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 an 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}

RELATIONS AND FUNCTIONS

Class 12, NCERT Chapter 1,  Exercise 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} 

SolutionGiven, A={1,2,3,4,5,6,7,8,9,10,11,12,13,14}and 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)∉ RR is not symmetric as (1,3)∈R, but (3,1)∉R R is not transitive  as (1,3), (3,9)∈ R, but (1,9)∉R Hence R is neither Reflexive, nor symmetric, nor transitive.

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}

Solution

R = {(x, y) : y = x + 5 and x < 4} = {(1,6), (2,7), (3,8)}

R is not reflexive as (1,1)∉R

R is not symmetric as (1,6)∈R, but (6,1)∉R

Since there is no pair in R such that (x,y) and (y,z)∈R, then (x,z) cannot belong t R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive. 

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} 

Soltion

A={1,2,3,4,5,6}

R = {(x,y): y is divisible by x}

As we know that any number (x) is divisible by itself.

So, (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. therefore z is also divisible by x.

∴ R is Transitive.

Hence, R is  reflexive and transitive but not symmetric 

(iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer} 

Solution

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

 therefore, 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) E R

∴R is transitive.

 Hence, R is reflexive, symmetric, and transitive. 

(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} 

Solution

R = {(x, y): x and y work at the same place}

⇨(x, x) ∈ R                                                        [as x and x work at the same place] 

∴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} 

Solution

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}

Solution

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.                                         [Since y is 7 cm smaller 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 wife of y}

Solution

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 symmetric. 

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 father of y}

Solution

R = {(x, y): x is the father of y} 

(x, x) ∉R

As x cannot be the father of himself. 

∴R is not reflexive. 

Now, let (x, y) E R. 

⇨x is the father of y. 

⇨y cannot be the father of x. 

Indeed, y is the son or the daughter of y. 

∴(y, x) R 

∴R is not symmetric. 

Now, 

let (x, y) E R and (y, z) 0 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.