Practicing ML Aggarwal Class 12 Solutions Chapter 1 Relations and Functions Ex 1.1 is the ultimate need for students who intend to score good marks in examinations.
ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.1
Question 1.
Determine whether each of the following relations are reflexive, symmetric and transitive :
(i) Relation R in the set A = {1, 2, 3, ………, 10} defined by R = {(x, y) : 2x – y = 0}.
(ii) Relation R in the set Z of all integers defined by
R = {(x, y): x – y is an integer} (NCERT)
Solution:
(i) Relation R defined on set A = {1, 2, 3, ………………, 10} by R = {(x, y) = 2x – y = 0}
1 ∈ A but 2 . 1 – 1 = 1 ≠ 0
∴ (1, 1) ∈ R
Thus R is not reflexive on set A.
for any 1, 2 ∈ A s.t 2 . 1 – 2 = 2 – 2 = 0
∴ (1, 2) ∈ R
but 2 . 2 – 1 = 4 – 1 = 3 ≠ 0
∴ (2, 1) ∉ 2 R
Thus R is not symmetric on A.
Since, (1, 2) ∈ R as 2 . 1 – 2 = 0
(2, 4) ∈ R as 2 . 2 – 4 = 0
But (1, 4) ∉ R [∵ 2 . 1 – 4 = 2 – 4 = – 2 ≠ 0]
Thus for any 1, 2, 4 ∈ A s.t. (1, 2), (2, 4) ∈ R but (1, 4) ∉ R
Thus R is not transitive on A.
Hence A is neither reflexive nor symmetric nor transisitve.
(ii) Relation R on set Z defined by R = {(x, y): x – y is an integer}
Reflexive :∀ x ∈ Z, x – x = 0 be an integer
∴. (x, x) ∈ R
∴ R is reflexive on Z.
Symmetric : ∀ x, y ∈ Z, (x, y) ∈ R
∴ x – y be an integer
i.e. x – y = n, where n ∈ Z
⇒ y – x = – n, where – n ∈ Z
∴ (y, x) ∈ R
Thus R is symmetric.
Transitive : ∀ x, y, t ∈ Z s.t (x, y) ∈ R and (y, t) ∈ R
Now (x, y) ∈ R
⇒ x – y be an integer
∴ x – y = n, where n ∈ Z
Also, (y, t) ∈ R
⇒ y – t be an integer
∴ y – t = m, where m ∈ Z
Now, x – y + y – t = n + m
⇒ x – t = n + m ∈ Z
[since n, m ∈ Z ⇒ n + m ∈ Z]
∴ x – t be an integer
∴ (x, t) ∈ R
∀ x, y t ∈ Z s.t (x, y) ∈ R, (y, t) ∈ R. Then (x, t) ∈ R
Thus, R is transitive on Z.
Hence R is reflexive, symmetric and transitive on Z.
(iii) Relation R on the set A = {1, 2, 3, 4, 5, 6} defined by R = {{a, b) : b = a + 1}.
Solution:
Given relation R on set A = {1, 2, 3, 4, 5, 6} defined by R {(a, b) ; b = a + 1}
when a = 1
∴ b = a + 1 = 1 + 1 = 2
∴ (1, 2) ∈ R when a = 2
∴ b = a + 1 = 2 + 1 = 3 ⇒ (2, 3) ∈ R when a = 3
∴ b = a + 1 = 3 + 1 = 4 ⇒ (3, 4) ∈ R when o = 4
∴ b = a + 1 = 4 + 1 = 5 ⇒ (4, 5) ∈ R when a = 5
∴ b = a + 1 = 5 + 1 = 6 ⇒ (5, 6) ∈ R
Thus, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Clearly 1 ∈ A but (1, 1) R
∴ R is not reflexive.
Now (1, 2) ∈ R but (2, 1) g R
∴ R is not symmetric
Now (1, 2), (2, 3) ∈ R but (1, 3) ∉ R
∴ R is not transitive.
Thus R is neither reflexive nor symmetric nor transitive.
Question 2.
If the relation R in the set A, where A = {1, 2, 3, 4, 5, 6}, is defined by R = {(x, y) : y is divisible by x :}, then express R in the roster form. Also determine whether the relation R is
(i) reflexive
(ii) symmetric
(iii) transitive.
Solution:
Given A = {1, 2, 3, 4, 5, 6},
R = {(x, y): y is divisible by x} on A
As x is divisible by x
∴ (x, x) ∈ R ∀ x ∈ A
Thus R is reflexive.
As (x, y) ∈ R
⇒ x is a factor of y
y is a factor of x
∴ (y, x) ∈ R.
Thus R is not symmetric.
∀ (x, y), (y, z) ∈ R where x, y, z ∈ A
⇒ (x, z) ∈ R
[∵ If y is divisible x and z is divisible by y Then z is divisible by x]
Thus R is transitive on A.
Question 3.
If R is the relation defined on the set of natural numbers N as follows :
R = {(x, y); x, y ∈ N, 2x + y = 41},
find the domain and the range of the relation R.
Determine whether the relation is reflexive, symmetric and transitive. (NCERT Exemplar)
Solution:
Given R be the relation defined on set of natural numbers N as follows:
R = {(x, y) : x, y ∈ N, 2x + y = 41}
Given 2x + y = 41
⇒ y = 41 – 2x
When x = 1
∴ y = 41 – 2 = 39 ⇒ (1, 39) ∈ R
When x = 2
∴ y = 41 – 4 = 37 ⇒ (2, 37) ∈ R
When x = 3
∴ y = 41 – 6 = 35 ⇒ (3, 35) ∈ R
and so on
When x = 19
∴ y = 41 – 38 = 3 ⇒ (19, 3) ∈ R
When x = 20
∴ y = 41 – 40 = 1 ⇒ (20, 1) ∈ R
When x = 21
∴ y = 41 – 42 = – 1 ∉ N
For other values of x, we do not get) ∈ N
∴ R = {(1, 39), (2, 37), ………….. (20, 1)}
Thus domain of R = { 1, 2, 3, ………………., 20}
Range of R = {39, 37, ………………., 1}
Clearly (1, 1) ∉ R as 2.1 + 1 = 3 ≠ 41
∴ R is not reflexive clearly (1, 39) ∈ R but (39, 1) ∉ R
[∵ 2 × 39 + 1 = 79 ≠ 41]
Thus R is not symmetric.
Also, (11, 19) ∈ R since 2.11 + 19 = 22 + 19 = 41
and (19, 3) ∈ R Since 2.19 + 3 = 38 + 3 = 41
but (11, 3) ∉ R [∵ 2 . 11 +3 = 25 ≠ 41]
∴ R is not transitive.
Hence R is neither reflexive, nor symmetric and nor transitive.
Question 4.
Determine whether each of the following relations in the set A of human beings in a city at a particular time are reflexive, symmetric and transitive :
(i) R = {(x, y) : x and y work at the same place). (NCERT)
(ii) R = {(x, y) : x and y live in the same locality). (NCERT)
(iii) R= {(x, y) : x is exactly 7 cm taller ttiany}.
(iv) R = ((x, y) : x and y live within 2 kilometres).
Solution:
(i) Clearly R is reflexive. Since x and x work at the same place. Now If x and y work at same place then y and x also work at same place .. R is symmetric also R is transitive.
If (x, y) ∈ R, (y, z) ∈ R
i.e. x andy work at same place also y and z work at same place.
Thus x and z work at same place
∴ (x, z) ∈ R
(ii) Clearly person x and x live in same locality
∴ (x, x) ∈ R ∀ x ∈ A.
Thus R is reflexive.
∀x, y ∈ A such that (x, y) ∈ R
i.e. x andy both persons live in same locality
∴ y and x also live in same locality.
Thus, (y, x) ∈ R
∴ R is symmetric.
∀x, y, z ∈ A s.t. (x y) ∈ R and (y, z) ∈ R
Now (x, y) ∈ R
⇒ both x and y persons live in same locality.
Now (y, z) ∈ R
⇒ y and z both persons live in same locality.
Thus clearly both persons x ad z live in same locality.
∴ (x, z) ∈ R.
Hence R is transitive.
(iii) Given A {all human beings in a city at a particular time)
R be a relation defined on A,
R = {(x, y) ; x ¡s exactly, 7 cm taller than y)
Since x ∈ a ∴ be exactly 7 cm taller than x itself.
∴ (x, x) ∉ R
∴ R is not reflexive.
For any x, y ∈ A s.t (x, y) ∈ R
i.e. x is exactly 7 cm taller than y.
∴ y is not exactly 7 cm taller than x
Since y is exactly 7 cm smaller than x.
∴ (y, x) ∉ R.
Thus R is not symmetric.
For any x, y, z ∈ A s.t (x, y) ∈ R and (y, z) ∈ R
Now (x, y) ∈ R x is exactly 7 cm taller than y
and (y, z) ∈ R y is exactly 7 cm taller than z
∴ x is exactly 14 cm taller than z.
∴ (x, z) ∉ R
Thus R is not transitive.
Hence R is neither reflexive nor symmetric and nor transitive.
(iv) Given R = {(x, y) : x and y live within 2 kilometer}
Reflexive : for any x ∈ A, x and x live within 2 kilometers
∴ (x, x) ∈ R
⇒ R is reflexive on A.
Symmetric :
∀ x, y ∈ A s.t (x, y) ∈ R
⇒ x and y live within 2 kilometers
∴ y and x live within 2 kilometers
⇒ (y, x) ∈ R
⇒ R is symmetric.
Transitive :
∀ x, y, z ∈ A s.t (x, y) ∈ R and (y z) ∈ R
Now (x, y) ∈ R
⇒ x and y live within 2 kilometers
and (y, z) ∈ R
⇒ y and z live within 2 kilometers
Then x and z may not be always live within 2 kilometers.
∴ (x, z) ∉ R
⇒ R is not transitive.
Hence R is reflexive, symmetric but not transitive.
Question 5.
Determine whether each of the following relations in the set A of students at a particular time are symmetric and transitive :
(i) R = {(x, y) : x, y ∈ A, x and y are honest}
(ii) R = {(x, y) : x, y ∈ A, x and y are obedient}
(iii) R = {(x, y) : x, y ∈ A, x and y are hardworking}
What are the advantages of students being honest, obedient and hardworking? (Value Based)
Solution:
(i) Given A be the set of all students at a particular time.
R = {(x, y) : x, y ∈ A, x and y are honest}
Symmetric : ∀ x, y ∈ A s.t. (x, y) ∈ R
⇒ x and y are honest.
⇒ y and x are honest
⇒ (y, x) ∈ R
⇒ R is symmetric.
Transitive : ∀ x, y, z ∈ A s.t (x, y) ∈ R,
(y, z) ∈ R
Now (x, y) ∈ R
⇒ x and y are honest students
and (y, z) ∈ R
⇒y and z are honest students.
∴ x and z are honest students.
⇒ (x, z) ∈ R
⇒ R is transitive
Thus R is symmetric and transitive.
(ii) Given R = {(x, y) ; x, y ∈ A, x and y are obedient}
Reflexive : ∀ x ∈ A, x and x itself is obedient student.
⇒ (x, x) ∈ R
Symmetric : ∀ x, y ∈ A s.t (x, y) ∈ R
⇒ x andy are obedient students
⇒ y and x are obedient students.
∴ (y, x) ∈ R
⇒ R is symmetric
Transitive : ∀ x, y, z ∈ A s.t (x, y), (y z) ∈ R
Now (x, y) ∈ R
⇒ x and y are obedient students.
(y, z) ∈ R
⇒ y and z are obedient students
∴ x and z are obeident students
⇒ (x, z) ∈ R
Thus R is transitive on A.
Hence R is symmetric and transitive on A.
(iii) Given R = {(x, y) : x, y ∈ A, x and y are hardworking}
Symmetric : ∀ x, y ∈ A s.t (x, y) ∈ R
⇒ x and y are hardworking
⇒ y and x are hardworking.
∴ (y, x) ∈ R
⇒ R is symmetric.
Transitive : ∀ x, y, z ∈ A s.t (x, y), (y z) ∈ R
Now (x, y) ∈ R
⇒ x and y are hardworking
and (y, z) ∈ R
⇒ y and z are hardworking students
∴ x and z are hardworking.
⇒ (x, z) ∈ R ⇒ R is transitive.
Hence R is symmetric and transitive.
If the students are honest, obedient and hardworking then they will definitely rise in their life and will becomes ideal citizen of the country. Hence the values promoted are honesty, obedience and hard working.
Question 6.
Show that the relation R in the set
A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive. (NCERT)
Solution:
Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} on A.
Clearly R is reflexive,
As (1, 1), (2, 2), (3, 3) ∈ R
Now, (1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric.
As (1, 2), (2, 3) ∈ R but (1,3) ∉ R
Thus R is not transitive.
Thus R is reflexive but neither symmetric nor transitive.
Question 7.
Let R be the relation on W (set of whole numbers) defined by
R = {(x, y) : x, y ∈ W, 3x + 2y = 12}.
Check the relation R for reflexivity, symmetricity and transitivity.
Solution:
Given R be the relation on W (set of whole numbers) defined by
R = {(x, y) : x, y ∈ W, 3x + 2y = 12}
Since 2y = 12 – 3x
⇒ y = \(\frac{12-3 x}{2}\)
When x = 0 ;
y = \(\frac{12-0}{2}\) = 6 ∈ W
∴ (0, 6) ∈ R
When x = 1 ;
y = \(\frac{12-3}{2}\) = \(\frac{9}{2}\) ∉ W
When x = 2 ;
y = \(\frac{12-6}{2}\) = 3 ∈ W
∴ (2, 3) ∈ R
When x = 3 ;
y = \(\frac{12-9}{2}\) = \(\frac{3}{2}\) ∉ W
Whenx = 4 ;
y = \(\frac{12-12}{2}\) = 0 ∈ W
∴ (4, 0) ∈ R
For other values of x we donot get values of y that belongs to set of whole numbers.
∴ R = {(0, 6), (2, 3), (4, 0)}
Clearly (1, 1) ∉ R ;
∴ R is not reflexive on W.
Further (0, 6) ∈ R but (6, 0) ∉ R
∴ R is not symmetric on W.
Clearly (4, 0), (0, 6) ∈ R but (4, 6) ∉ R
∴ R is not transitive on W.
Hence R is neither reflexive, symmetric nor transitive.
Question 8.
Let R be the relation defined on Q0 (set of non-zero rational numbers) defined by R = {(a, b) : a, b ∈ Q0, b = \(\frac{1}{a}\)}
Check the relation R for reflexivity, symmetricity and transitivity.
Solution:
Let R be the relation defined on Q0 (set of non-zero rational numbers) defined by
R = {{a, b); a, b ∈ Q0, b = \(\frac{1}{a}\)}
Clearly (1, 1) ∈ R as 1 = \(\frac{1}{1}\)
but (2, 2) ∉ R as 2 ∉ \(\frac{1}{2}\)
∴ R is not reflexive on Q0.
i.e. ∀ a ∈ Q0 s.t. (a, a) ∈ R ⇒ a = \(\frac{1}{a}\)
⇒ a2 = 1
⇒ a = ± 1
for any other ordered pair reflexivity does not holds.
Symmetry :
∀ a, b ∈ Q0 s.t (a, b) ∈ R ,
⇒ b = \(\frac{1}{a}\)
⇒ a = \(\frac{1}{b}\)
⇒ (b, a) ∈ R
∴ R is symmetric on Q0
Transitivity :
∀ a, b, c ∈ Q0 s.t (a, b) ∈ R, (b, c) ∈ R
since (a, b) ∈ R ⇒ b = \(\frac{1}{a}\)
and (b, c) ∈ R ⇒ c = \(\frac{1}{b}\)
∴ c = \(\frac{1}{1/a}\) = a
i.e. c ≠ \(\frac{1}{a}\)
⇒ (a, c) ∉ R
i.e. R is not transitive on Q0.
[Clearly (2, \(\frac{1}{2}\)), (\(\frac{1}{2}\), 2) ∈ R but (2 ,2) ∉ R
Thus R is neither reflexive nor transitive but symmetric.
Question 9.
Let R be the relation on R (set of real numbers) defined by R = {(a, b): a, b ∈ R, a2 + b2 = 1}. Show that R is symmetric but neither reflexive nor transitive.
Solution:
Given R be the relation defined by R = {(a, b) : a, b ∈ R, a2 + b2 = 1}
Reflexivity : (a, a) ∈ R ∀ a ∈ R
⇒ a2 + a2
⇒ a2 = \(\frac{1}{2}\)
a = ± \(\frac{1}{\sqrt{2}}\)
also 1 ∈ R but 12 + 12 ≠ 1
∴ (1, 1) ∈ R
∴ R is not reflexive.
Symmetry : ∀ a, b ∈ R s.t (a, b) ∈ R
⇒ a2 + b2 = 1
⇒ a2 + a2 = 1
[commutative law holds under addition in R]
⇒ (b, a) ∈ R
∴ R is symmetric on real number system.
Transistivity : ∀ a, b, c ∈ R s.t (a, b), (b, c) ∈ R.
Then (a, c) ∈ R.
Now (1, 0), (0, 1) ∈ R but (1, 1) ∉ R
since 12 + 12 = 2 ≠ 1
Thus R is not transitive on real number system.
Hence R is symmetric but neither reflexive nor transitive.
Question 10.
Determine whether each of the following relations are reflexive, symmetric and transitive :
(i) R = {(a, b) : a, b ∈ N, a < b}
(ii) R = {(a, b) : a, b ∈ R, a < b} (NCERT)
Solution:
(i) Given relation R = {(a, b) ; a, b ∈ N, a < b)
Since 1 ∈ N and 1 ≮ 1
∴ (1, 1) ∉ R
⇒ R is not reflexive.
Since 1 < 2
∴ (1, 2) ∈ R
but 2 ≯ 1
∴ (2, 1) ∉ R
⇒ R is not symmetric.
∀ a, b, c ∈ R s.t {a, b), (b, c) ∈ R
Now, (a, b) ∈ R ⇒ a < b and
(b, c) ∈ R ⇒ b < c
∴ a < c ⇒ (a, c) ∈ R
Thus R is transitive.
Hence R is neither reflexive nor symmetric but transitive.
(ii) Given R = {(a, b) : a < b, a, b are real numbers}
As, a ≤ a (always) ∀ a ∈ R
⇒ (a, a) ∈ R
Thus R is reflexive.
If (a, b) ∈ R ⇒ a ≤ b ……………….(1)
and (b, c) ∈ R ⇒ b ≤ c ………………(2)
From (1) and (2) ; we have
a ≤ b ≤ c ⇒ a ≤ c ⇒ (a, c) ∈ R,
Thus R is transitive.
If (a, b) ∈ R ⇒ a ≤ b ⇏ b ≤ a
∴ (b, a) ∉ R
Thus R is not symmetric.
Thus R is reflexive, transitive but not symmetric.
Question 11.
Let T be the set of all triangles drawn in a plane with R as a relation in T given by R = {(T1, T2) : T1 ≅ T2}. Show that R is an equivalence relation.
Solution:
Let R be a relation on T defined by aRb
if a is congruent to b ∀ a, b ∈ T
where T be the set of all triangles in the Euclidean plane.
∀ a ∈ T, we know that every triangle is congruent to itself.
∴ a is congruent to a.
∴ aRa
∴ R is reflexive on T.
Let (a, b) ∈ R ∀ a, b ∈ T
∴ a is congruent to b.
Thus triangle b is congruent to triangle a.
(b, a) ∈ R i.e. bRa
Hence R is symmetric on T.
Let (a, b), (b, c) ∈ R ∀ a, b, c ∈ T
since (a, b) ∈ R ⇒ Triangle a is congruent to triangle b.
and (b, c) ∈ R ⇒ triangle b is congruent to triangle c.
∴ triangle a is congruent to triangle c,
∴ R is transitive on T.
Hence R is reflexive, symmetric and transitive on T.
Therefore R is equivalence relation on set T.
Question 12.
Show that the relation R in the set A of all books in a library of a school, given by R = {(x, y) : x and y have same number of pages}, is an equivalence relation. (NCERT)
Solution:
Set A = Set of all books in the library of a college
R = {(x, y) : x and y have the same number of pages}
Now,
R is reflexive since (x, x) ? R as x and x has the same number of pages.
Let (x, y) ∈ R
x and y have the same number of pages. y and x have the same number of pages.
⇒ (y, x) ∈ R
Hence, R is symmetric.
Let (x, y) ∈ R and (y, z) ∈ R
x and y and have the same number of pages and y and z have the same number of pages.
Therefore, x and z have the same number of pages ⇒ (x, z) e R
Hence, R is transitive.
Thus, R is an equivalence relation.
Question 13.
Show that the relation R on the set I of all integers defined by R = {{a, b) : a + b is divisible by 2, a, b ∈ 1} is an equivalence relation. Also, write the equivalence class of 0.
Solution:
Since a + a = 2a, which is divisible by 2
⇒ {a, a) ∈ R
⇒ R is reflexive.
Symmetry : (a, b) ∈ R
⇒ a + b is divisible by 2
⇒ b + a is divisible by 2
[∵ a + b = b + a ∀ a, b ∈ I]
⇒ (b, a) ∈ R
∴ R is symmetric on I.
Transitivity : (a, b), (b, c) ∈ R
Now (a, b) ∈ R
⇒ a + b is divisible by 2
⇒ a + b = 2k
(b, c) ∈ R
⇒ b + c is divisible by 2
⇒ b + c = 2k
⇒ a + c + 2b = 2k + 2K
⇒ a + c = 2(k + k’ – b) where k, k’, b ∈ I
⇒ (a, c) ∈ R
Thus, R is transitive on I.
∴ R is reflexive, symmetric and transitive on I.
⇒ R is equivalence relation on I.
Consider a ∈ I s.t (a, 0) ∈ R
⇒ a + 0 is multiple of 2
⇒ a is multiple of 2, where a ∈ Z
∴ a = ………….. – 4, – 2, 0, 2, 4, ……………
Thus equivalence class of 0 = {0} = {……………. – 4,- 2, 0, 2, 4, ……………..}
Question 13 (old).
Show that the relation R on the set I of all integers defined by R = {(a, b) : a – b is divisible by 3, a, b ∈ I} is an equivalence relation.
Solution:
Given relation R = {(a, b) : a – b is divisible by 3; a, b ∈ Z}
Reflexivity : ∀ a ∈ Z, we have a – a = 0 = 0 × 3
Thus, a – a is divisible by 3
∴ (a, a) ∈ R
Thus R is reflexive.
Symmetry : ∀ a, b ∈ Z such that (a, b) ∈ R Since, (a, b) ∈ R
⇒ a – b is divisible by 3
∴ 3/a – b
⇒ a – b = 3k where k ∈ Z
b – a = – 3k – 3(- k), where – k ∈ Z
⇒ 3/b – a
⇒ b – a is divisible by 3.
∴ (b, a) ∈ R
Thus, R is symmetric.
Transitivity : Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ Z
Since, (a, b) ∈ R
⇒ a – b is divisible by 3
⇒ 3/a – b
⇒ a – b = 3k1 where k1 ∈ Z
Also, (b, c) ∈ R
⇒ b – c is divisible by 3
⇒ 3/b – c
⇒ b – c = 3k2, where k2 ∈ Z
Thus, a – b + b – c = 3(k1 + k2)
⇒ a – c = 3k, where k = k1 + k2 ∈ Z
[Since sum of two integers will be an integer]
∴ 3/a – c
⇒ a – c is divisible by 3.
⇒ (a, c) ∈ R
Thus, R is transitive.
Hence, R is reflexive, symmetric and transitive on Z. Therefore R be an equivalence relation on Z.
Question 14.
Prove that the relation R in the set A = {5, 6, 7, 8, 9}, given by R = {{a, b) : |a – b | is divisible by 2}, is an equivalence relation. Find all elements related to element 6.
Solution:
Given set A = {5, 6, 7, 8, 9}
and R = {(a, b) : | a – b | is divisible by 2}
Reflexive : ∀ a ∈ A, since | a – a | = 0 = 2 – 0
⇒ | a – a | is divisible by 2 ⇒ (a, a) ⇒ R
∴ R is reflexive.
Symmetric : ∀ a, b ∈ A s.t (a, b) ∈ R
⇒ | a – b | is divisible by 2
⇒ a – b is divisible by 2
⇒ a – b = 2m, where m ∈ Z
⇒ b – a = – 2m = 2(- m), where – m ∈ Z
∴ b – a is divisible by 2
⇒ | b – a | is divisible by 2
⇒ (b, a) ∈ R
⇒ R is symmetric
Transitive : ∀ a, b, c ∈ A s.t (a, b) (b, c) ∈ R
Now (a, b) ∈ R
⇒ | a – b | is divisible by 2
⇒ a – b is divisible by 2
⇒ a – b = 2 m1 ………………(1) and (b, c) ∈ R
⇒ | b – c | is divisible by 2
⇒ b – c is divisible by 2
⇒ b – c = 2 m2 ………………(2)
where m1, m2 ∈ Z
On adding (1) and (2); we have
a – b + b – c = 2 (m1 + m2)
⇒ a – c = 2 m, where m ∈ Z
⇒ a – c is divisible by 2
⇒ | a – c | is divisible by 2.
⇒ (a, c) ∈ R
⇒ R is transitive.
Thus R is reflexive, symmetric and transitive.
Hence R be an equivalence relation on A.
To find elements related to 6 :
Consider any a ∈ A s.t (a, 6) ∈ R
⇒ | a – 6 | is divisible by 2
⇒ a- 6 is divisible by 2
⇒ a – 6 = 2m, m ∈ I
⇒ a = 2m + 6, m ∈ I
putting m = 0, 1, 2, 3, ……………., – 1, – 2, – 3, ……………..
a = 6, 8, 10, 12, ………….., 4, 2, 0, …………… but a ∈ A = {5, 6, 7, 8, 9}
⇒ a = 6, 8
∴ The set of elements related to 6 are {6, 8}.
Question 15.
Give examples of relations which are
(i) symmetric but neither reflexive nor transitive.
(ii) transitive but neither reflexive nor symmetric.
(iii) symmetric and transitive but not reflexive.
(iv) symmetric and reflexive but not transitive.
(v) reflexive and transitive but not symmetric. (NCERT)
Solution:
Let A= {2, 3, 4}
(i) Let R = {(2, 3), (3, 2), (2, 4), (4. 2), (3, 4), (4, 3), (2, 2), (3, 3)} on A
Clearly R is symmetric, here (4, 4) ∈ R thus R is not reflexive.
Also (4, 2), (2, 4) ∈ R but (4, 4) ∉ R
∴ R is not transitive.
Thus R is symmetric but neither reflexive nor transitive.
(ii) Let R = {(2, 3), (2, 4), (4, 2). (2, 2), (4,4)} on A
Clearly R is transitive.
Here (3, 3) ∉ R thus R is not reflexive.
Also (2, 3) ∈ R but (3, 2) ∈ R
∴ R is not symmetric but R is transitive on A.
(iii) Let R = {(2, 3), (3, 2), (2, 2), (3, 3)} on A
Clearly R is transitive and symmetric
As (4, 4) ∉ A
∴ A is not reflexive.
(iv) Let R ={(2, 2), (3, 3), (4, 4), (2, 3), (3, 2), (2, 4), (4, 2)} on A
Clearly R is reflexive and symmetric
As (4, 2), (2, 3) ∈ R but (4, 3) ∉ R
∴ R is not transitive.
(v) Let R = {(2, 2), (3, 3), (4, 4), (2, 3)} on A
Clearly R is reflexive and transitive
Here, (2, 3) ∈ R but (3, 2) ∉ R
∴ R is not symmetric.
Question 16.
Give an example of a relation R on A = {(a, b, c) which is
(i) neither reflexive nor symmetric but transitive.
(ii) neither symmetric nor transitive but reflexive.
(iii) neither transitive nor reflexive but symmetric.
Solution:
Given set A = {a, b, c}
(i) Let R = {(a, a), (a, b)}
Clearly (b, b), (c, c) ∉ R
⇒ R is not reflexive
(a, b) ∈ R but (b, a) ∉ R
⇒ R is not symmetric.
∀ a, b ∈ A s.t (a, a), (a, b) ∈ R
⇒ (a, b) ∈ R ⇒ R is transitive
Hence R is neither reflexive nor symmetric but transitive.
(ii) Let R = {(a, a), (b, b), (c, c), (a, b), (b, c)}
∀ a, b, c ∈ A.
(a, a), (b, b), (c, c) ∈ R
⇒ R is reflexive
∀ a, b ∈ A s.t (a, b) ∈ R but (b, a) ∉ R
⇒ R is not symmetric.
∀ a, b, c ∈ A s.t (a, b), (b, c) ∈ R but (a, c) ∉ R
∴R is not transitive.
Thus R is reflexive but neither symmetric nor transitive.
(iii) Let R = {(a, b), (b, a)}
∀ a ∈ A we have (a, a) ∉ R
⇒ R is not reflexive.
∀ a, b ∈ A s.t (a, b) ∈ R
⇒ (b, a) ∈ R
∴ R is symmetric.
∀ a, b ∈ A s.t (a, b), (b, a) ∈ R but (a, a) ∉ R
∴ R is not transitive.
Hence, R is neither transitive nor reflexive but symmetric.
Question 17.
Show that the relation S in the set A = {x : x ∈ Z : 0 ≤ x ≤ 12} given by S = {(a, b) : a, b e A, | a – b | is divisible by 4} is an equivalence relation.
Find the set of elements related to 1.
Solution:
Given A = {x : x ∈ Z, 0 < x < 12}
and R = {(a, b) ; | a – b | is divisible by 4}
Clearly |a – a| i.e. 0 is divisible by 4
∴ (a, a) ∈ R ⇒ R is reflexive.
Let (a, b) ∈ R
⇒ | a – b | is divisible by 4
⇒ a – b is divisible by 4
⇒ a – b = 4 m, where m ∈ Z
⇒ b – a = 4 (- m), where – m ∈ Z
b – a is divisible by 4
⇒ | b – a | is also divisible by 4.
⇒ R is symmetric.
Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a – b | and | b – c | are divisible by 4
a – b = 4m1 and b – c = 4m2,
where m1, m2 ∈ Z
∴ a – b + b – c = 4 (m1 + m2)
⇒ a – c = 4 m, where m = m1 + m2 ∈ Z
⇒ | a – c | is divisible by 4
⇒ (a, c) ∈ R
∴ R is transitive.
Thus R is reflexive, symmetric and transitive.
Hence R is an equivalence relation.
Now | 1 – 5 | = | – 4| = 4 is divisible by 4
∴ (1, 5) ∈ R
Thus 1 related with 5.
Also | 1 – 9| = | – 8| = 8is divisible by 4
∈ (1, 9) ∈ R
Thus 1 related with 9.
Similarly | 1 – 1 | = 0 is divisible by 4
Thus elements related to 1 are {1, 5, 9}.
Aliter : The find elements related to 1 consider any a ∈ A s.t (a, 1) ∈ R
| a – 1 | is divisible by 4
a – 1 is divisible by 4
⇒ a – 1 = 4m, m ∈ 1
=> a = 4m + 1, m ∈ I
Putting m = 0, 1, 2, 3, ………………. – 1, – 2, – 3, …………..
∴ a= 1, 5, 9, 13, – 3, – 7,- 11, ….
but a e A = {0, 1, 2, 3, ………….., 12}
∴ a = 1, 5, 9
Thus, the set of elements related to 1 are {1, 5, 9}.
Question 18.
Show that the relation R in the set A = (x ∈ W, 0 ≤ x ≤ 17} given by
(i) R = {(a, b) : | a – b | is a multiple of 5}
(ii) R = {(a, b) : a = b}
are equivalence relations. Find the set of all elements related to 2 in each case.
Solution:
(i) Given A = {x ∈ W, 0 < x < 17}
and R = {(a, b): |a – b| is a multiple of 5}
Reflexive : ∀ a ∈ A, since | a – a | = 0 = 0 – 5
⇒ | a – a | is multiple of 5.
⇒ (a, a) ∈ R
⇒ R is reflexive.
Symmetric :
∀ a, b ∈ A s.t (a, b) ∈ R
⇒ | a – b | is multiple of 5
⇒ a – b is multiple of 5
⇒ (b – a) is multiple of 5
⇒ | b – a | is a multiple of 5
⇒ (b, a) ∈ R
∴ R is symmetric.
Transitive :
∀ a, b, c ∈ A s.t (a, b) (b, c) ∈ R (a, b) ∈ R
⇒ | a – b | is multiple of 5
⇒ a – b = 5m1, where m1 ∈ I ………………(1)
and (b, c) ∈ R
⇒ | b – c | is a multiple of 5.
⇒ b – c is multiple of 5
⇒ b – c = 5m2, where m2 ∈ I ……………..(2)
On adding (1) and (2); we have
a – b + b – c = 5 (m1 + m2) = 5 m
⇒ a – c = 5 m, where m = m1 + m2 ∈ I
⇒ a – c is a multiple of 5
⇒ | a – c | is a multiple of 5.
(a, c) ∈ R
∴ R is transitive on A.
Hence R is reflexive, symmetric and transitive.
∴ R is an equivalence relation on A.
To find elements related to 2 :
Consider any a e A s.t (a, 2) ∈ R
⇒| a – 2 | is multiple of 5.
⇒ a – 2 is a multiple of 5.
⇒ a – 2 = 5m, where m ∈ I
⇒ a = 5m + 2, m ∈ I
putting m = 0, 1, 2, 3, …………….., – 1, – 2, ……………., we have
a = 2, 7, 12, 17, ………………, – 3, – 8, …………. but a ∈ A = {x ∈ W ; 0 ≤ x ≤ 17}
∴ a = 2, 7, 12, 17
Thus, the set of all elements related to 2 = {2, 7, 12, 17}
(ii) Given relation R = {(a, b) : a = b)
Reflexive : ∀ a ∈ A, a = a (always)
∴ (a, a) ∈ R
Thus R is reflexive.
Symmetric :
∀ a, b ∈ A s.t (a, b) ∈ R
⇒ a = b
⇒ b = a
⇒ (b, a) ∈ R
∴ R is symmetric.
Transitive :
∀ a, b, c ∈ A s.t (a, b), (b, c) ∈ R
Now {a, b) ∈ R
⇒ a = b and (b, c) ∈ R
⇒ b = c
∴ a = b = c
⇒ a = c
⇒ (a, c) ∈ R
∴ R is transitive.
Thus, R is reflexive, symmetric and transitive on A.
Hence R be an equivalence relation on A.
To find all elements related to 2 :
Consider any a ∈ A s.t (a, 2) ∈ R
⇒ a = 2 ∈ A
∴ set of all elements related to 2 be {2}.
Question 19.
Show that the relation R defined by (a, b) R (c, d) ⇒ a + d = b + c on A x A where A = {1, 2, 3, 10} is an equivalence relation. Hence, write the equivalence class [(3, 4)].
Solution:
Given R be the relation defined by (a, b) R (c, d)
⇒ a + d = b + c on A × A where A = {1, 2, 3, ……………, 10}
Reflexive :
∀ a, b ∈ A
Now (a, b) R (a, b)
⇒ a + b = b + a
[since commutative law holds under addition on A]
∴ R is reflexive.
Symmetry :
∀ a, b, c, d ∈ A s.t. (a, b) R (c, d)
⇒ a + d= b + c
⇒ d + a = c + b
⇒ (c + b) = d + a
⇒ (c, d) R (a, b)
∴ R is symmetric on A.
Transistivity :
∀ a, b, c, d, e, f ∈ A s.t {a, b) R (c, d) and (c, d) R (e, f)
Now (a, b) R (c, d)
⇒ a + d = b + c …………..(1)
(c, d) R (e, f) ⇒ c + f – d + e ……………..(2)
=> a + d + c + f = b + c + d + e
[On adding (1) and (2)]
⇒ a + f = b + e
⇒ (a, b) R (e, f)
∴ R is transitive on A.
Thus R is reflexive, symmetric and transitive on A.
Hence R be an equivalence relation on A.
To find equivalence class [(3, 4)] i.e. class of element (3, 4).
Let (a, b) R (3, 4)
⇒ a + 4 = b + 3
⇒ b = a + 1
but a, b ∈ A = {1, 2, ………………….., 10}
For taking a = 1, 2, 3, …. 9 we get the corresponding values of b = 2, 3, 4, …, 10
∴ equivalence class [(3, 4)] = {(4. 2). (2, 3), >) (3, 4), …, (9, 10)}
Question 20.
Show that the relation R in the set N × N defined by (a, b) R (c, d) iff a2 + d2 = b2 + c2 for all a, b, c, d ∈ N is an equivalence relation.
Solution:
Given relation R on set N × N defined by (a, b) R (c, d)
⇔ a2 + d2 = b2 + a2
Reflexive :
Since a2 + b2 = b2 + a2 ∀ a, b ∈ N because commutative law holds under addition on N.
⇒ (a, b) R (a, b)
∴ R is reflexive on N × N
Symmetric :
Now (a, b) R (c, d) ∀ a,b, c, d ∈ N
⇒ a2 + d2 = b2 + c2
⇒ c2 + b2 = d2 + a2
⇒ (c, d) R (a, b)
R is reflexive on N × N.
Transitivity :
(a, b) R (c, d) ∀ a, b, c, d ∈ N
⇒ a2 + d2 = b2 + c2 ………………(1)
and (c, d) R (e, f) ∀ c, d, e, f ∈ N
⇒ c2 + f2 = d2 + e2 ……………….(2)
On adding (1) and (2);
⇒ a2 + d2 + c2 + f2 = b2 + c2 + d2 + e2
⇒ a2 + f2 = b2 + e2
[∵ d2 + c2 = c2 + c2]
⇒ (a, b) R (e, f)
Thus, R is transitive on N × N.
R is reflexive, symmetric and Transitive on N × N.
R forms on equivalence relation on N × N.
Question 20 (old).
If R1 and R2 are equivalence relations on a set A, show that R1 ∪ R2 is reflexive symmetric but not transitive.
Solution:
Reflexivity : Since R1 and R2 are equivalence relations on set A
∴ (a, a) ∈ R1, (a, a) ∈ R2 V a ∈ A
∴ (a, a) ∈ R1 ∪ R2
Thus, R1 ∪ R2 is reflexive on A.
Symmetry :
∀ a, b ∈ A s.t (a, b) e R1 ∪ R2
∴ (a, b) ∈ R1 or (a, b) ∈ R2
Since R1 and R2 both are equivalence relation on set A.
Thus, R1, R2 are symmetric on A
∴ (b, a) ∈ R1 or (b, a) ∈ R2
⇒ (b, a) ∈ R1 ∪ R2,
Thus R1 ∪ R2 is symmetric on A.
Transitive :
Let R1 = ((1,2), (2, 1),(1, 1), (2, 2)} and
R2 = {(1, 3), (3, 1), (1, 1), (3, 3)}
both R1 and R2 are equivalence relation on A = {1, 2, 3}
Here. R1 ∪ R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}
i.e. (2, 1), (1, 3) ∈ R1 ∪ R2 but (2, 3) ∉ R1 ∪ R2
∴ R1 ∪ R2 is not transitive on A.
Thus R1 ∪ R2 is symmetric, reflexive but not transitive on A.
Question 21.
If A = {- 1, 1, 3}, then what is the number of relations on A ?
Solution:
Given A = {- 1, 1, 3}
∴ O (A) = 3
∴ Total number of relations on A
= 2O (A × A)
= 2O(A) × 0 (A)
= 23 × 3
= 29
= 512
Question 22.
(i) If R = {(x, y) : x + 2y = 8} is a relation on N, write the range of R.
(ii) Let R = {(a, a3) : a is a prime number less than 5}. Find the range of R.
Solution:
(i) Given R = {(x, y) : x + 2y = 8}
Since, x + 2y = 8, where x, y ∈ N 8 – x
∴ 2y = 8 – x
⇒ y = \(\frac{8-x}{2}\)
When x = 1
⇒ y = \(\frac{8-1}{2}\) = \(\frac{7}{2}\) ∉ N
When x = 2
⇒ y = 3
∴ (2, 3) ∈ R
When x = 4
⇒ y = \(\frac{4}{2}\) = 2
∴ (4, 2) ∈ R
When x = 6
=> y = \(\frac{2}{2}\) = 1
∴ (6, 1) ∈ R
When x = 8
=> y = 0 ∉ N
∴ R = {(2, 3), (4, 2), (6, 1)}
Thus, range of R = {1, 2, 3}.
(ii) Given R = {(a, a3) : a is a prime number less than 5]
since 2 and 3 be the prime numbers less than 5
∴ (2, 8) ∈ R and (3, 27) ∈ R
R = {(2, 8), (3, 27)}
Thus range of R = {8, 27}
Question 23.
Let A be any non-empty set. State true or false :
(i) Identity relation on A is reflexive.
(ii) Every reflexive relation on A is identity relation on A.
(iii) Identity relation on A is symmetric.
(iv) Identity relation on A is an equivalence relation.
(v) Universal relation on A is an equivalence relation.
Solution:
(i) True, Identity relation IA = {(a, a); ∀ a ∈ A}
Clearly it is a reflexive relation on A.
(ii) False, e.g. R = {(1, 1), (2, 2), (3, 3), (1, 2)} be reflexive on A = {1, 2, 3} but not an identity relation on A
∵ (1, 2) ∈ R.
(iii) True, IA = {(a, a) ; ∀ a ∈ A} is an identity relation on A.
Clearly it is symmetric relation on A.
∀ a ∈ A, (a, a) ∈ A
⇒ (a, a) ∈ A
(iv) True, IA = {{a, a) ; ∀ a ∈ A} is an identity relation on set A.
IA is reflexive, symmetric and also transitive.
So IA be an equivalence relation on A.
(v) True, by def. of universal relation, since A × A ⊂ A × A
∴ A × A be a relation on A.
Question 24.
State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive.
Solution:
Let A = {1, 2, 3} and R be a relation defined on A be {(1, 2), (2, 1)}
Clearly (1, 2), (2, 1) ∈ R but (1, 1) ∉ R
Clearly R is not transitive on A = {1, 2, 3}.
Question 25.
Write the smallest equivalence relation on A where A = {2, 3, 4}. ( NCERT Exemplar)
Solution:
Given set A = {2, 3, 4}
Let relation R on set A = {(2, 2), (3, 3), (4, 4)}
Clearly (2, 2), (3, 3), (4, 4) ∈ R
⇒ R is reflexive
∀ a, b ∈ A s.t (a, b) ∈ R
⇒ (b, a) ∈ R
∴ R is symmetric.
∀ a, b, c ∈ A s.t (a, b), (b, c) ∈ R.
Then (a, c) ∈ R
∴ R is transitive.
Thus R is reflexive, symmetric and transitive on A.
∴ R be forms an smallest equivalence relation on A.
Question 26.
If A = {(0, 1, 2, ………………., 9} and the relation R on A is defined by R = {(x, y) : x, y ∈ A, y = 2x +1}, then determine whether the relation R is (i) reflexive
(ii) symmetric
(iii) transitive.
Solution:
Given A = {0, 1, 2, ………………., 9}
and relation R = {(x, y); x, y ∈ A, y = 2x + 1}
(i) reflexive :
Clearly 1 ≠ 2 x 1 + 1 = 3
∴ (1, 1) ∉ R
∴ R is not reflexive on A
(ii) Symmetric :
Clearly (1, 3) ∈ R,
Since 3 = 2 . 1 + 1
but (3, 1) ∉ R
∵ 1 ≠ 2 × 3 + 1 = 7
R is not symmetric.
(iii) Transitive :
Clearly (1, 3) ∈ R
∵ 3 = 2 . 1 + 1 = 3
and (3, 7) ∈ R
∵ 7 = 2 × 3 + 1 = 7
but (1, 7) ∉ R
∵ 7 ≠ 2 × 1 + 1 = 3
∴ for any 1, 3, 7 ∈ A s.t (1, 3), (3, 7) ∈ R but (1, 7) ∉ R
Thus R is not transitive on A.
Question 27.
Is the relation R on the set R of real numbers defined by R = {(a, b) : a, b ∈ R, 1 + ab > 0} transitive ? Justify your answer.
Solution:
Given relation R on set of real numbers is given by
R = {(a, b) ; a, b ∈ R, 1 + ab ≥ 0}
Now (1, – 1) ∈ R
∵ 1 + 1 × (- 1) ≥ 0
⇒ 0 ≥ 0, which is true
and (- 1, – 2) ∈ R
∵ 1 + (- 1) (- 2) = 1 + 2 = 3 > 0
But (1, – 2) ≠ R
∵ 1 + 1 × (- 2) = 1 – 2 = – 1 < 0
∴ R is not transitive on set of real numbers.
Question 28.
Is the relation R on the set Q of rational numbers defined by R = {(x, y) : x, y ∈ Q, x < y2}, symmetric? Justify your answer.
Solution:
Given relation R = {(x, y); x, y ∈ Q, x < y2}
∀ x, y ∈ Q s.t (x, y) ∈ R
⇒ x < y2 ⇏ y2 < x
i.e. (y, x) ∉ R
e.g. : (1, 2) ∈ R, since 1 < 22 = 4
but (2, 1) ∉ R
∵ 22 = 4 > 1 i.e. 4 ≮ 1
Thus R is not symmetric on Q.
Question 29.
If the relation R on the set A = {1, 2, 3} is defined by R = {(1, 1), (2, 2), (3, 3), (2. 1), (3, 2)}, then determine whether the relation R is
(i) reflexive
(ii) symmetric
(iii) transitive.
Solution:
Given set A = {1, 2, 3}
and R = {(1, 1), (2, 2), (3, 3), (2, 1), (3, 2)}
(i) reflexive :
Clearly 1, 2, 3 ∈ A and (1, 1), (2, 2), (3, 3) ∈ R
∴ Thus R is reflexive on A.
(ii) symmetric:
Clearly (2, 1) ∈ R but (1, 2) ∉ R
∴ R is not symmetric on A.
(iii) Transitive :
Clearly (3, 2), (2, 1) ∈ R but (3, 1) ∉ R
For transitivity, we have ∀ a, b, c ∈ A
s.t (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R
∴ R is not transitive on A.
Question 30.
How many reflexive relations are possible on a Set A of order 3?
Solution:
26
Question 31.
If R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}, then determine whether
(i) R is reflexive and symmetric but not transitive.
(ii) R is reflexive and transitive but not symmetric.
(iii) R is symmetric and transitive but not reflexive.
Solution:
Let A = {1, 2, 3, 4} and relation R on set A be defined by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
Reflexive:
Clearly (1, 1), (2, 2), (3, 3), (4, 4) ∈ R
∴ R is reflexive on A.
Symmetric :
Clearly (1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric on A.
Transitive :
Clearly (1, 2), (2, 2) ∈ R ⇒ (1, 2) ∈ R
(1, 1), (1, 2) ∈ R ⇒ (1, 2) ∈ R
(1, 1), (1, 3) ∈ R ⇒ (1, 3) ∈ R
(1, 3), (3, 3) ∈ R ⇒ (1, 3) ∈ R
(1,3), (3, 2) ∈ R ⇒ (1, 2) ∈ R
(3, 3), (3, 2) ∈ R ⇒ (3, 2) ∈ R
(3, 2), (2, 2) ∈ R ⇒ (3, 2) ∈ R
Thus ∀ a, b, c ∈ A s.t (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R
Thus R is transitive on A.
Hence R is reflexive, transitive but not symmetric.
(i) No
(ii) Yes
(iii) No