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Class 11 RD Sharma Solutions – Chapter 2 Relations – Exercise 2.3 | Set 2
  • Last Updated : 25 Jan, 2021
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Question 12. Let A = {x, y, z} and B = {a, b}. Find the total number of relations from A into B.

Solution:

The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. 

If n(A) = p and n(B) = q, then n(A × B) = pq. 

So, the total number of relations is 2pq.

n(A) = 3 and n(B) = 2 ⇒ n(A × B) = 3 × 2 = 6



∴ Total number of relations = 26 = 64

Question 13. Let R be a relation from N to N defined by R= {(a, b): a, b ∈ N and a = b2}.

Are the following statements true?

(i) (a, a) ∈ R for all a ∈ N

(ii) (a, b) ∈ R ⇒(b, a) ∈ R

(iii) (a, b) ∈ R and (b, c) ∈ R  ⇒(a, c) ∈ R

Solution:

Given:  R = {(a, b): a, b ∈ N and a = b2}.

(i) (a, a) ∈ R for all a ∈ N

Here, take b = 2

a = b2 = 22 = 4

∴ (4, 2) R but (2, 2) ∉ R

As, 22 ≠ 2



So, 

No, the statement is false.

(ii) (a, b) ∈ R ⇒ (b, a) ∈ R

Here, take b = 2

a = b2 = 22 = 4

∴ (4, 2) ∈ R but (2, 4) ∉ R

As, 42 ≠ 2

So,

No, the statement is false.

(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Here, take b = 4

a = b2 = 42 = 16

(16, 4)∈ R

Now, b = c2

4 = c2

c = -2 ∉ N 

(4, 2) ∈ R

But (16, 2) ∉ R

As, 22 ≠ 16

So,

No, the statement is false.

Question 14. Let A= {1, 2, 3,….,14}. Define a relation on a set A by R= {(x, y): 3x – y = 0, where x, y ∈ A}. Depict this relationship using an arrow diagram. Write down its domain, co-domain, and range.

Solution:

Given: R = {(x, y): 3x – y = 0, where x, y ∈ A}

A = {1, 2, 3,…,14}

As, y = 3x

∴ R = {(x, 3x): where x, 3x ∈ A}

R = {(1, 3×1), (2, 3×2), (3, 3×3), (4, 3×4)}

Note: We cannot include (5, 3×5) as 15 ∉ A

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

So,

Domain of relation R = {1, 2, 3, 4}

Co-Domain of relation R = {1, 2, 3,…,14} = A

Range of relation R = {3, 6, 9, 12}

Question 15: Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural less than 4, x, y ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range of R.

Solution:

R = {(x, y): y = x + 5, x is a natural less than 4, x, y ∈ N}

Natural numbers less than 4 are 1, 2 and 3.

On putting x = 1, y = 1 + 5 = 6

On putting x = 2, y = 2 + 5 = 7

On putting x = 3, y = 3 + 5 = 8

(i) R = {(1, 6), (2, 7), (3, 8)}

(ii) The arrow diagram represent the relation(R):

So,

Domain of relation R = {1, 2, 3}

Range of relation R = {6, 7, 8}

Question 16: A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd, x ∈ A, y ∈ B}. Write R in Roster form.

Solution:

Given:

Relation R from A to B by R = {(x, y): the difference between x and y is odd, x ∈ A, y ∈ B}

A = {1, 2, 3, 5} and B = {4, 6, 9}

For x = 1:

y – x = 4 – 1 = 3 which is odd ⇒ 4 ∈ y

y – x = 6 – 1 = 5 which is odd ⇒ 6 ∈ y

y – x = 9 – 1 = 8 which is even ⇒ 8 ∉ y

For x = 2:

y – x = 4 – 2 = 2 which is even ⇒ 4 ∉ y

y – x = 6 – 2 = 4 which is even ⇒ 6 ∉ y

y – x = 9 – 2 = 7 which is odd ⇒ 8 ∈ y

For x = 3:

y – x = 4 – 3 = 1 which is odd ⇒ 4 ∈ y

y – x = 6 – 3 = 3 which is odd ⇒ 6 ∈ y

y – x = 9 – 3 = 6 which is even ⇒ 8 ∉ y

For x = 5:

x – y = 5 – 4 = 1 which is odd ⇒ 4 ∈ y

y – x = 6 – 5 = 1 which is odd ⇒ 6 ∈ y

y – x = 9 – 5 = 4 which is even ⇒ 8 ∉ y

∴ R = {(1, 4), (1, 6), (2, 8), (3, 4), (3, 6), (5, 4), (5, 6)}

Domain of relation R = {1, 2, 3, 5}

Range of relation R = {4, 6, 8}

Question 17. Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.

Solution:

Given: R = {(x, x3): x is a prime number less than 10}

Prime numbers less than 10 are 2, 3, 5 and 7

∴ R = {(2, 23), (3, 33), (5, 53), (7, 73)}

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

So,

Domain of relation R = {2, 3, 5, 7}

Range of relation R = {8, 27, 125, 343}

Question 18. Let A = {1, 2, 3, 4, 5, 6}. Let R be a relation on A defined by

R = {(a, b): a, b ∈ A, b is exactly divisible by a}

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

Solution:

Given:

R= {(a, b): a, b ∈ A, b is exactly divisible by a}

A= {1, 2, 3, 4, 5, 6}

Here,

6 is exactly divisible by 1, 2, 3 and 6

5 is exactly divisible by 1 and 5

4 is exactly divisible by 1, 2 and 4

3 is exactly divisible by 1 and 3

2 is exactly divisible by 1 and 2

1 is exactly divisible by 1

(i) R = {(1, 1), (2, 1), (2, 2), (3, 1), (3, 3), (4, 1), (4, 2),

             (4, 4), (5, 1), (5, 5), (6, 1), (6, 2), (6, 3), (6, 6)}

(ii) Domain of relation R = {1, 2, 3, 4, 5, 6}

(iii) Range of relation R = {1, 2, 3, 4, 5, 6}

Question 19. The figure shows a relationship between the sets P and Q. Write this relation in

(i) Set builder form

(ii) Roster form

(iii) What is its domain and range?

Solution:

(i) Since 5 – 3 = 6 – 4 = 7 – 5 = 2

∴ x – y = 2 where x ∈ P and y ∈ Q

So,

R = {(x, y): x – y = 2, x ∈ P, y ∈ Q}

(ii) Now, R = {(5, 3), (6, 4), (7, 5)}

(iii) Domain of relation R = {5, 6, 7}

Range of relation R = {3, 4, 5}

Question 20: Let R be the relation on Z defined by R = {(a, b) Z, a – b is an integer}. Find the domain and range of R.

Solution:

Given: R = {(a, b) ∈ Z, a – b is an integer}

Z denotes integer, and here a and b both are integers

We know that difference of two integers is always an integer

∴ a and b can be any integer in relation R

The domain of relation R = Z (as a ∈ Z)

The range of relation R = Z (as b ∈ Z)

Question 21: For the relation R1 defined on R by the rule (a, b) ∈ R1 <=>1 + ab > 0. Prove that: (a, b) ∈ R1 and (b, c) ∈ R1 ⇒(a, c) ∈ R1 is not true for all a, b, c ∈ R.

Solution:

To prove: (a, b) ∈ R1 and (b, c) ∈ R1 ⇒(a, c)∈ R1 is not true for all a, b, c R.

Given: R1 = {(a, b)∈ R <=>1 + ab > 0}

Let a = 1, b = -0.5, c = -4

Here, (1, -0.5) ∈ R1 [∵ 1+(1×-0.5) = 0.5 > 0]

And, (-0.5, -4) ∈ R1 [∵ 1+(-0.5×-4) = 3 > 0]

But, (1, -4) ∉ R1 [∵ 1+(1×-4) = -3 < 0]

∴ (a, b) ∈ R1 and (b, c) ∈ R1 ⇒(a, c) R1 is not true for all a, b, c ∈ R

Hence, Proved.

Note: Here, R1 is a relation whereas R denotes a real number.

Question 22. Let R be a relation on N x N defined by (a, b) R (c, d)<=> a + d = b + c for all (a, b), (c, d) ∈ N x N.

Show that:

i. (a, b) R (a, b) for all (a, b) ∈ N x N

ii. (a, b) R (c, d) ⇒(c, d) R (a, b) for all (a, b), (c, d) ∈ N x N

iii. (a, b) R (c, d) and (c, d) R (e, f) ⇒(a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N

Solution:

Given:

(a, b) R (c, d) a + d = b + c for all (a, b), (c, d) ∈ N x N

(i) (a, b) R (a, b)

a + b = b + a for all (a, b) ∈ N x N

∴ (a, b) R (a, b) for all (a, b) ∈ N x N

(ii) (a, b) R (c, d)

a + d = b + c ⇒ c + b = d + a

(c, d) R (a, b) for all (c, d), (a, b) ∈ N x N

(iii) (a, b) R (c, d) and (c, d) R (e, f)

a + d = b + c and c + f = d + e

a + d + c + f = b + c + d + e

a + f = b + c + d + e – c – d

a + f = b + e

(a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N

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