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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter Relations and Functions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 1: Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) 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}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}.
With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
(i) A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.
[3(1) − 9 ≠ 0]
Hence, R is neither reflexive, nor symmetric, nor transitive.
(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R.
∴ R is not reflexive.
(1, 6) ∈R
But,
(1, 6) ∉ R.
∴ R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(iii) A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
(x, x) ∈R
∴ R is reflexive.
Now,
(2, 4) ∈R [as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4]
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
∴ z is divisible by x.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) R = {(x, y): x − y is an integer}
Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer.
∴ R is reflexive.
Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.
⇒ −(x − y) is also an integer.
⇒ (y − x) is an integer.
∴ (y, x) ∈ R
∴ R is symmetric.
Now,
Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z.
⇒ (x − y) and (y − z) are integers.
⇒ x − z = (x − y) + (y − z) is an integer.
∴ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(v) (a) R = {(x, y): x and y work at the same place}
(x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
⇒ y and x work at the same place.
⇒ (y, x) ∈ R.
∴ R is symmetric.
Now, let (x, y), (y, z) ∈ R
⇒ x and y work at the same place and y and z work at the same place.
⇒ x and z work at the same place.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y live in the same locality and y and z live in the same locality.
⇒ x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y): x is exactly 7 cm taller than y}
Now, (x, x) ∉ R
Since human being x cannot be taller than himself.
∴ R is not reflexive.
Now, let (x, y) ∈R.
⇒ x is exactly 7 cm taller than y.
Then, y is not taller than x.
∴ (y, x) ∉R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
∴R is not symmetric.
Now,
Let (x, y), (y, z) ∈ R.
⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
⇒ x is exactly 14 cm taller than z .
∴ (x, z) ∉R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y): x is the wife of y}
Now, (x, x) ∉ R
Since x cannot be the wife of herself.
∴R is not reflexive.
Now, let (x, y) ∈ R
⇒ x is the wife of y.
Clearly y is not the wife of x.
∴ (y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
∴ R is not transitive.
Let (x, y), (y, z) ∈ R
⇒ x is the wife of y and y is the wife of z.
This case is not possible. Also, this does not imply that x is the wife of z.
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
Now (x, x) ∉ R
As x cannot be the father of himself.
∴R is not reflexive.
Now, let (x, y) ∈R.
⇒ x is the father of y.
⇒ y cannot be the father of y.
Indeed, y is the son or the daughter of y.
∴ (y, x) ∉ R
∴ R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x is the father of y and y is the father of z.
⇒ x is not the father of z.
Indeed x is the grandfather of z.
∴ (x, z) ∉ R
∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 1: Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relat....
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Comments
Very helpful
Thanks for answer for Question no.1
Thanks
Thanks
Really good.
Thanks , Sant Kumar Hooda
2 is not correct
solution of A={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15} S={(x,y)/y=5x ; xâËËA, yâËËA} find S is equivalence relation or not?
Very helpful