RD Sharma Class 12 Ex 3.1 Solutions Chapter 3 Binary Operations

Here we provide RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter3
Exercise3.1
CategoryRD Sharma Solutions

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

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

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

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

(iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a × 6 b = Remainder when a b is divided by 6.

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

=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}

(vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N

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

Solution:

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

Let a, b ∈ N. Then,

ab ∈ N      [∵ ab≠0 and a, b is positive integer]

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

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

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

⇒ a * b = 3-1

\frac{1}{3}  ∉ Z

Thus, * is not a binary operation on Z.

(iii)  Given ‘*’ 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 ∉ N

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

So, * is not a binary operation on N.

(iv) Given ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a ×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.

⇒ For a = 2 and b = 3,

a ×6 b = 2 ×6 3 = remainder when 6 divided by 6 = 0 ≠ S

Thus, ×6 is not a binary operation on S.

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

=\begin{cases}a+b,\ if\ a+b<6\\a+b-6,\ if\ a+b\ge 6\end{cases}

Consider the composition table,

+6012345
0012345
1123450
2234501
3345012
4450123
5501234

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

⇒ For a = 2 and b = 3,

a ×6 b = 2 ×6 3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×6 is not a binary operation on S.

(vi) Given ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N

Let a, b ∈ N. Then,

ab, ba ∈ N

⇒ ab + ba ∈ N      [∵Addition is binary operation on N]

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii) Given ‘*’ 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 = \frac{(a - 1)}{ (b + 1)}

\frac{(2 - 1)}{ (- 1 + 1)}

\frac{1}{0}  [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 − 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 ∉ Z+ [because Z+ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z+

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

(ii) Given Z+, define * by a*b = a b

Let a, b ∈ Z+

⇒ a, b ∈ Z+

⇒ a * b ∈ Z+

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab2

Let a, b ∈ R

⇒ a, b2 ∈ R

⇒ ab2 ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

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

Let a, b ∈ Z+

⇒ | a – b | ∈ Z+

⇒ a * b ∈ Z+

Therefore,

a * b ∈ Z+, ∀ a, b ∈ Z+

Thus, * is a binary operation on Z+.

(v) Given on Z+ define * by a * b = a

Let a, b ∈ Z+

⇒ a ∈ Z+

⇒ a * b ∈ Z+

Therefore, a * b ∈ Z+ ∀ a, b ∈ Z+

Thus, * is a binary operation on Z+.

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

Let a, b ∈ R

⇒ a, 4b2 ∈ R

⇒ a + 4b2 ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ 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 − 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 ∉ {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 n^{n^2}

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is 3^{3^2}=3^9

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 2^{2^2}=2^4=16

Question 7. Prove that the operation * on the set

M=\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈  R=\{0\}\right\}  defined by A + B = AB is a binary operation.

Solution: 

We have,

\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b ∈  R=\{0\}\right\}  and

A + B = AB for all A, B ∈ M

Let A =\\begin{bmatrix}a&0\\0&b\end{bmatrix}∈M  and B = \begin{bmatrix}c&0\\0&d\end{bmatrix}∈M

Now, AB = \begin{bmatrix}a&0\\0&b\end{bmatrix}\begin{bmatrix}c&0\\0&d\end{bmatrix}=\begin{bmatrix}ac&0\\0&bd\end{bmatrix}

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

⇒ ac ∈ R and bd ∈ R

⇒ \begin{bmatrix}ac&0\\0&bd\end{bmatrix}∈M

⇒ 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 \frac{m}{n}  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\frac{m}{n}  where m ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

⇒ ab = \frac{35}{6}∈S

Therefore, a * b ∉ S

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

Question 9. The binary operation & : R × R → R is defined as a*b = 2a + b

Solution:

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

Now,

(2*3) = 2 × 2 + 3

         = 4 + 3

(2*3)*4 = 7*4 = 2 × 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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.