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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 1 |

Exercise | 1 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 1.1 Solutions Chapter 1 Relations**

**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.

**Question 11. Is it true every relation which is symmetric and transitive is also reflexive? Give reasons.**

**Solution:**

We will verify this by taking an example.

Consider a set A = {1, 2, 3} and a relation R on A such that R = { (1, 2), (2,1), (2,3), (1,3) }

The relation R over the set A is symmetric and transitive.

But, it is not reflexive.

(1,1),(2, 2) and (3,3) ∉ R.

Therefore, R is not a reflexive relation.

Hence, it not true that every relation which is symmetric and transitive is also reflexive because it is possible that all pairs of type (x, x) is not present in the relation.

**Question 12. An integer m is said to be related to another integer n if m is multiple of n. Check if the relation is symmetric, reflexive and transitive.**

**Solution:**

Let us define a relation such that

R = {m, n : m, n ∈ Z, m = k×n} where, k ∈ N (natural number)

Let m be an element of R.

Then, m = k×m is true for k=1

(m, m) ∈ R.

So, R is reflexive.

Let (m, n) ∈ R

⇒ m = k×n for some k ∈ N

and according to transitivity, n = (1/k)×m for some k ∈ N but, 1/k ∉ N.

So, R is not a symmetric relation.

Let m, n, o be any elements of R then, (m, n) and (n, o) ∈ R

⇒ m = k1×n and n = k2×o for some k1, k2 ∈ N

⇒ m = (k1×k2)×o

⇒ (m, o) ∈ R.

So, R is a transitive relation.

**Question 13. Show that the relation “**≥**” on the set R of all real number is reflexive, transitive but not symmetric.**

**Solution:**

Let us define a relation R as

R = { (a, b) a, b ∈ R ; a ≥ b }

Let a be an element of R.

⇒ a ∈ R

⇒ a ≥ a , which is always true.

⇒ (a, a) ∈ R

Hence, R is a reflexive relation.

Let (a, b) ∈ R.

⇒ a ≥ b

⇒ b ≥ a [according to transitivity]

but it is not always true except when a=b.

⇒ (b, a) ∉ R

Hence, R is not a symmetric relation.

Let (a, b) and (b, c) ∈ R

⇒ a ≥ b and b ≥ c

⇒ a ≥ b ≥ c

⇒ a ≥ c

⇒ (a, c) ∈ R

Hence, R is a transitive relation.

**Question 14. Give an example of a relation. Which is**

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

**Solution:**

Reflexive Relation:

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.Symmetric Relation:

A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A.Transitive Relation:

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 A be a set as,

A = {1, 2, 3J

(i)Let R be the relation on A such thatR = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3}

Thus, R is reflexive and symmetric, but not transitive.

(ii)Let R be the relation on A such that AR= { (1,1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3) }Clearly, the relation R on A is reflexive and transitive, but not symmetric.

(iii)Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3,1), (2, 3) }We see that the relation R on A is symmetric and transitive, but not reflexive.

(iv)Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3, 1) }The relation R on A is symmetric, but neither reflexive nor transitive.

(v)Let R be the relation on A such that R = { (1, 2), (2, 3) ,(1, 3) }The relation R on A is transitive, but neither symmetric nor reflexive.

**Question 15. Given the relation R={(1, 2), (2, 3)} on the set A={1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmetric, reflexive**,** and transitive.**

**Solution:**

We have given the relation,

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

For R to be reflexive it must have (1,1), (2, 2), (3,3).

For R to be a symmetric relation, all the ordered pairs upon interchanging the elements must be present in the relation R. Therefore, R must contain (2,1 ) and (3, 2), (3,1), (1,3).

And for to be a transitive relation, it must contain (1,3).

Hence, the number of ordered pairs to be added to R is 7, i.e. {(1,1), (2, 2), (3,3), (1,3), (3,1), (2,1), (3, 2)}.

**Question 16. Let A={1, 2, 3} and R={(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs to be added in R so that it may become a transitive relation on A.**

**Solution:**

The relation R on A is given such that R = {(1, 2), (1,1), (2,3)}

For the relation R to be transitive, we must have (1, 2) ∈ R, since (2, 3) ∈ R

⇒ (1,3) ∈ R

Therefore, the minimum number of ordered pairs need to be added to relation R is 1, i.e. (1, 3) to make it a transitive relation on A.

**Question 17. Let A={a, b, c} and a relation R to be defined on A as follows**

**R={(a, a), (b, c), (a, b)}. Then write minimum number of ordered pairs to be added in R to make it reflexive and transitive.**

**Solution:**

We have given a set A = {a, b, c} and a relation R={(a, a), (b, c), (a, b)}.

For relation to be reflexive, it should contain (b, b) and (c, c).

And for relation to be transitive R should contain (a, c) since (a, b) ∈ R and (b, c) ∈ R

Therefore, the minimum number of ordered pair to be added to relation R is (b, b) , (c, c) and (a, c) i.e. 3.

**Question 18. Each of the following defines a relation on N **

**(i) x > y, x, y ∈ N****(ii) x + y =10, x, y ∈ N****(iii) xy is a square of integer,** **x, y ∈ N****(iv) x + 4y =10, x, y ∈ N**

**Determine which of the above relations are symmetric, reflexive**,** and transitive.**

**Solution:**

(i)We have given the relation defined as

R = {(x > y), x, y ∈ N}if (x, x) ∈ R then, x > x, which is not true.

⇒ (x, x) ∉ RSo, the relation is not a reflexive relation.

Let (x, y) ∈ R, then x R y

⇒ x > y

and according to symmetric property, (y, x) ∈ R

⇒ y > x, but it is not true since x > y

⇒ (x, y) ∈ R but (y, x) ∉ R

So, the relation is not a symmetric relation.

Let (x, y) ∈ R and (y, z) ∈ R

⇒ x > y and y > z

⇒ x > z

⇒ (x, z) ∈ R

So, R is a transitive relation as well.

(ii)We have given the relation defined as

R = { x + y =10, x, y ∈ N }Clearly, the relation will be R = { (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) }

We can see that, (1, 1) ∉ R.

So, R is not a reflexive relation.

By observing the above relation, we can say that ∀ (x, y) ∈ R, (y, x) ∈ R.

So, R is a symmetric relation.

In the relation, (1, 9) ∈ R and (9, 1) ∈ R but (1, 1) ∉ R

So, R is not a transitive relation.

(iii)We have given the relation as

R = { xy is a square of integer, x, y ∈ N }Clearly, (x, x) ∈ R ∀ x ∈ N

since, x

^{2}is square of an integer for any x ∈ N.Hence, R is a reflexive relation.

Let (x, y) ∈ R

⇒ xy is a square of an integer

⇒ yx is also a square of the same integer since, xy = yx

⇒ (y, x) ∈ R

So, the relation is a symmetric relation.

Let (x, y) ∈ R and (y, z) ∈ R

⇒ xy is an square of an integer and yz is an square of integer

Then let xy = m

^{2}and yz = n^{2}for some m, n ∈ Z⇒ x = m

^{2}/y and z = n^{2}/y⇒ xz = (m

^{2}n^{2})/y^{2 }, which is also a square of an integer⇒ (x, z) ∈ R

So, R is a transitive relation.

(iv)We have given the relation as

R = { x + 4y =10, x, y ∈ N }Clearly, the relation will be R = {(2, 2), (6, 1)} [since x, y ∈ N]

(1, 1) ∉ R

So, the relation R is not reflexive.

We can see that, (1, 6) R but (6, 1) R

So, R is not a symmetric relation.

From the definition, we can see that R is a transitive relation.

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