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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Problem 1 Solution**. Solutions for other problems are available at Exercise 1.1 SolutionsExercise 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 a ∈ A.

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_2 ∈ A

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_3 ∈ A

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 a ∈ A

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_2 ∈ A

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_3 ∈ A

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 a ∈ A.

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_2 ∈ A

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_3 ∈ A

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 a ∈ A.

In this case, for any element x, {x - x = 0}. As 0 is also an integer, this property is true for every element x ∈ Z. In otherwords, (x, x) ∈ R

∴ R is Reflexive

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

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}

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

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

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}

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

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

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

As a person x can never be taller than oneself, (x, x) ∉ R.

∴ R is not Reflexive

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

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}

As a person x can not be wife of oneself. So, (x, x) ∉ R.

∴ R is not Reflexive

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

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}

As a person x can not be father of oneself. So, (x, x) ∉ R.

∴ R is not Reflexive

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

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.