Class 12 RD Sharma Solutions - Chapter 3 Binary Operations - Exercise 3.2

Question 1. Let ‘*’ be a binary operation on N defined by a * b = 1.c.m. (a, b) for all a, b ∈ N

(i) Find 2 * 4, 3 * 5, 1 * 6

Solution:

We are given that a * b = L.C.M. (a, b) 

⇒ 2 * 4 = L.C.M. (2, 4) = 4

and, 3 * 5 = L.C.M. (3, 5) = 15

now, 1 * 6 = L.C.M. (1, 6) = 6

Hence, 2 * 4 = 4, 3 * 5 = 15 and 1 * 6 = 6.

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

Solution:

For Commutativity:

Let a, b ∈ N

a * b = L.C.M. (a, b) = L.C.M. (b, a) = b * a

Therefore, a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

For Associativity:

Let a, b, c ∈ N

⇒ a * (b * c) = a * L.C.M. (b, c) = L.C.M. (a, (b, c)) = L.C.M. (a, b, c)

And, (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, ∀ a, b, c ∈ N

Thus, * is associative on N.

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

(i) * on N defined by a * b = 1 for all a, b ∈ N

Solution:

For commutativity:

Let a, b ∈ N

a * b = 1 and b * a = 1

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

Thus * is commutative on N.

For associativity:

Let a, b, c ∈ N

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

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

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

Thus, * is associative on N.

Hence, * is both commutative and associative on N.

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

Solution:

For Commutativity:

Let a, b ∈ N

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

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

For Associativity:

Let a, b, c ∈ 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) ≠ (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

and, (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 ∈ N such that a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

Hence * is commutative on N but not associative on N.

Question 3. Let A be any set containing more than one element. Let ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?

Solution:

For Commutativity:

Let a, b ∈ A.

Then, a * b = b

⇒ b * a = a

Therefore, a * b ≠ b * a

Thus, * is not commutative on A.

Now we have to check associativity:

Let a, b, c ∈ A

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

Therefore, a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Thus, * is associative on A.

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

(i) ‘*’ on Z defined by a * b = a + b + a b for all a, b ∈ Z

Solution:

For Commutativity:

Let a, b ∈ Z

Then a * b = a + b + ab = b + a + ba = b * a

Therefore, a * b = b * a, ∀ a, b ∈ Z

Hence, * is commutative on Z.

For Associativity:

Let a, b, c ∈ 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

Now, (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

Clearly, a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) ‘*’ on N defined by a * b = 2^ab for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N

a * b = 2^ab = 2^ba = b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

For Associativity:

Let a, b, c ∈ N

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

and, (a * b) * c = (2^ab) * c = 2ab^2c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

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

and, (a * b) * c = (a – b) * c = a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(iv) ‘⊙’ on Q defined by a ⊙ b = a² + b² for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a ⊙ b = a² + b² = b² + a² = b ⊙ a

Clearly, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b² + c²)

= a² + (b² + c²)2

= a² + b⁴ + c⁴ + 2b²c²

(a ⊙ b) ⊙ c = (a² + b²) ⊙ c

= (a² + b²)² + c²

= a⁴ + b⁴ + 2a²b² + c²

Clearly, (a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

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

Clearly, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q.

For Associativity:

Let a, b, c ∈ 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

and, (a o b) o c = (ab/2) o c = [(ab/2) c] /2 = (a b c)/4

Clearly, a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi) ‘*’ on Q defined by a * b = ab² for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = ab²

b * a = ba²

Clearly, * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to check associativity of *

Let a, b, c ∈ Q, then

a * (b * c) = a * (bc²)

= a (bc²)²

= ab² c⁴

(a * b) * c = (ab²) * c

= ab²c²

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(vii) ‘*’ on Q defined by a * b = a + ab for all a, b ∈ Q

Solution:

For commutative:

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba = b + ab

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

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

= a + a (b + bc)

= a + ab + abc

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

= (a + ab) + (a + ab)c

= a + ab + ac + abc

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R

Solution:

For Commutativity:

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7 = b * a

Clearly, a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R.

For Associativity:

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

and, (a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

Clearly, a * (b * c ) = (a * b) * c, for all a, b, c ∈ R

Thus, * is associative on R.

(ix) ‘*’ on Q defined by a * b = (a – b)² for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = (a – b)²

= (b – a)²

= b * a

Clearly, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)²

= a * (b² + c² – 2bc)

= (a – b² – c² + 2bc)²

(a * b) * c = (a – b)² * c

= (a² + b² – 2ab) * c

= (a² + b² – 2ab – c)²

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x) ‘*’ on Q defined by a * b = ab + 1 for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Clearly, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

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

= a (bc + 1) + 1

= abc + a + 1

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

= (ab + 1) c + 1

= abc + c + 1

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi) ‘*’ on N defined by a * b = a^b for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

Clearly, a * b ≠ b * a

Thus, * is not commutative on N.

For Associativity:

a * (b * c) = a * (bc) = {a^b}^c

and, (a * b) * c = (ab) * c = (a^b)^c = a^bc

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

(xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z

Solution:

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

(xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

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

Thus, * is commutative on Q

For Associativity:

Let a, b, c ∈ Q, then

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

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

= (a b c/16)

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

= [(ab/4) c]/4

= abc/16

Clearly a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Thus, * is associative on Q.

(xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z

Solution:

For Commutativity:

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

Clearly, a * b = b * a, for all a, b ∈ Z

Thus, * is commutative on Z.

For Associativity:

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – bc)

= a + b + c- b c – ab – ac + abc

(a * b) * c = (a + b – ab) c

= a + b – ab + c – (a + b – ab) 

= a + b + c – ab – ac – bc + a b c

Clearly, a * (b * c) = (a * b) * c, for all a, b, c ∈ Z

Thus, * is associative on Z.

(xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

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

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ N

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

= gcd (a, b, c)

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

= gcd (a, b, c)

Clearly, a * (b * c) = (a * b) * c, for all a, b, c ∈ N

Thus, * is associative on N.

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

Solution:

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – b = boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}.

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

Solution:

Let a, b ∈ Z

a * b = 3a + 7b

and, b * a = 3b + 7a

Clearly, a * b ≠ b * a for all a, b ∈ Z.

Example, Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2 = 3 + 14 = 17

2 * 1 = 3 × 2 + 7 × 1 = 6 + 7 = 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

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

Solution:

Let a, b, c ∈ 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

Clearly, a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Z

Thus, * is not associative on Z.

Question 8. Let S be the sum of all real numbers except −1 and let * be an operation defined by a * b = a + b + ab for all a,b ∈ S. Determine whether * is a binary operation on S. If yes, check its commutativity and associativity. 

Solution:

Given: a * b = a + b + ab, a, b ∈ S = R − {−1}

Let a, b ∈ S.

Thus, ab ∈ S and hence, a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

For Commutativity:

a * b = a + b + ab = b +a + ba = b * a

Hence, * is commutative.

For Associativity:

Let a, b, c ∈ Z, Then,

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

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

= a + b + c + ab + ac + bc + abc      .....(a)

Now, a * (b * c) = a * (b + c + bc)

= a + b + c + bc + ac +ab +abc        .....(b)

From (a) and (b), it is clear that a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.

Question 9. On Q, the set of rational numbers, * is defined by a * b = (a - b)/2, show that * is not associative.

Solution:

Let a, b, c ∈ Q. Then,

(a * b) * c = \frac{a − b}{2}  * c = \frac{\frac{a-b}{2}-c}{2}  = \frac{a-b-2c}{4}         .......(a)

Now, a * (b * c) = a * \frac{b - c}{2}  = \frac{2a − b + c}{4}                        ..........(b)

From (a) and (b), it is clear that a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

Question 10. Let binary operation * : R×R⇥R is defined as a * b = 2a + b. Find (2 * 3) * 4

Solution:

Given, a * b = 2a + b

⇒ (2 * 3) * 4 = (2 × 2 + 3) * 4 = 7 * 4 = (2 × 7 + 4) = 18

Hence, (2 * 3) * 4 = 18.

Question 11. On Z, the set of integers, a binary operation * is defined as a * b = a + 3b − 4. Prove that * is neither commutative nor associative on Z.

Solution:

For Commutativity:

a * b = a + 3b − 4 ≠ b + 3a − 4 = b * a

⇒ a * b ≠ b * a

Hence * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + 3b − 4) * c 

= a + 3b − 4 + 3c − 4 

= a + 3b + 3c − 8     .......(a)

Now, a * (b * c) = a + 3(b + 3c − 4) − 4

= a + 3b + 9c − 16    ......(b)

From (a) and (b),  we get a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

Question 12. On the set Q of all rational numbers if a binary operation * is defined as 

a * b = ab/5, prove that * is associative on Q.

Solution:

Let a, b, c ∈ Z, then,

(a * b) * c = ab/5 * c = abc/25      .....(a)

and, a * (b * c) = a * bc/5 = abc/25    ....(b)

From eq (a) and (b), we have 

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

Hence, * is associative on Q.

Question 13. The binary operation * is defined as a * b = ab/7 on the set Q of rational numbers. Prove that * is associative on Q.

Solution:

Let a, b, c ∈ Z, then,

(a * b) * c = ab/7 * c = abc/49      .....(a)

and, a * (b * c) = a * bc/7 = abc/49   ....(b)

From eq(a) and (b), we have 

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

Hence, * is associative on Q.

Question 14. On Q, the set of all rational numbers, a binary operation * is defined as (a + b)/2 . Show that * is not associative on Q.

Solution:

Let a, b, c ∈ Z, then,

(a * b) * c = \frac{a + b}2   * c = \frac{\frac{a-b}{2}-c}{2}  = \frac{a-b-2c}{4}        ...(a)

a * (b * c) = a * \frac{b - c}{2}  = \frac{a+\frac{b+c}{2}}{2}  = \frac{2a+b+c}4     ...(b)

From eq(a) and (b), we have, 

a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

Question 15. Let S be the sum of all real numbers except 1 and let * be an operation defined by a * b = a + b − ab for all a, b ∈ S. Prove that:

(i) * is a binary operation on S. 

Solution:

Let a, b ∈ S

Thus, ab ∈ S and hence, 

a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

(ii) is commutative and associative. 

Solution:

For Commutativity:

a * b = a + b − ab = b + a − ba = b * a

Hence, * is commutative.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + b − ab) * c

= a + b − ab + c + (a + b − ab)c

= a + b + c − ab − ac − bc + abc      .....(a)

Now, a * (b * c) = a * (b + c − bc)

= a + b + c − bc − ac − ab +abc        .....(b)

From eq(a) and (b), it is clear that 

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

Hence, * is associative on Q.

Summary

Exercise 3.2 in Chapter 3 of RD Sharma's Class 12 mathematics textbook likely focuses on more advanced aspects of binary operations. This exercise typically covers topics such as identifying and proving properties of binary operations (like associativity, commutativity, and distributivity), finding identity and inverse elements, and exploring binary operations in specific mathematical structures. It may also include problems related to isomorphisms between algebraic structures and the analysis of more complex binary operations.