AP Inter 2nd Year Maths Exercise 1a Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions Exercise 1a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Relations and Functions Solutions Exercise 1a

I. Give an example of a relation. Which is

Question 1.
Symmetric but neither reflexive nor transitive:
Solution:
Let A = {1, 2}. On A × A consider a relation R = {(1, 2),(2, 1)}
(i) (1, 1),(2, 2) ∈ R . So, R is not reflexive.
(ii) (1, 2) ∈ R and (2, 1) ∈ R. So, R is symmetric
(iii) (1, 2), (2, 1) ∈ R,but (1, 1) ∉ R. So, R is not transitive.
∴ This relation R is symmetric but not reflexive or transitive.

Question 2.
Transitive but neither reflexive nor symmetric:
Solution:
Let A = {1, 2, 3}. On A × A consider a relation R = {(a, b): a < b}
(i) (1, 1) ∉ R. So, R is not reflexive, [v 1 cannot be less than itself ]
(ii) (1, 2) ∈ R. But (2, 1) ∉ R.So, R is not symmetric. [∵ 2 is not less than 1]
(iii) (1, 2),(2, 3) ∈ R ⇒ (1, 3) ∈ R. So, R is transitive. [∵ 1 < 2 and 2 < 3 ⇒ 1 < 3]
∴ This relation R is transitive but not reflexive and symmetric.

Question 3.
Reflexive and symmetric but not transitive:
Solution:
LetA = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
(i) (a, a) ∈ R ∀ a ∈ A, . So, R is reflexive.
(ii) (1, 2) ∈ R ⇒ (2, 1) ∈ R for a, b ∈ R. So, R is symmetric.
(iii) (1, 2), (2, 3) ∈ R ⇒ (1, 3) ∉ R. So, R is not transitive.
∴ This relation R is reflexive and symmetric but not transitive.

Question 4.
Reflexive and transitive hut not symmetric.
Solution:
Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(i) (a, a) ∈ R ∀ a ∈ A, . So, R is reflexive.
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R . So, R is not symmetric.
(iii) (1, 2) ∈ R, (2, 3) ∈ R ⇒ (1, 3) ∈ R . So, R is transitive.
∴ This relation R is reflexive and transitive but not symmetric.

Question 5.
Symmetric and transitive but not reflexive.
Solution:
Let A = {1, 2} and R = {(1, 2), (2, 1), (1, 1)}
(i) (2, 2) ∉ R . So, R is not reflexive.
(ii) (1, 2), (2, 1) ∈ R .So, R is symmetric.
(iii) (1, 2), (2, 1) ∈ R ⇒ (1, 1) ∈ R . So, R is transitive.
∴ This relation R is not reflexive but symmetric and transitive.

Question 6.
Define reflexive relation
Solution:
Reflexive Relation: A relat ion R on a set A is reflexive if (a, a) ∈ R for all a ∈ A.
Ex : The relation equality (=) on reals.

Question 7.
Define symmetric relation
Solution:
Symmetric Relation: A relation R on a set A is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R
Ex: The relation ‘parallel of lines (||) in a plane.

Question 8.
Define transitive relation
Solution:
Transitive Relation: A relation R on a set A is transitive if(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
Ex: The relation greater than ( > ) on reals.

Question 9.
Define an equivalence relation
Solution:
Equivalence Relation:
A relation R on a set A is called an equivalence relation if R is reflexive, symmetric and transitive.
Ex: The relation equality (=) on reals.

II.

Question 1.
Determine whether the Relation R in the set A = {I, 2, 3, 13, 14} defined as R = {(x, y): 3x – y = 0} is reflexive, symmetric and transitive.
Solution:
Given that A = {1, 2, 3, …, 13, 14} andR= {(x, y): 3x – y = 0}.
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12), (14, 52)}
(i) (1, 1), (2, 2), …. ∉ R. So, R is not reflexive.
(ii) (1, 3) ∈ R, but (3, 1) ∉ R. So, R is not symmetric.
(iii) (1, 3), (3, 9) ∈ R, but(1, 9) ∉ R. So, R is not transitive.

Question 2.
Determine whether the Relation R in the set N of natural numbers defined as,
R = {(x, y): y = x + 5 and x < 4} is reflexive, symmetric and transitive.
Solution:
Given N = {1, 2, 3, …} and R = {(x, y): y = x + 5 and x < 4}
∴ R = {(1, 6), (2, 7), (3, 8)}
(i) (1, 1) ∉ R. So, R is not reflexive.
(ii) (1, 6) ∈ R but(6, 1) ∉ R So, R is not symmetric.
(iii) We find no 3 ordered pairs such that (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R .
Among the three ordered pairs in R, no chaining is possible vacously true. So, R is transitive.

Question 3.
Determine whether the Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x} is reflexive, symmetric and transitive.
Solution:
Given that A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
(i) (x, x) ∈ R ∀ x ∈ A. So, R is reflexive.
[∵ Any number other than 0 is divisible by itself. ]

(ii) (x, y) ∈ R then (y, x)need not belong to R. So, R is not symmetric.
Let (2, 4) ∈ R [because 4 is divisible by 2]
But (4, 2)∉ R[since 2 is not divisible by 4]

(iii) (x, y) and (y, z) ∈ R ⇒(x, z) ∈ R. So, R is transitive.
Let (1, 3) ∈ R [because 3 is divisible by 1]
and (3, 6) ∈ R[since 6 is divisible by 3 ]
Then (1, 6) ∈ R[since 6 is divisible by 1]

Question 4.
Determine whether, the Relation R in the set Z of all integers defined as R = {(x, y): x – y is an integer} is reflexive, symmetric and transitive.
Solution:
Given that x, y ∈ Z and R = {(x, y): x – y is an integer)}
(i) (x, x) ∈R x ∈ Z. So, R is reflexive.
[∵ x – x = 0 and 0 is an integer.]

(ii) (x, y) ∈ R, =(y, x) ∈ R. So, R is symmetric.
[∵ if (x – y) is an integer then (y – x) is also an integer.]

(iii) (x, y), (y, z) ∈ R =(x, z) ∈ R where x, y, z ∈ Z. So, R is transitive.
[∵ if (x – y) and (y – z) are integers then (x – z) = (x – y) + (y – z) is also an integer]

Question 5.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b²) is neither reflexive nor symmetric nor transitive.
Solution:
Given that a, b ∈ R and R =(a, b): a ≤ b²}
(i) (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R. So, R is not reflexive. [∵ \(\frac{1}{2}\) > (\(\frac{1}{2}\))²]
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R. So, R is not symmetric.
[∵ (1, 2) ∈ R ⇒ 1< 2² = 4, (2, 1) ∈ R ⇒ 2 < 1² = 1, which is absurd.]

(iii) (3, 2), (2, 1.5) ∈ R ⇒ (3, 1.5) ∉ R So, R is not transitive.
[(3, 2) ∈ R⇒ 3 < 2² = 4 and also 2 < (1.5)² = 2.25 .But 3 < (1.5)² = 2.25, which is absurd.]
∴ R is not reflexive, not symmetric, not transitive.

Question 6.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Solution:
Given that A = {1, 2, 3, 4, 5, 6} and R = {(a, b):b = a + 1}
∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
(i) (1, 1), (2, 2) ∉ R . So, R is not reflexive.
(ii) (1, 2) ∈ R, but (2, 1) ∉ R. So, R is not symmetric.
(iii) (1, 2), (2, 3) ∈ R but (1, 3) ∉ R. So, R is not transitive.
∴ R is not reflexive, not symmetric, not transitive.

Question 7.
Show that the relation R in R defined as R = {(a, b): a ≤ b} is reflexive and transitive but not symmetric.
Solution:
Let a, b ∈ R and R = {(a, b): a ≤ b}
(i) (a, a)∈ R. So, R is reflexive [∵ a ≤ a]
(ii) (a, b) ∈ R . But (b, a) ∉ R. So, R is not symmetric.
[∵ (2, 4) ∈ R(as 2 < 4) and(4, 2) ∉ R(as 4 > 2)]
(iii) (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R . So, R is transitive, [∵ a ≤ b and b ≤ c ⇒ a ≤ c]
∴ R is reflexive and transitive but not symmetric.

Question 8.
Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.
Solution:
Given that a, b ∈ R and R = {(a, b):a ≤ b³}
(i) (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R. So, R is not reflexive. [∵ \(\frac{1}{2}\) > (\(\frac{1}{2}\))³]
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R. So, R is not symmetric.
[∵ (1, 2) ∈ R ⇒ 1 < 2³ = 8.(2, 1) ∈ R ⇒ 2 < 1³ = 1, which is absurd.]
(iii) Consider (3, 1.5), (1.5, 1.2), (3, 1.2) . Here R is not transitive [ (1.2)³ = 1.728, (1.5)³ = 3.75]
[∵ 3 ≤ (1.5)³ ⇒ (3, 1.5) ∈ R; 1.5 ≤ (1.2)³ ⇒ (1.5, 1.2) ∈ R. But(1.5)³ >3 ⇒ (3, 1.5) ∉ R]
∴ R is not reflexive, not symmetric and not transitive.

Question 9.
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Solution:
Given that A = {1, 2, 3}; R = {(1, 2), (2, 1)}
(i) (1, 1), (2, 2), (3, 3) ∉ R. So, R is not reflexive
(ii) (1, 2) ∈ R and (2, 1) ∈ R. So, R is symmetric.
(iii) (1, 2) ∈ R and (2, 1) ∈ R, (1, 1) ∉ R. So, R is not transitive.
∴ R is not reflexive, not transitive but R is symmetric.

Question 10.
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation. .
Solution:
Set A contain various library books and x, y related to the number of pages in any two books.
Given that R = {(x, y): x and y have same number of pages}.
(i) (x, x) ∈ R. So, R is reflexive
∵ x and x have same number of pages.]

(ii) (x, y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[ ∵ If x and y have same number of pages then y and x also have same number of pages.]

(iii) (x, y) ∈ R, (y, z) ∈ R ⇒ (z, x) ∈ R. So, R is transitive.
[∵ If x and y have same number of pages, and y and z have same number of pages, then x and z have same number of pages. ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to ally element of {2, 4}.
Solution:
GiyenthatA={1, 2, 3, 4, 5}givenbyR={(a, b):|a-b| is even},
(i) (a, a) ∈ R ∀ a ∈ A. So, R is reflexive. [∵ |a-a| = 0 (which is even)]
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R. So, R is symmetric. [∵ if |a-b| is even then |-(a-b)| =|b-a| is also even]
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. So, R is transitive.
[∵ if |a—b| , |b—c| are even then (a-b), (b-c) are also even .
Now (a-c) = (a+b) +(b-c) is even ⇒ |a-c| is even ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, all elements of {1, 3, 5} are related to each other because they are all odd.
So, the modulus of the difference between any two elements is even.
Similarly, all elements {2, 4} are related to each other because they are all even.
No element of {1, 3, 5}is related to any elements of {2, 4} as all elements of {1, 3, 5}are odd and all elements of {2, 4} are even. So, the modulus of the difference between the two elements will not be even.

Question 12.
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤, x ≤ 12}, given by R = {(a, b): |a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 .
Solution:
Given that A = {x ∈ Z: 0 ≤ x ≤ 12} ⇒ A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a, b):|a – b|is a mutiple of 4}
(i) (a, a) ∈ R ∀ a ∈ A. So, R is reflexive, [∵ |a – a| = 0 and 0 is a multiple of 4]

(ii) (a, b) ∈ R ⇒ (b, a) ∈ R . So, Ris symmetric.
[∵ |a – b| is a multiple of 4 ⇒ |-(a -b)| = |b – a| is also a multiple of 4]

(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R So, Ris transitive.
[∵If |a-b|, |b-c| are multiples of 4 ⇒ (a-b), (b-c) are multiples of 4.
Now (a-c) = (a-b) + (b-c) is a multiple of 4 ⇒ |a – c| is a multiple of 4 ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation. Further, the set of elements related to 1 is {1, 5, 9}
∵ |1 – 1| = 0 is a multiple of 4; |5 – 1| = 4 is a multiple of 4., |9 -1| = 8 is a multiple of 4

Question 13.
Show that each of the relation R in the set A = {x ∈ Z : 0≤ x ≤ 12}, given by R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1.
Solution:
Given that R = {(a, b):a = b}
(i) (a, a) ∈ R ∀ a ∈ A . So, R is reflexive, [∵ a = a ∀a]
(ii) (a, b) ∈ R ⇒ a = b => b = a ⇒ (b, a)e R . So, R is symmetric.
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. So, R is transitive. [∵ a = b and b = c ⇒ a = c]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the set of elements related to 1 are those elements of A which are equal to 1.
∴ the required set is {1}.

III.

Question 1.
Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as center.

Solution:
Given that A is the set of points in a plane and
R = {(P, Q): Distance of the point P from the origin = the distance of the point Q from the origin}
(i) Clearly (P, P) ∈ R. So, R is reflexive. [∵ OP = OP]
(ii) (P, Q) ∈ R ⇒ (Q, P) ∈ R . So, R is symmetric. [∵ PQ = QP]
(iii) (P, Q), (Q, S) ∈ R ⇒ (P, S) ∈ R . So, R is transitive
[∵ (P, Q), (Q, S) ∈ R ⇒ OP = OQ and OQ = OS then OP = OS ⇒ (P, S) ∈ R]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the set of points related to P ≠ (0, 0) will be those points whose distance from origin is same as distance of P from the origin.
Set of points forms a circle with the centre as brigin and this circle passes through P.

Question 2.
Show that the relation R defined in the set A of all triangles as
R = {(T₁, T₂): T₁ is similar to T₂}, is an equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

Solution:
Given that R = {(T₁, T₂):T₁ is similar to T₂}
(i) (T₁, T₁) ∈ R ∀ T₁ ∈ A . So, R is reflexive.
[∵ Every triangle is similar to itself. ]

(ii) If (T₁, T₂) ∈ R ⇒ (T₂, T₁) ∈ R. So, R is symmetric.
[∵ If T₁ is similar to T₂ then T₂ is similar to T₁]

(iii) (T₁, T₂), (T₂, T₃) ∈ R ⇒ (T₁, T₃) ∈ R. So, R is transitive.
[∵ If T₁ is similar to T₂ and T₂ is similar to T₃ then T₁ is similar to T₃]
The given 3 triangles with sides (3, 4, 5), (5, 12, 13) and (6, 8, 10) are right angled triangles.
But among T₁, T₂, T₃ the Corresponding sides of T₁ and T₃ are in the same ratio.
∵ \(\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\left(\frac{1}{2}\right)\)
So, triangle T₁ is similar to triangle T₃. Hence, T₁ is related to T₃.

Question 3.
Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?
Solution:
Given that A is the set of all polygons and R = {(P₁, P₂):P₁ and P₂ have same number of sides}
(i) (P₁, P₁) ∈ R. So, R is reflexive.
[∵ Same polygon has same number of sides.]

(ii) (P₁, P₂) ∈ R ⇒ (P₂, P₁) ∈ R. So, R is symmetric.
[∵ if P₁ and P₂ have same number of sides then P₂ and P₁) have same number of sides.]

(iii) (P₁, P₂), (P₂, P₃) ∈ R ⇒ (P₁, P₃) ∈ R. So, R is transitive.
[∵ (P₁, P₂), (P₂, P₃) ∈ R ⇒ P₁, P₂ and P₂ P₃ have same number of sides.
Then P₁, P₃ have same number of sides ⇒ (P₁, P₃) ∈ R. ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the elements in A related to right-angled triangle (T) with sides 3, 4, 5 are those polygons which have three sides.
Set of all elements in a related to triangle T is the set of all triangles.

Question 4.
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Solution:
Given that L is the set of all lines and R = {(L₁, L₂): L₁ is parallel to L₂}.
(i) (L₁, L₁) ∈ R. So, R is reflexive ‘
[∵ Any line L₁ is parallel to itself. ]

(ii) If (L₁, L₂) ∈ R ⇒ (L₂, L₁) ∈ R. So, R is symmetric.
[∵ If L₁ is parallel to L₂ then L₂ is parallel to L₁]

(iii) (L₁, L₂), (L₂, L₃) ∈ R ⇒ (L₁, L₃) ∈ R. So, R is transitive.
[∵ If L₁ is parallel to L₂ and L₂ is parallel to L₃ then L₁ is parallel to L₃.]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Also, set of all lines related to the line y = 2x + 4 is the set of all lines parallel to y = 2x + 4
Slope of the line y = 2x + 4 is m = 2.
Line parallel to the given line is in the form y = 2x + c, where c ∈ R .
Set of all lines related to the given line is given by y = 2x + c, where c ∈ R

Question 5.
Determine whether each of the following relations are reflexive, symmetric and transitive. Relation R in the set A of human beings in a town at a particular time given by a) R = {(x, y): x and y work at the same place)
b) R {(x, y): x and y live in the same locality)
Solution:
Given that A is the set of all human beings in a town
a) Given that R = {(x,y): x and y work at the same place)
(i) (x, x) ∈ R ∀ x ∈ A. So, R is reflexive.
[∵ Every person x, he himself works at the same place]

(ii) (x, y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[∵ If x and y work at the same place then y and x also work at the same place.]

(iii) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R. So, R is transitive.
[∵ If x and y work at the same place, and y and z also work at the same place, then x and z also work at the same place] .

b) Given that R={(x, y): x and y live in the same locality)
(i) (x, x) ∈ R. So, R is reflexive.[∵ Every person x, he himself lives at the same locality]
(ii) (x,y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[∵ If x and y live in the same locality then y and x also live in the same locality.]
(iii) If(x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R. So, R is transitive
∵ If x and y live in the same locality, and y and z also live in the same locality, then x and z also live in the same locality.]

Question 6.
Determine whether each of the following relations are reflexive, symmetric and transitive. Relation R in the set A of human beings in a town at a particular time given by
a) R = {(x, y): x is exactly 7 cm taller than y}
b) R = {(x, y): x is wife of y} .
c) R = {(x, y): x is father of y}
Solution:
a) Given that R = {(x, y): x is exactly 7 cm taller than y}
(i) (x, x) ∉ R. So, R is not reflexive
[∵ A person cannot be 7 cm taller than himself.]

(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric .
[∵ If x is exactly 7cm taller than y then y is clearly not taller than x.]

(iii) If(x, y), (y, z) ∈ R then (x, z) ∉ R. So, R is not transitive
[∵ If x is exactly 7 cm taller than y and y is.exactly 7cm taller than z, then x is not exactly 7 cm taller than z.]

b) Given that R = {(x, y): x is wife of y}
(i) (x, x) ∉ R. So, R is not reflexive
[∵ A women can never be a wife to herself]
(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric.
[∵ If x is the wife of y then y can never be the wife of x.]
(iii) If (x, y) ∈ R and (y, z)e R then (x, z) ∉ R. So, R is not transitive
[∵ If x is wife of y and y is wife of z (this itself is a contradiction, so we need not go for the next step)]

c) Given that R = {(x, y): x is father of y}
(i) (x, x) ∉ R So, R is not reflexive
[∵ A man can never be a father to himself.]
(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric .
[∵ if x is the father of y then y can never be a father of x]
(iii) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∉ R. So, R is not transitive
[∵ if x is father of y and y is father of z (this itself is a contradiction, so we need not go for the next step)]

Question 7.
If R₁ and R₂ are equivalence relations in a set A, show that R₁ ∩ R₂ is also an equivalence relation.
Solution:
Given that R₁ and R₂ are equivalence relations .
(i) ∀ a ∈ A, (a, a) ∈ R₁, and (a, a) ∈ R₂ ⇒ (a, a) ∈ R₁ ∩ R₂. So R₁ ∩ R₂ is reflexive.
(ii) (a, b) ∈ R₁ ∩ R₂ ⇒ (a, b) ∈ R₁ and (a, b) ∈ R ₂ ⇒ (b, a) ∈ R₁ and (b, a) ∈ R₂
⇒ (b, a) ∈ R₁ ∩ R₂. So; R₁ ∩ R₂ is symmetric.
(iii) (a, b) ∈ R₁ ∩ R₂ and (b, c) ∈ R₁ ∩ R₂ ⇒ (a, c) ∈ R₁ and (a, c) ∈ R₂ ⇒ (a, c) ∈ R₁ ∩ R₂.
So, R₁ ∩ R₂ is transitive.
Thus, R₁ ∩ R₂ is reflexive, symmetric, transitive. Hence it is an equivalence relation.

Question 8.
Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y)R(u, v) if and only if xv = yu. Show that R is an equivalence relation.
Solution:
(i) (x, y) R(x, y), [∵ xy = yx, ∀ (x, y) ∈ A]. So, R is reflexive.
(ii) (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx ⇒ (u, v) R (x, y). So, R is symmetric.
(iii) (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and ub = va; Now we show that xb = ya
xv = yu ⇒ y = \(\frac{x v}{u}\) and ub = va ⇒ a = \(\frac{u b}{v}\)
Now ya = \(\left(\frac{\mathrm{xv}}{\mathrm{u}}\right)\left(\frac{\mathrm{ub}}{\mathrm{v}}\right)\). Hence, we have xb = ya
⇒ (x, y) R (a, b). So, R is transitive.
Thus, R is reflexive, symmetric, transitive. Hence R is an equivalence relation.

Question 9.
Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R₁ be a relation in X given by
R₁ = {(x, y): x – y is divisible by 3} and R₂ be another relation on X given by R₂ = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R₁ = R₂.
Solution:
R₁ Contains (x, y) such that x – y should be divisible by 3
So, elements of X are grouped according to their remainders when divided by 3:
Remainder 1: {1, 4, 7}; Remainder2: {2, 5, 8}; Remainder 0: {3, 6, 9}
Thus, R₁ relates all ordered pairs where both elements come from the same group.
Step 1: Ordered pairs in R₁
Frpm {1, 4, 7}: (1, 1), (1, 4), (1, 7), .(4, 1), (4, 4), (4, 7), (7, 1), (7, 4), (7, 7)
From {2, 5, 8}: (2, 2), (2, 5), (2, 8), (5, 2), (5, 5), (5, 8), (8, 2), (8, 5), (8, 8)
From {3, 6, 9}: (3, 3), (3, 6), (3, 9), (6, 3), (6, 6), (6, 9), (9, 3), (9, 6), (9, 9)

Step 2: Ordered pairs in R₂
By definition, R₂ consists of all ordered pairs (x, y) such that: both x, y ∈ {1, 4, 7} or both x, y ∈ {2, 5, 8} or both x, y ∈ (3, 6, 9}
This gives exactly the same ordered pairs listed above for R₁. Hence R₁ = R₂

Question 10.
Let f : X → Y be a function. Define a relation R in X given by
R = {(a, b): f(a) = f(b)}. Examine whether R is an equivalence relation or not.
Solution:
(i) ∀a ∈ X, (a, a) ∈ R, since f(a) = f(a). So, R is reflexive.
(ii) (a, b) ∈ R ⇒ f(a) = f(b) ⇒ f(b) = f(a) ⇒ (b, a) ∈ R. So, R is symmetric.
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ f(a) = f(b) and f(b) = f(c)
⇒ f(a) = f(c) ⇒ (a, c) ∈ R, So, R is transitive.
Thus, R is reflexive, symmetric, transitive. Hence R is an equivalence relation.

Question 11.
Find the number of all one-one functions from set A = {1, 2, 3} to itself.
Solution:
Total number of one-one function from {1, 2, 3} to itself is
= Total number of permutations on{1, 2, 3} = ³P₃ = 3! = 3 × 2 × 1 = 6.

Question 12.
Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.
Solution:
The smallest relation R₁ containing (1, 2) and (2, 3) which is reflexive and transitive but not
symmetric is R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.
Now, if we add the pair (2, 1) to R₁ we get R₂, which is reflexive, transitive but not symmetric. Similarly, if we add the pair (3, 2) we get R₃ by adding (3, 2) to R₁ to get the desired relation. Thus, the total number of desired relations is three.(R₁, R₂, R₃)
[Note: We can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R₁ at a time, as by doing so, we need to add the remaining pair in order to maintain transitivity. Then the relation will become symmetric which is not required.]

Question 13.
Show that the number of equivalence relation in the set {1, 2, 3} containing (1, 2) and (2, 1) is two.
Solution:
The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is
R₁ = {(1, 1), (2, 2), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one pair (2, 3) to R₁, then for symmetry we must add (3, 2) and for transitivity we need to add (1, 3) and (3, 1). So R₂ = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3) , (3, 1), (3, 3)}.
Thus, the only equivalence relation bigger than R₁ is this universal relation.
This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two.

Question 14.
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B, Is R an equivalence relation on P(X)? Justify your answer.
Solution:
(i) ARA for all A e P(X) [ every set is a subset of itself]. So, R is reflexive.
(ii) Let ARB ⇒ A ⊂ B . This does not imply that B ⊂ A. So, R is not symmetric.
(iii) If ARB and BRC, then A ⊂ B and B ⊂ C ⇒ A ⊂ C ⇒ ARC . So, R is transitive.
R is not an equivalence relation as it is not symmetric.