RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

RELATIONS Exercise 2D, 2E, 2C

Exercise 2D

Q. 1. What do you mean by a binary relation on a set A?
Define the domain and range of relation on A.

Answer:  A binary relation on a set A is a subset of the Cartesian product A x A.
It’s a rule or way of associating elements of a set with other elements within the same set. 
▶️ The domain of a relation on A is the set of all first elements in the ordered pairs of the relation,
▶️ The range of a relation on A is the set of all second elements in those ordered pairs. 

Q. 2. Let A = {2, 3, 5} and R = {(2, 3), (2, 5), (3, 3), (3, 5)}. 
Show that R is a binary relation on A.
Find its domain and range.

Solution: Given: A = {2, 3, 5}
∴ A×A = {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}
∵ R = {(2, 3), (2, 5), (3, 3), (3, 5)}.
Since, R is a subset of A × A, it’s a binary relation on A.
The domain of R is the set of first co-ordinates of R
∴ Dom(R) = {2, 3}
The range of R is the set of second co-ordinates of R.
∴ Range(R) = {3, 5}

Q. 3. Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8} and let R = {(a, b) : a, b ∈ A and 2a + 3b = 12}.
Express R as a set of ordered pairs.
Show that R is a binary relation on A.
Find its domain and range.

Solution: A = {0, 1, 2, 3, 4, 5, 6, 7, 8}
2a + 3b = 12
⇒ 3b = 12 – 2a
⇒ b = ^{12 – 2a}/₃
If a = 0 then b = 4
a = 3 then b = 2
and a = 6 then b = 0
∴  R = {(0, 4), (3, 2), (6, 0)}
Since, R is a subset of A × A, it is a binary relation to A.
∴  Dom(R) = {0, 3, 6}
∴  Range(R) = {4, 2, 0}

Q. 4. If R is a binary relation on a set A define R^–1 on A.
Let R = {(a, b) : a, b ∈ W and 3a + 2b = 15} and 3a + 2b = 15}, where W is the set of whole numbers.
Express R and R^–1 as sets of ordered pairs.
Show that (i) dom (R) = range (R^–1)
(ii) range (R) = dom (R^–1)

Solution: R = {(a, b) : a, b ∈ W and 3a + 2b = 15} and 3a + 2b = 15}
3a + 2b = 15
⇒ 2b = 12 – 3a
⇒ b = ^{12 – 3a}/₂
If a = 1 then b = 6
a = 3 then b = 3
and a = 5 then b = 0
∴  R = {(1, 6), (3, 3), (5, 0)}
∴  Dom(R) = {1, 3, 5} and
Range(R) = {6, 3, 0}
Now
R^–1 = {(6, 1), (3, 3), (0, 5)}
∴  Dom(R^–1) = {6, 3, 0} and
Range(R^–1) = {1, 3, 5}
Thus,
Dom (R) = {1, 3, 5} = Range (R^–1) (Proved)
Range (R) = {6, 3, 0} = Dom (R^–1)  (Proved)

Q. 5. What is an equivalence relation?
Show that the relation of ‘similarity’ on the set S of all triangles in a plane is an equivalence relation.

Solution: An equivalence relation is one which possesses the properties of reflexivity, symmetry and transitivity.
Let S be a set of all triangles in a plane.
Reflexivity: A relation R on A is said to be reflexive if (a, a) ∈ R for all a ∈ A.
   Since every triangle is similar to itself, it is reflexive.

Symmetry: A relation R on A is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R for all (a, b) ∈ A.
   If one triangle is similar to another triangle, it implies that the other triangle is also similar to the first triangle.
Hence, it is symmetric.

Transitivity: A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all (a, b, c) ∈ A.
   If one triangle(Δ₁) is similar to a triangle(Δ₂) and another triangle(Δ₃) is also similar to that triangle(Δ₂), then all the three triangles are similar.
So the first triangle(Δ₁) is also similar to the last triangle(Δ₃).
Hence, it is transitive.

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Q. 6. Let R = {(a, b) : a, b ∈ Z and (a – b) is even}.
Then, show that R is an equivalence relation on Z.

Solution: R = {(a, b) : a, b ∈ Z and (a – b) is even}
Reflexivity: Let a ∈ Z,
∴ a – a = 0 ∈ Z which is also even.
Thus, (a, a) ∈ R for all value of a ∈ Z.
Hence, it is reflexive
Symmetry: Let (a, b) ∈ R,
(a, b) ∈ R
∴ a – b is even
∴ -(b – a) is even
⇒ (b – a) is even
∴ (b, a) ∈ R
Thus, (a, b) ∈ R ⇒ (b, a) ∈ R
So it is symmetric.

Transitivity: Let (a, b) ∈ R and (b, c) ∈ R
(a – b) is even and (b – c) is even.
∴ [(a – b) + (b – c)] is even
= (a – c) is even.
Thus (a, c) ∈ R.
So it is transitive.
Since, the given relation possesses the properties of reflexivity, symmetry and transitivity,
So it is an equivalence relation.

Q. 7. Let A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and |a² – b²| ≤ 5.
Write R as a set of ordered pairs.
Mention whether R is (i) reflexive (ii) symmetric (iii) transitive.
Give reason in each case.

Solution: Given: A = {1, 2, 3} and R = {(a, b) : a, b ∈ A and |a² – b²| ≤ 5.
If a = 1, b = 1 then |1² – 1²| = 0 < 5
∴ (1, 1) ∈ R
If a = 1, b = 2 then |1² – 2²| = 3 < 5
∴ (1, 2) ∈ R
If a = 1, b = 3 then |1² – 3²| = 8 > 5
∴ (1, 3) ∉ R

If a = 2, b = 1 then |2² – 1²| = 3 < 5
∴ (2, 1) ∈ R
If a = 2, b = 2 then |2² – 2²| = 0 < 5
∴ (2, 2) ∈ R
If a = 2, b = 3 then |2² – 3²| = 5 = 5
∴ (2, 3) ∈ R
If a = 3, b = 1 then |3² – 1²| = 8 > 5
∴ (3, 1) ∉ R
If a = 3, b = 2 then |3² – 2²| = 5 = 5
∴ (3, 2) ∈ R
If a = 3, b = 3 then |3² – 3²| = 0 < 5
∴ (3, 3) ∈ R
∴ R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2

(i) R is reflexive if (a, a) ∈ R for all value of a ∈ Z.
|a² – a²| = 0 ≤ 5.
Thus, it is reflexive.
(ii) R is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all value of a, b ∈ Z.
Let (a, b) ∈ R
(a, b) ∈ R
⇒ |a² – b²| ≤ 5
∴ |b² – a²| ≤ 5
∴ (b, a) ∈ R
Hence, it is symmetric

(iii) R is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of a, b, c ∈ Z.
Putting a = 1 , b = 2 , c = 3.
|1² – 2²| = 3 ≤ 5 and
|2² – 3²| = 5 ≤ 5
but |1² – 3²| = 8 > 5
Thus, it is not transitive.

Q. 8. Let R = {(a, b) : a, b ∈ Z and b = 2a – 4}.
If (a, –2} ∈ R and (4, b²) ∈ R. Then, write the values of a and b.

Solution: b = 2a – 4……(ii)
Putting  b = -2 in equation (i) ,
-2 = 2a – 4
⇒ 2a = 3
⇒ a = 1
Putting  a = 4 in equation (i)
b = 2×4 – 4
⇒ b = 8 – 4 = 4
Ans: Value of a = 1 , b = 4

Q. 9. Let R be a relation on Z, defined by (x, y) ∈ R ⇔ x² + y² = 9.
Then, write R as a set of ordered pairs.
What is its domain?

Solution: x² + y² = 9
(x, y) ∈ R
If x = 0 , y = ±3 , 0² + (±3)² = 9
If x = ±3 , y = 0 , (±3)² + 0² = 9
∴ R = {(0, 3) , (0, -3) , (3, 0) , (-3, 0)}
∴ Dom(R) = {-3 , 0 , 3}

Q. 10. Let A be the set of first five natural numbers and let R be a relation on A,
defined by (x, y) ∈ R ⇔ x ≤ y.
Express R and R^–1 as sets of ordered pairs.
Find: dom (R^–1) and range (R).

Solution:  A = {1, 2, 3, 4, 5}
Since, x ≤ y
∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}

R^–1 = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4), (5, 5)}
∴ Dom(R^–1) = {1, 2, 3, 4, 5}
∴ Range(R) = {1, 2, 3, 4, 5}

Q. 11. Let R = (x, y) : x, y ∈ Z and x² + y² = 25}.
Express R and R^–1 as sets of ordered pairs.
Show that R = R^–1.

Solution: x² + y² = 25
(x, y) ∈ R
If x = 0 , y = ±5 , 0² + (±5)² = 25
If x = ±3 , y = ±4 , (±3)² + (±5)² = 25
x = ±4 , y = ±3 , (±4)² + (±3)² = 25
If x = ±5 , y = 0 , (±5)² + (0)² = 25
∴ R = {(-5, 0), (-4, -3), (-4, 3), (-3, -4), (-3, 4), (0, -5), (0, 5), (3, -4), (3, 4), (4, -3), (4, 3), (5, 0)}
∴ R^–1 = {(0, -5), (-3, -4), (3, -4), (-4, -3), (4, -3), (-5, 0), (5, 0), (-4, 3), (4, 3), (-3, 4), (3, 4), (0, 5)}
Since, all ordered pair of R is in R^–1.
Hence, R = R^–1 (Proved)

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2
Exercise 2D

Q. 12. (i) Find R^–1, when
R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)}

Solution: R = {(1, 2), (1, 3), (2, 3), (3, 2), (4, 5)}
∴ R^–1 = {(2, 1), (3, 1), (3, 2), (2, 3), (5, 4)}

Q. 12. (ii) Find R^–1, when
R = {(x, y) : x, y ∈ N, x + 2y = 8}.

Solution: R = {(x, y) : x, y ∈ N, x + 2y = 8}.
Put x = 2, y = 3
∴ x + 2y = 2 + 2.3 = 8
Put x = 4, y = 2
∴ x + 2y = 2 + 2×3 = 8
Put x = 6, y = 1
∴ x + 2y = 6 + 2×1 = 8
∴ R = {(2, 3), (4, 2), (6, 1)}
R^–1 = {(3, 2), (2, 4), (1, 6)}

Q. 13. Let A = {a, b}. List all relation on A and find their number.

Solution:  Any relation on A is a subset of A×A
∴ A×A = {(a, a), (a, b), (b, a), (b, b)}
Number of relation on A is the number of subsets of A×A
∴ All relation on A are:
{}, {(a, a)}, {(a, b)}, {(b, a)}, {(b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)}, {(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, a), (b, a), (b, b)}, {(a, b), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}
Thus, there are 16 total relations.

Q. 14. Let R = {(a, b) : a, b, ε N and a < b}.
Show that R is a binary relation on N, which is neither reflexive nor symmetric.
Show that R is transitive.

Solution: N is the set of all the natural numbers.
∴ N = {1, 2, 3, 4, 5, 6, 7…..}
R = {(a, b) : a, b, ∈ N and a < b}
R = {(1, 2), (1, 3), (1, 4) …. (2, 3), (2, 4), (2, 5) ……}
For reflexivity,
A relation R on N is said to be reflexive if (a, a) ∈ R for all value of a ∈ N.
Here R = {(a, b) : a, b, ∈ N and a < b}
Hence a < b,
So the two co-ordinates of the ordered pair are never equal.
Thus, the relation is not reflexive.

For symmetry,
A relation R on N is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R for all value of a, b ∈ N
Here R = {(a, b) : a, b ∈ N and a < b}
(a, b) ∈ R
⇒ a < b
⇒ b > a
∴ (b, a) ∉ R 
 Thus, it is not symmetric.
For transitivity,
A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of (a, b, c) ∈ N.
Let’s take three values a, b and c such that a < b < c.
So, (a, b) ∈ R and (b, c) ∈ R
⇒ (a,c) ∈ R.
Thus, it is transitive.

Exercise 2E

Q. 1. (i) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2. n(A ∪ B)

Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
We know,
  n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ n(A ∪ B) = 5 + 3 – 2 = 6

Q. 1. (ii) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2.  n(A × B)

Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
We know, 
n(A × B) = n(A) × n(B)
∴ n(A × B) = 5 × 3 = 15

Q. 1. (iii) Let A and B be two sets such that n(A) = 5, n(B) = 3 and n(A ∩ B) = 2. n(A × B) ∩ (B × A)

Solution: Given: n(A) = 5, n(B) = 3 and n(A ∩ B) = 2
∴ n(A × B) ∩ (B × A)
= 2^{n(A ∩ B)}
= 2² = 4

Q. 2 Find a and b when (a – 2b, 13) = (7, 2a – 3b)

Solution: Since the ordered pairs are equal,
Comparing corresponding elements,
∴ a – 2b = 7 …… (ii)  and
2a – 3b = 13 …… (ii)
(i)×2 – (ii),
2a – 4b – 2a + 3b = 14 – 13
⇒ -b = 1
⇒ b = -1
Putting b = -1 in equation (i), 
a – 2(-1) = 7
⇒ a = 7 – 2
⇒ a = 5
Hence a = 5; b = -1

Q. 3. If A = {1, 2}, find A × A × A

Solution: Given: A = {1, 2}
∴ A × A
= {1, 2} × {1, 2}
= {(1, 1), (1, 2), (2, 1), (2, 2)}

∴ A × A × A
= {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

Q. 4. If A = {2, 3, 4} and B = {4, 5}, draw an arrow diagram represent (A × B}.

Solution: Given: A = {2, 3, 4} and B = {4, 5},
Arrow Diagram:

ARROW DIAGRAM:

Q. 5. If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A × (B × C).

Solution: Given: A = {3, 4}, B = {4, 5} and C = {5, 6}
B × C
= {4, 5} × {5, 6}
= {(4, 5), (4, 6), (5, 5), (5, 6)}

∴ A × (B × C)
= {3, 4} × {(4, 5), (4, 6), (5, 5), (5, 6)}
= {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}

Q. 6. If A ⊆ B, prove that A × C = B × C

Solution: Given: A ⊆ B
Let any arbitrary element (a, c) ∈ A × C
(a, c) ∈ A × C
⇒ a ∈ A and c ∈ C
⇒ a ∈ B and c ∈ C …… [∵ A ⊆ B]
∴ (a, c) ∈ B × C
∴ A × C = B × C [Proved]

Q. 7. Prove that A × B = B × A ⇒ A = B.

Solution: Let A and B be any two sets such thatA × B = {(a, b): a ∈ A, b ∈ B}
Now, B × A = {(b, a): b ∈ A, a ∈ B}
Since A × B = B × A
∴ (a, b) = (b, a)
This is possible only when the ordered pairs are equal.
Therefore,a = b and b = a
Hence  A = B. (Proved)

Q. 8. If A = {5} and B = {5, 6}, write down all possible subsets of A × B.

Solution: Given: A = {5}, B = {5, 6}
∴ A × B
= {(5, 5), (5, 6)}
∴ All  possible subsets of A × B are:
{}, {(5, 5)}, {(5, 6)}, {(5, 6), (5, 6)}

Q. 9. Let R = {(x, x²): x is a prime number less than 10}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: (i) {(x, x²): x is a prime number less than 10}.
Prime number less than 10 are 2, 3, 5 and 7
∴ Roster form:
R = {(2, 4), (3, 9), (5, 25), (7, 49)}
(ii) Domain of R is the set of first coordinates of R
∴ Dom(R)
= {2, 3, 5, 7}
Range of R is the set of second coordinates of R
Range(R) = {4, 9, 25, 49}

Q. 10. Let A = (1, 2, 3} and B = {4}
How many relations can be defined from A to B.

Solution: Given: A = (1, 2, 3} and B = {4}
∴ n(A) = 3 and n(B) = 1
∴ n(A×B)
= n(A) × n(B)
= 3 × 1= 3
The number of relations from set A to set B
= 2^n(A×B)
= 2³ = 8
Total number of relations = 8

Q. 11. Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and a < b}
(i) Write R in roster form.
(ii) Find: dom (R) and range (R)
(iii) Write R^–1 in roster form.

Solution: (i) Given: A = {3, 4, 5, 6} and R = {(a, b) : a, b ∈ A and a > b}
∴ Roster form: R = {(4, 3), (5, 3), (5, 4), (6, 3), (6, 4), (6, 5)}
(ii) Domain of R is the set of first coordinates of R
Dom(R) = {4, 5, 6}
Range of R is the set of second coordinates of R
Range(R) = {3, 4, 5}
(iii) R^–1 = = {(3, 4), (3, 5), (4, 5), (3, 6), (4, 6), (5, 6)}

Q. 12. Let R = {(a, b) : a, b ∈ N and a < b}.
Show that R is a binary relation on N which is neither reflexive nor symmetric.
Show that R is transitive.

Solution:  A binary relation R on sets A and B is a subset of the Cartesian product A×B
Given: R = {(a, b) : a, b ∈ N and a < b}
Since a, b ∈ N
So the ordered pair of (a, b) is an element of the Cartesian product N×N
∴ R is a subset of the Cartesian product N×N
Therefore R is a binary relation on N. (Proved)

For reflexivity,
A relation R on N is said to be reflexive if (a, a) ∈ R for all value of a ∈ N.
Here R = {(a, b) : a, b, ∈ N and a < b}
Hence a < b, so the two co-ordinates of the ordered pair are never equal.
Thus, the relation is not reflexive.
For symmetry,
A relation R on N is said to be symmetrical if (a, b) ∈ R ⇒ (b, a) ∈ R
Here R = {(a, b) : a, b ∈ N and a < b}
(a, b) ∈ R
∴ a < b
⇒ b > a
∴ (b, a) ∉ R 
Thus, it is not symmetric.

For transitivity,
A relation R on A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all value of (a, b, c) ∈ N.
Let’s take three values a, b and c such that a < b < c.
So, (a, b) ∈ R and (b, c) ∈ R
⇒ (a,c) ∈ R.
Thus, it is transitive.