Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.1

Question 1. Determine whether the following operation define a binary operation on the given set or not:

(i) β€˜*’ on N defined by a * b = ab for all a, b ∈ N.

(ii) β€˜O’ on Z defined by a O b = ab for all a, b ∈ Z.

(iii) β€˜*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

(iv) β€˜Γ—6β€˜ on S = {1, 2, 3, 4, 5} defined by a Γ— 6 b = Remainder when a b is divided by 6.

(v) β€˜+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b

(vi) β€˜βŠ™β€™ on N defined by a βŠ™ b= ab + ba for all a, b ∈ N

(vii) β€˜*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

Solution:

(i) Given β€˜*’ on N defined by a * b = ab for all a, b ∈ N.

Let a, b ∈ N. Then,

ab ∈ N      [∡ abβ‰ 0 and a, b is positive integer]

β‡’ a * b ∈ N

Therefore,

a * b ∈ N, βˆ€ a, b ∈ N

Thus, * is a binary operation on N.

(ii) Given β€˜O’ on Z defined by a O b = ab for all a, b ∈ Z.

Both a = 3 and b = -1 belong to Z.

β‡’ a * b = 3-1

=  βˆ‰ Z

Thus, * is not a binary operation on Z.

(iii)  Given β€˜*’ on N defined by a * b = a + b – 2 for all a, b ∈ N

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 βˆ‰ N

Thus, there exist a = 1 and b = 1 such that a * b βˆ‰ N

So, * is not a binary operation on N.

(iv) Given β€˜Γ—6β€˜ on S = {1, 2, 3, 4, 5} defined by a Γ—6 b = Remainder when a b is divided by 6.

Consider the composition table,

X612345
112345
224024
330303
442042
554321

Here all the elements of the table are not in S.

β‡’ For a = 2 and b = 3,

a Γ—6 b = 2 Γ—6 3 = remainder when 6 divided by 6 = 0 β‰  S

Thus, Γ—6 is not a binary operation on S.

(v) Given β€˜+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b

Consider the composition table,

+6012345
0012345
1123450
2234501
3345012
4450123
5501234

Here all the elements of the table are not in S.

β‡’ For a = 2 and b = 3,

a Γ—6 b = 2 Γ—6 3 = remainder when 6 divided by 6 = 0 β‰  Thus, Γ—6 is not a binary operation on S.

(vi) Given β€˜βŠ™β€™ on N defined by a βŠ™ b= ab + ba for all a, b ∈ N

Let a, b ∈ N. Then,

ab, ba ∈ N

β‡’ ab + ba ∈ N      [∡Addition is binary operation on N]

β‡’ a βŠ™ b ∈ N

Thus, βŠ™ is a binary operation on N.

(vii) Given β€˜*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

If a = 2 and b = -1 in Q,

a * b = 

=  [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

Question 2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.

(i) On Z+, defined * by a * b = a – b

(ii) On Z+, define * by a*b = ab

(iii) On R, define * by a*b = ab2

(iv) On Z+ define * by a * b = |a βˆ’ b|

(v) On Z+ define * by a * b = a

(vi) On R, define * by a * b = a + 4b2

Here, Z+ denotes the set of all non-negative integers.

Solution:

(i) Given On Z+, defined * by a * b = a – b

If a = 1 and b = 2 in Z+, then

a * b = a – b

= 1 – 2

= -1 βˆ‰ Z+ [because Z+ is the set of non-negative integers]

For a = 1 and b = 2,

a * b βˆ‰ Z+

Thus, * is not a binary operation on Z+.

(ii) Given Z+, define * by a*b = a b

Let a, b ∈ Z+

β‡’ a, b ∈ Z+

β‡’ a * b ∈ Z+

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab2

Let a, b ∈ R

β‡’ a, b2 ∈ R

β‡’ ab2 ∈ R

β‡’ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z+ define * by a * b = |a βˆ’ b|

Let a, b ∈ Z+

β‡’ | a – b | ∈ Z+

β‡’ a * b ∈ Z+

Therefore,

a * b ∈ Z+, βˆ€ a, b ∈ Z+

Thus, * is a binary operation on Z+.

(v) Given on Z+ define * by a * b = a

Let a, b ∈ Z+

β‡’ a ∈ Z+

β‡’ a * b ∈ Z+

Therefore, a * b ∈ Z+ βˆ€ a, b ∈ Z+

Thus, * is a binary operation on Z+.

(vi) Given On R, define * by a * b = a + 4b2

Let a, b ∈ R

β‡’ a, 4b2 ∈ R

β‡’ a + 4b2 ∈ R

β‡’ a * b ∈ R

Therefore, a *b ∈ R, βˆ€ a, b ∈ R

Thus, * is a binary operation on R.

Question 3. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b βˆ’ 3. Find the value of 3 * 4.

Solution:

Given:

a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

Question 4. Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.

Solution:

LCM12345
112345
2226410
33531215
44412420
551015205

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 βˆ‰ {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

Question 5. Let S = {a, b, c}. Find the total number of binary operations on S.

Solution:

Number of binary operations on a set with n elements is 

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is 

Question 6. Find the total number of binary operations on {a, b}.

Solution: 

We have,

S = {a, b}

The total number of binary operation on S = {a, b} in 

Question 7. Prove that the operation * on the set

M= defined by A + B = AB is a binary operation.

Solution: 

We have,

 and

A + B = AB for all A, B ∈ M

Let A =\ and B = 

Now, AB = 

Therefore, a ∈ R, b ∈ R, c ∈ R and d ∈ R

β‡’ ac ∈ R and bd ∈ R

β‡’ 

β‡’ A * B ∈ M

Hence, the operator * defines a binary operation on M

Question 8. Let S be the set of all rational numbers of the form  where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation

Solution: 

S = set of rational numbers of the form where m ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

β‡’ ab = 

Therefore, a * b βˆ‰ S

Hence, the operator * does not defines a binary operation on S

Question 9. The binary operation & : R Γ— R β†’ R is defined as a*b = 2a + b

Solution:

It is given that, a*b = 2a + b

Now,

(2*3) = 2 Γ— 2 + 3

         = 4 + 3

(2*3)*4 = 7*4 = 2 Γ— 7 + 4

            = 14 + 4

            = 18

Question 10. Let * be a binary operation on N given by a*b = LCM(a, b) for all a, b ∈ N. Find 5*7.

Solution:

It is given that a*b = LCM (a, b)

Now,

5*7 = LCM (5, 7)

       = 35