# Class 12 RD Sharma Solutions – Chapter 1 Relations – Exercise 1.1 | Set 1

**Question 1. Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:**

**(i) R = {(x. y) x and y work at the same place} ****(ii) R = {(x. y) x and y live in the same locality}****(iii) R = {(x. y) x is wife of y} ****(iv) R = {(x. y) x is father of y}** ** **

**Solution:**

(i)Given the relation R = {(x, y): x and y work at the same place}Now we need to check whether the relation is reflexive or not.

Check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Let x be any element of R.

Then, x ∈ R

⇒ x and x work at the same place is true since they are same.

⇒ (x, x) ∈ R [condition for reflexive relation]

So, R is a reflexive relation.

Now we have to check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b) ∈ A.

Let (x, y) ∈ R

⇒ x and y work at the same place [since it is given in the question]

⇒ y and x work at the same place [same as “x and y work at the same place”]

⇒ (y, x) ∈ R

So, R is a symmetric relation also.

Now we have to check whether the given relation is Transitive relation or not. Relation R is said to be Transitive over set A if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let (x, y) ∈ R and (y, z) ∈ R.

Then, x and y work at the same place. [Given]

y and z also work at the same place. [(y, z) ∈ R]

⇒ x, y and z all work at the same place.

⇒ x and z work at the same place.

⇒ (x, z) ∈ R

Therefore, R is a transitive relation also.

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

reflexive relation,symmetric relation and transitive relationas well.

(ii)Given the relation R = {(x, y): x and y live in the same locality}Now we have to check whether the relation R is reflexive, symmetric and transitive or not.

Check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Let x be any element of relation R.

Then, x ∈ R

It is given that x and x live in the same locality is true since they are the same.

So, R is a reflexive relation.

Now we have to check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b) ∈ A.

Let (x, y) ∈ R

⇒ x and y live in the same locality [ it is given in the question ]

⇒ y and x live in the same locality [if x and y live in the same locality, then y and x also live in the same locality]

⇒ (y, x) ∈ R

So, R is a symmetric relation as well.

Now we have to check whether the given relation is Transitive relation or not. Relation R is said to be Transitive over set A if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let x, y and z be any elements of R and (x, y) ∈ R and (y, z) ∈ R.

Then,

x and y live in the same locality and y and z live in the same locality

⇒ x, y and z all live in the same locality

⇒ x and z live in the same locality

⇒ (x, z) ∈ R

So, R is a transitive relation also.

So the relation R = {(x, y): x and y live in the same locality} is a

reflexive relation, symmetric relation and transitive relationas well.

(iii)Given R = {(x, y): x is wife of y}Now we have to check whether the relation R is reflexive, symmetric and transitive relation or not.

Check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Let x be an element of R.

Then, x is wife of x cannot be true. [since the same person cannot be the wife of herself]

⇒ (x, x) ∉ R

So, R is not a reflexive relation.

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b) ∈ A.

Let (x, y) ∈ R

⇒ x is wife of y

⇒ x is female and y is male

⇒ y cannot be wife of x as y is husband of x

⇒ (y, x) ∉ R

So, R is not a symmetric relation.

Check whether the given relation is Transitive relation or not. Relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let (x, y) ∈ R, but (y, z) ∉ R

Since x is wife of y, but y cannot be the wife of z, since y is husband of x.

⇒ x is not the wife of z.

⇒(x, z)∈ R

So, R is a transitive relation.

Hence the given relation R = {(x, y): x is wife of y} is a

transitive relation but not a reflexive and symmetric relation.

(iv)Given the relation R = {(x, y): x is father of y}Now we have to check whether the relation R is reflexive, symmetric and transitive or not.

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Let x be an arbitrary element of R.

Then, x is father of x cannot be true. [since no one can be father of himself]

So, R is not a reflexive relation.

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (x, y)∈ R

⇒ x is the father of y.

⇒ y is son/daughter of x.

⇒ (y, x) ∉ R

So, R is not a symmetric relation.

Now, check whether the given relation is Transitive relation or not. Relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let (x, y)∈ R and (y, z)∈ R.

Then, x is father of y and y is father of z

⇒ x is grandfather of z

⇒ (x, z)∉ R

So, R is not a transitive relation.

Hence, the given relation R = {(x, y): x is father of y} is

not a reflexive relation, not a symmetric relation and not a transitive relationas well.

**Question 2. Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:**

**R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}****R2 = {(a, a)}****R3 = {(b, c)}****R4 = {(a, b), (b, c), (c, a)}.**

**Find whether or not each of the relations R1, R2, R3, R4 on A is (i) reflexive (ii) symmetric and (iii) transitive.**

**Solution:**

i)Considering the relation R1, we haveR1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}

Now we have check R1 is reflexive, symmetric and transitive or not.

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

Given (a, a), (b, b) and (c, c) ∈ R1 [since each element maps to itself]

So, R1 is reflexive.

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

We see that for every ordered pair (x, y), there is a pair (y, x) present in the relation R1.

So, R1 is symmetric.

Transitive: A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

In the relation, (a, b) ∈ R1, (b, c) ∈ R1 and also (a, c) ∈ R1

So, R1 is transitive.

Therefore,

R1 isreflexive relation, symmetric relation and transitive relationas well.

(ii)Considering the relation R2, we haveR2 = {(a, a)}

First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A.

We can see that (a, a) ∈ R2. [since each element maps to itself]

So, R2 is a reflexive relation.

Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

We can see that (a, a) ∈ R

⇒ (a, a) ∈ R.

So, R2 is symmetric.

Check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

R2 is clearly a transitive relation. [since there is only one element in it]

Therefore,

R2 is reflexive relation, symmetric relation and transitive relationas well.

(iii)Considering the relation R3, we haveR3 = {(b, c)}

In the relation, (a, a)∉ R3, (b, b)∉ R3 neither (c, c) ∉ R3.

So, R3 is not reflexive. [since all pairs of type (x, x) should be present in the relation]

In the relation, (b, c) ∈ R3, but (c, b) ∉ R3

So, R3 is not symmetric.

Check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

R3 has only two elements.

Hence, R3 is transitive.

Therefore,

R2 is transitive relation but not a reflexive relation and not a symmetric relation also.

(iv)Considering the relation R4, we haveR4 = {(a, b), (b, c), (c, a)}

In the relation, (a, a) ∉ R4, (b, b) ∉ R4 (c, c) ∉ R4

So, R4 is not a reflexive relation.

Here, (a, b) ∈ R4, but (b, a) ∉ R4.

So, R4 is not symmetric

Check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

In the relation, (a, b) ∈ R4, (b, c) ∈ R4, but (a, c) ∉ R4

So, R4 is not a transitive relation.

Therefore,

R2 is not a reflexive relation, not a symmetric relation and neither a transitive relationas well.

**Question 3. Test whether the following relation R**_{1}, R_{2}, and R_{3} are (i) reflexive (ii) symmetric and (iii) transitive:

_{1}, R

_{2}, and R

_{3}are (i) reflexive (ii) symmetric and (iii) transitive:

**(i) R _{1} on Q0 defined by (a, b) ∈ R_{1} ⇔ a = 1/b.**

**(ii) R**

_{2}on Z defined by (a, b) ∈ R_{2}⇔ |a – b| ≤ 5**(iii) R**

_{3}on R defined by (a, b) ∈ R_{3}⇔ a2 – 4ab + 3b2 = 0.**Solution:**

i)Given R_{1}on Q_{0}defined as (a, b) ∈ R_{1}⇔ a = 1/b.Let a be an element of R1.

Then, a ∈ R1

⇒ a ≠1/a ∀ a ∈ Q

_{0}So, R1 is not reflexive.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (a, b) ∈ R1

Then, (a, b) ∈ R1

Therefore, we can write ‘a’ as a =1/b

⇒ b = 1/a

⇒ (b, a) ∈ R1

So, R1 is symmetric.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Here, (a, b) ∈ R1 and (b, c) ∈ R2

⇒ a = 1/b and b = 1/c

⇒ a = 1/ (1/c) = c

⇒ a ≠ 1/c

⇒ (a, c) ∉ R1

So, R1 is not a transitive relation.

(ii)Given R2 on Z defined as (a, b) ∈ R2 ⇔ |a – b| ≤ 5Now we have to check whether R2 is reflexive, symmetric and transitive or not.

Let a be an element of R2.

Then, a ∈ R2

On applying the given condition we will get,

⇒ | a−a | = 0 ≤ 5

So, R1 is a reflexive relation.

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (a, b) ∈ R2

⇒ |a−b| ≤ 5 [Since, |a−b| = |b−a|]

⇒ |b−a| ≤ 5

⇒ (b, a) ∈ R2

So, R2 is a symmetric relation.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let (1, 3) ∈ R2 and (3, 7) ∈ R2

⇒|1−3|≤5 and |3−7|≤5

But |1−7| ≰ 5

⇒ (1, 7) ∉ R2

So, R2 is not a transitive relation.

(iii)Given R3 on R defined as (a, b) ∈ R3 ⇔ a^{2}– 4ab + 3b^{2}= 0.Now we have to check whether R2 is reflexive, symmetric and transitive or not.

Let a be an element of R3.

Then, a ∈ R3.

⇒ a

^{2}− 4a × a+ 3a^{2}= 0So, R3 is reflexive

Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.

Let (a, b) ∈ R3

⇒ a

^{2}−4ab+3b^{2}=0But b

^{2}−4ba+3a^{2 }≠ 0 ∀ a, b ∈ RSo, R3 is not symmetric.

Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A.

Let (1, 2) ∈ R3 and (2, 3) ∈ R3

⇒ 1 − 8 + 6 = 0 and 4 – 24 + 27 = 0

But 1 – 12 + 9 ≠ 0

So, R3 is not a transitive relation.

**Question 4. Let A = {1, 2, 3}, and let R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R2 = {(2, 2), (3, 1), (1, 3)}, R3 = {(1, 3), (3, 3)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.**

**Solution:**

Considering the relation R1, we have

R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

Here, (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R

So, R1 is reflexive.

In the given relation, (2, 1) ∈ R1 but (1, 2) ∉ R1

So, R1 is not symmetric.

In the relation, (2, 1) ∈ R1 and (1, 3) ∈ R1 but (2, 3) ∉ R1

So, R1 is not transitive.

Therefore, the relation

R1 is reflexive but not symmetric and transitive relation.Now considering the relation R2, we have

R2 = {(2, 2), (3, 1), (1, 3)}

Clearly, (1, 1) and (3, 3) ∉ R2

So, R2 is not a reflexive relation.

In the relation, (1, 3) ∈ R2 and (3, 1) ∈ R2

So, R2 is a symmetric relation.

In the relation, (1, 3) ∈ R2 and (3, 1) ∈ R2 but (3, 3) ∉ R2

So, R2 is not a transitive relation.

Therefore, the relation

R2 is symmetric but not a reflexive and transitive relation.Considering the relation R3, we have

R3 = {(1, 3), (3, 3)}

In the relation, (1, 1) ∉ R3

So, R3 is not reflexive.

In the relation, (1, 3) ∈ R3, but (3, 1) ∉ R3

So, R3 is not symmetric.

Here, (1, 3) ∈ R3 and (3, 3) ∈ R3

Also, (1, 3) ∈ R3

So, R3 is transitive.

Therefore, the relation

R3 is transitive but not a reflexive and symmetric relation.

**Question 5. The following relation is defined on the set of real numbers.**

**(i) aRb if a – b > 0****(ii) aRb iff 1 + a b > 0****(iii) aRb if |a| ≤ b.**

**Find whether relation is reflexive, symmetric or transitive.**

**Solution:**

(i)Consider the relation defined as aRb if a – b > 0Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

Let a be an element of R.

Then, a ∈ R

But a − a = 0 ≯ 0

So, this relation is not a reflexive relation.

Let (a, b) ∈ R

⇒ a − b > 0

⇒ − (b − a) > 0

⇒ b − a < 0

So, the given relation is not a symmetric relation.

Let (a, b) ∈ R and (b, c) ∈ R.

Then, a − b > 0 and b − c > 0

Adding the two, we will get

a – b + b − c > 0

⇒ a – c > 0

⇒ (a, c) ∈ R.

So, the given relation is a transitive relation.

(ii)Consider the relation defined as aRb iff (read as “if and only if”) 1 + a b > 0Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

Let a be an element of R.

Then, a ∈ R

⇒ 1 + a × a > 0

i.e. 1 + a

^{2}> 0 [since, square of any number is positive]So, the given relation is a reflexive relation.

Let (a, b) ∈ R

⇒ 1 + a b > 0

⇒ 1 + b a > 0

⇒ (b, a) ∈ R

So, the given relation is symmetric.

Let (a, b) ∈ R and (b, c) ∈ R

⇒1 + a b > 0 and 1 + b c >0

But 1+ ac ≯ 0

⇒ (a, c) ∉ R

So, the given relation is not a transitive relation.

(iii)Consider the relation defined as aRb if |a| ≤ b.Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

Let a be an element of relation R.

Then, a ∈ R [Since, |a|=a]

⇒ |a| ≮ a

So, R is not a reflexive relation.

Let (a, b) ∈ R

⇒ |a| ≤ b

⇒ |b| ≰ a ∀ a, b ∈ R

⇒ (b, a) ∉ R

So, R is not a symmetric relation.

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a| ≤ b and |b| ≤ c

Multiplying the corresponding sides, we will get

|a| × |b| ≤ b c

⇒ |a| ≤ c

⇒ (a, c) ∈ R

Thus, R is a transitive relation.

**Question 6. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.**

**Solution:**

Given R = {(a, b): b = a + 1}

Now, for this relation we have to check whether it is reflexive, transitive and symmetric or not.

Let a be an element of R.

Then, a = a + 1 cannot be true for all a ∈ A.

⇒ (a, a) ∉ R

So, R is not a reflexive relation over the given set.

Let (a, b) ∈ R

⇒ b = a + 1

⇒ −a = −b + 1

⇒ a = b − 1

So, (b, a) ∉ R

Thus, R is not a symmetric relation over the given set.

Let (1, 2) and (2, 3) ∈ R

⇒ 2 = 1 + 1 and 3

2 + 1 is true.

But 3 ≠ 1+1

⇒ (1, 3) ∉ R

So, R is not a transitive relation over the given set.

**Question 7. Check whether the relation R on R defined as R = {(a, b): a ≤ b**^{3}} is reflexive, symmetric or transitive.

^{3}} is reflexive, symmetric or transitive.

**Solution:**

We have given the relation R = {(a, b): a ≤ b

^{3}}First let us check whether the given relation is reflexive or not.

It can be observed that (1/2, 1/2) in R as 1/2 > (1/2)

^{3}= 1/8So, R is not a reflexive relation.

Now, check for whether the relation is symmetric or not

(1, 2) ∈ R (as 1 < 2

^{3}= 8)But,

(2, 1) ∉ R (as 2 > 1

^{3}= 1)So, R is not a symmetric relation.

We have (3, 3/2), (3/2, 6/5) in “R as” 3 < (3/2)

^{3}and 3/2 < (6/5)^{3}But (3, 6/5) ∉ R as 3 > (6/5)

^{3}So, R is not a transitive relation.

Hence,

R is neither reflexive, nor symmetric, nor transitive.

**Question 8. Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.**

**Solution:**

We will verify this by taking example.

Let A be a set.

Then, Identity relation IA=I

_{A}is reflexive, since (a, a) ∈ A ∀ a ∈ A.The converse of this need not be necessarily true.

Now, consider the set A = {1, 2, 3}

Here, relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.

But, R is not an identity relation.

Hence proved, that every identity relation on a set is reflexive but the converse is not necessarily true.

**Question 9. If A = {1, 2, 3, 4} define relations on A which have properties of being**

**(i) Reflexive, transitive but not symmetric****(ii) Symmetric but neither reflexive nor transitive.****(iii) Reflexive, symmetric and transitive. **

**Solution:**

(i)We have given the set A = {1, 2, 3, 4}The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R [satisfies the reflexivity property]

and (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R [satisfies the transitivity property]

However, (2, 1) ∈ R, but (1, 2) ∉ R [does not satisfies the symmetric property]

(ii)We have given the set A = {1, 2, 3, 4}The relation on A having properties of being reflexive, transitive, but not symmetric is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R [satisfies the reflexivity property]

And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R [satisfies the transitivity property]

However, (2, 1) ∈ R, but (1, 2) ∉ R [does not satisfies the symmetric property]

(iii)We have given the set A = {1, 2, 3, 4}The relation on A having properties of being symmetric, reflexive and transitive is

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}

Relation R satisfies reflexivity, symmetricity and transitivity.

⇒ (1, 1), (2, 2), (3, 3) ∈ R [satisfies the reflexivity property]

⇒ (1, 1) ∈ R and (2, 1) ∈ R [satisfies the symmetric property]

⇒ (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R [satisfies the transitivity property]

**Question 10. Let R be a relation defined on the set of natural number N as R={(x, y): x, y ∈ N, 2x + y = 41}. Find the domain and range of R. Also verify whether R is (i) reflexive (ii) symmetric (iii) transitive. **

**Solution:**

We have given,

{(x, y) : x, y ∈ N, 2x + y = 41}

Now,

2x + y = 41

⇒ y = 41 – 2x

Put the value of x one by one to form the relation R.

The relation we will after putting x = 1, 2, 3, ……. ,20 is:

[we can’t put x=21 since y = 41 – 2(2) < 0, which is not a natural number]R = {(1, 39), (2, 37), (3, 35)………….., (20, 1)}

So the domain of R is

Domain(R) = {1, 2, 3, ……… ,20}And the range of R is

Range(R) = {39, 37, 33, ……. ,1} and can be rearranged as {1, 3, 5, ……….. ,39}Now for this relation we have to check whether it is reflexive, transitive and symmetric or not.

Let x be an any element of relation R.

Since, (2, 2) ∉ R

So, R is not a reflexive relation.

Since, (1, 39) ∈ R but (39, 1) ∉ R.

So, R is not symmetric.

Since, (15,11) ∈ R and (11,19) ∈ R but (15,19) ∉ R.

Thus, R is not transitive.