NCERT Solutions
Class 12 Maths
Relations and Functions
Ex. 1.1 Q1
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}
(i) Given, A = {1, 2, 3 ... 13, 14}
R = {(x, y): 3x − y = 0}
So, 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. [Since 3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R. [Since 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 clear that (1, 1) ∉ R.
So, R is not reflexive.
(1, 6) ∈ R But, (1, 6) ∉ R.
So, 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.
So, 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.
So, (x, x) ∈ R
Hence, R is reflexive.
Now,
(2, 4) ∈ R [as 4 is divisible by 2]
But, (4, 2) ∉ R [as 2 is not divisible by 4]
So, 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
So, 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.
So, 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
So, 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
So, 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 [as x and x work at the same place]
So, 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.
So, 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
So, 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.
So, 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
So, 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
So, 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.
So, R is not reflexive.
Now, let (x, y) ∈ R.
=> x is exactly 7 cm taller than y.
Then, y is not taller than x. [Since, y is 7 cm smaller than x]
So, (y, x) ∉ R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
So, 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
So, 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.
So, R is not reflexive.
Now, let (x, y) ∈ R
=> x is the wife of y.
Clearly y is not the wife of x.
So, (y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
So, 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
So, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
So, 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
So, 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
So, R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.