# Problem 1 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 1 Solution. Solutions for other problems are available at Exercise 1.1 Solutions
Exercise 1.1 Problem 1 Solution
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 \lt 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}
i. Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {(x, y) : 3x – y = 0}
3x - y = 0
3x = y
y = 3x
When {x = 1}, {y = 3 × x = 3 × 1 = 3}. As 3 ∈ A, (1, 3) ∈ R
When {x = 2}, {y = 3 × x = 3 × 2 = 6}. As 6 ∈ A, (2, 6) ∈ R
When {x = 3}, {y = 3 × x = 3 × 3 = 9}. As 9 ∈ A, (3, 9) ∈ R
When {x = 4}, {y = 3 × x = 3 × 4 = 12}. As 12 ∈ A, (3, 12) ∈ R
When {x = 5}, {y = 3 × x = 3 × 5 = 15}. As 15 ∉ A, (3, 15) ∉ R
These calculations can also be represented in the tabular form as follows:
x
y = 3x
(x, y) ∈ R?
1
3 × 1 = 3
2
3 × 2 = 6
3
3 × 3 = 9
4
3 × 4 = 12
5
3 × 5 = 15
For higher values of x, the value of y will be more than 14. As all the elements in A are less than 14, we can stop here.
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
In this case, (1, 1) ∉ R, (2, 2) ∉ R, (3, 3) ∉ R,…. (13, 13) ∉ R, (14, 14) ∉ R.
R is not Reflexive
Note: Ideally, we don’t have to check for all the elements. Even if one element for example (1,1) ∉ R, then we can say that R is not reflexive)
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
Here we see that (1, 3) ∈ R, but (3, 1) ∉ R
R is not Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
Here we see that (1, 3) ∈ R and also (3, 9) ∈ R but (1, 9) ∉ R
R is not Transitive
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 \lt 4}
As it is given that {x \lt 4}, x takes the values 1, 2 and 3.
When {x = 1}, {y = x + 5 = 1 + 5 = 6}. So, (1, 6) ∈ R
When {x = 2}, {y = x + 5 = 2 + 5 = 7}. So, (2, 7) ∈ R
When {x = 3}, {y = x + 5 = 3 + 5 = 8}. So, (3, 8) ∈ R
This can also be put in the tabular form as follows:
x
y = x + 5
(x, y) ∈ R?
1
1 + 5 = 6
2
2 + 5 = 7
3
3 + 5 = 8
R = {(1, 6), (2, 7), (3, 8)}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA
In this case, (1, 1) ∉ R, (2, 2) ∉ R and (3, 3) ∉ R.
R is not Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
In this case, we see that (1, 6) ∈ R, but (6, 1) ∉ R
R is not Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
In this case, we see that (1, 3) ∈ R and also (3, 8) ∈ R but (1, 8) ∉ R
R is not Transitive
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}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
As we know, every element x is divisible by itself i.e. x. In otherwords, (x, x)R
R is Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
In this case, we can see that 6 is divisible by 3. However, 3 is not divisible by 6. In otherwords, (3, 6) ∈ R but (6, 3) ∉ R.
R is not Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
Here we see that (1, 3) ∈ R as 3 is divisible by 1. Also, (3, 6) ∈ R as 6 is divisible by 3. Also we see that 6 is divisible by 1 i.e. (1, 6) ∈ R
R is Transitive
R is both 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}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
In this case, for any element x, {x - x = 0}. As 0 is also an integer, this property is true for every element xZ. In otherwords, (x, x)R
R is Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If {x - y = k}, where k is an integer.
Now, {y - x = -(x - y) = -k}. If k is an integer, then -k is also an integer. So, we see that if (x, y)R then (y, x)R.
R is Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
Here we see that if {(x - y)} is an integer and {(y - z)} is another integer, then {(x - z) = (x - y) + (y - z)} is also an integer, as sum of two integers is integer. So, we can say that if (x, y)R and (y, z)R then (x, z)R.
R is Transitive
R is reflexive, symmetric as well as transitive.
v (a). R = {(x, y) : x and y work at the same place}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
In this case, if x is working at a given place, then the same person x is also working at the same place. So, clearly (x, x)R.
R is Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If x and y are working at a given place, then it implies that y and x are also working at the same place. So, we can say that if (x, y)R then (y, x)R
R is Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
If x and y are working at the given place and also y and z are working at that place, then it implies that both x and z are also working at the same place. In otherwords, if (x, y)R, (y, z)R then (x, y)R.
R is Transitive
R is reflexive, symmetric as well as transitive.
v (b). R = {(x, y) : x and y live in the same locality}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
In this case, if x is living at a given locality, then the same person x is also living at the same locality. So, clearly (x, x)R.
R is Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If x and y are living at a given locality, then it implies that y and x are also living at the same locality. This implies that if (x, y)R then (y, x)R
R is Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
If x and y are living at a given locality and also y and z are living at that locality, then it implies that both x and z are also living at the same locality.
In otherwords, if (x, y)R, (y, z)R then (x, z)R.
R is Transitive
R is reflexive, symmetric as well as transitive.
v (c). R = (x, y) : x is exactly 7 cm taller than y
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
As a person x can never be taller than oneself, (x, x)R.
R is not Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If x is 7 cm taller than y, then we can not say that y is also taller than x. In fact, y will be shorter than x. So, in this case, if (x, y)R then (y, x)R
R is not Symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
If x is exactly 7 cm taller than x and y is exactly 7 cm taller than z, then x is (7 cm + 7 cm) = 14 cm taller than z. So, x is not exactly 7 cm taller than z. (Though x is taller than z but not exactly by 7 cm). In otherwords, if (x, y)R, (y, z)R then (x, z)R.
R is not Transitive
R is neither reflexive nor symmetric nor transitive.
v (d). R = {(x, y) : x is wife of y}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
As a person x can not be wife of oneself. So, (x, x)R.
R is not Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If x is wife of y then y can not be wife of x (y will be husband of x). So, if (x, y)R then (y, x)R.
R is not symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
If x is wife of y, then y can not be wife of any other person z (as y is husband). So, when (x, y)R, (y, z)R.
R is not Transitive
R is neither reflexive nor symmetric nor transitive.
v (d). R = {(x, y) : x is father of y}
To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a)R, for every aA.
As a person x can not be father of oneself. So, (x, x)R.
R is not Reflexive
To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2)R implies that (a_2, a_1)R, for all a_1, a_2A
If x is father of y then y can not be father of x (y will be child of x). So, if (x, y)R then (y, x)R.
R is not symmetric
To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2)R and (a_2, a_3)R implies that (a_1, a_3)R, for all a_1, a_2, a_3A
If x is father of y and y is father of z, then x is not father of z (x will be grandfather of z). So, if (x, y)R and (y, z)R then (x, z)R.
R is not Transitive
R is neither reflexive nor symmetric nor transitive.