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Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.1
  • Last Updated : 19 Jan, 2021

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,

X6 1 2 3 4 5
1 1 2 3 4 5
2 2 4 0 2 4
3 3 0 3 0 3
4 4 2 0 4 2
5 5 4 3 2 1

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,

+6 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4

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:

LCM 1 2 3 4 5
1 1 2 3 4 5
2 2 2 6 4 10
3 3 5 3 12 15
4 4 4 12 4 20
5 5 10 15 20 5

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

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