commutativity of *

Let a, b âˆˆ N, then

a * b = ab

b * a = ba

Therefore, a * b â‰  b * a

Thus, * is not commutative on N.

Now we have to check associativity of *

a * (b * c) = a * (bc)

=

(a * b) * c = (ab) * c

= (ab)c

= abc

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N

(xii) First we have to check commutativity of *

Let a, b âˆˆ Z, then

a * b = a â€“ b

b * a = b â€“ a

Therefore,

a * b â‰  b * a

Thus, * is not commutative on Z.

Now we have to check associativity of *

Let a, b, c âˆˆ Z, then

a * (b * c) = a * (b â€“ c)

= a â€“ (b â€“ c)

= a â€“ (b + c)

(a * b) * c = (a â€“ b) â€“ c

= a â€“ b â€“ c

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Z

(xiii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b c/4)

= [a (b c/4)]/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= [(ab/4) c]/4

= a b c/16

Therefore,

a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Thus, * is associative on Q.

(xiv) First we have to check commutativity of *

Let a, b âˆˆ Z, then

a * b = a + b â€“ ab

= b + a â€“ ba

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ Z

Thus, * is commutative on Z.

Now we have to check associativity of *

Let a, b, c âˆˆ Z

a * (b * c) = a * (b + c â€“ b c)

= a + b + c- b c â€“ ab â€“ ac + a b c

(a * b) * c = (a + b â€“ a b) c

= a + b â€“ ab + c â€“ (a + b â€“ ab) c

= a + b + c â€“ ab â€“ ac â€“ bc + a b c

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c âˆˆ Z

Thus, * is associative on Z.

(xv) First we have to check commutativity of *

Let a, b âˆˆ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ N

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c âˆˆ N

a * (b * c) = a * [gcd (a, b)]

= gcd (a, b, c)

(a * b) * c = [gcd (a, b)] * c

= gcd (a, b, c)

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c âˆˆ N

Thus, * is associative on N.

5. If the binary operation o is defined by a0b = a + b â€“ ab on the set Q â€“ {-1} of all rational numbers other than 1, show that o is commutative on Q â€“ [1].

Solution:

Let a, b âˆˆ Q â€“ {-1}.

Then aob = a + b â€“ ab

= b+ a â€“ ba

= boa

Therefore,

aob = boa for all a, b âˆˆ Q â€“ {-1}

Thus, o is commutative on Q â€“ {-1}

6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?

Solution:

Let a, b âˆˆ Z

a * b = 3a + 7b

b * a = 3b + 7a

Thus, a * b â‰  b * a

Let a = 1 and b = 2

1 * 2 = 3 Ã— 1 + 7 Ã— 2

= 3 + 14

= 17

2 * 1 = 3 Ã— 2 + 7 Ã— 1

= 6 + 7

= 13

Therefore, there exist a = 1, b = 2 âˆˆ Z such that a * b â‰  b * a

Thus, * is not commutative on Z.

7. On the set Z of integers a binary operation * is defined by a 8 b = ab + 1 for all a, b âˆˆ Z. Prove that * is not associative on Z.

Solution:

Let a, b, c âˆˆ Z

a * (b * c) = a * (bc + 1)

= a (bc + 1) + 1

= a b c + a + 1

(a * b) * c = (ab+ 1) * c

= (ab + 1) c + 1

= a b c + c + 1

Thus, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Z.