# Class 12 RD Sharma Solutions – Chapter 1 Relations – Exercise 1.1 | Set 2

**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)

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 m be an element of R.

Then, m = k×m is true for k=1

(m, m) ∈ R.

So, R 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.

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.

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 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 }

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 a be an element of R.

⇒ a ∈ R

⇒ a ≥ a , which is always true.

⇒ (a, a) ∈ R

Hence, R 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.

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.

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 (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}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.

if (x, x) ∈ R then, x > x, which is not true.

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

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 (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.

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.

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.

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 (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.