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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2

Question 1.

Discuss the following relations for reflexivity, symmetricity and transitivity:

(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.

Solution:

Let Z = {1, 2, 3, ……….}

R is a relation defined on the set of all positive integers by m R n if m divides n

R = { (m, n) : \(\frac{\mathrm{m}}{\mathrm{n}}\) for all m, n ∈ Z } n

(a) Reflexive:

m divides m for all m ∈ Z

∴ (m, m) ∈ R for all m ∈ Z

Hence R is reflexive

(b) Symmetric:

Let (m, n) ∈ R ⇒ m divides n

⇒ n = km for some integers k

But km need not divide m, ie. n need not divide m

∴ (n, m) ∉ R

Hence R is not symmetric.

(c) Transitive:

Let (m, n), (n, r) ∈ R

Then m divides n ⇒ n = km and

n divides r ⇒ r = k_{1}n

r = k_{1}(km) = (k_{1}k) m

m divides r

∴ (m, r) ∈ R

Hence R is transitive.

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”.

Solution:

P = {set of all straight lines in a plane}

lRm ⇒ l is perpendicular to m

(a) lRl ⇒ l is not perpendicular to l

⇒ It is not reflexive

(b) lRm ⇒ l is perpendicular to m

mRl ⇒ m is perpendicular to l

It is symmetric

(c) l perpendicular to m ⇒ m perpendicular to n ⇒ l is parallel to n It is not transitive

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”.

Solution:

Let F = Father,

M = Mother

G = Male child

H = Female child

A = { F, M, G, H }

The relation R is defined by

a R b if a is not a sister of b.

R = {(F, F), (F, M), (F, G), (F, H), (M, F), (M, M), (M, G), (M, H), (G, F), (G, M), (G, G), (G, H), (H, F), (H, M), (H, H)}

(a) Reflexive:

(F, F) , (M , M), (G, G), ( H, H ) ∈ R

∴ R is reflexive.

(b) Symmetric:

For (G, H) ∈ R, we have (H, G) ∉ R

∴ R is not symmetric.

(c) Transitive:

Suppose in a family if we take mother M , male child-G and female child-H.

H is not a sister of M ⇒ HRM, (H, M) ∈ R

M is not a sister of G ⇒ MRG, (M, G) ∈ R

But H is a sister of G ⇒ HRG, (H, G) ∉ R

Thus, for (H, M), (M, G) ∈ R

we have (H, G) ∉ R

∴ R is transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R

defined by “aRb if a is not a sister of b”.

Solution:

A = {set of all female members of a family}

(a) aRa ⇒ a is a sister of a

It is reflexive

(b) aRb ⇒ a is a sister of b

bRa ⇒ b is the sister of a

⇒ It is symmetric

(c) aRb ⇒ a is a sister of b bRc ⇒ b is a sister of c aRc ⇒ a can be the sister of c It is not transitive.

(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.

Solution:

x + 2y = 1 for x, y ∈ N

There is no x , y ∈ N satisfying x + 2y = 1

∴ The relation R is an empty relation.

An empty relation is symmetric and transitive.

∴ R is symmetric and transitive.

R is not reflexive

Question 2.

Let X = {a, b, c, d} and R = {(a, a), {b, b), (a, c)}. Write down the minimum number

of ordered pairs to be included to R to make it

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

X = {a, b, c, d}

R = {(a, a), (b, b), (a, c)}

(i) To make R reflexive we need to include (c, c) and (d, d)

(ii) To make R symmetric we need to include (c, a)

(iii) R is transitive

(iv) To make R reflexive we need to include (c, c)

To make R symmetric we need to include (c, c) and (c, a) for transitive

∴ The relation now becomes

R = {(a, a), (b, b), (a, c), (c, c), (c, a)}

∴ R is equivalence relation.

Question 3.

Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

Given A = {a, b, c }

R = { (a, a), (b, b),(a, c) }

(i) The minimum ordered pair to be included to R in order to make it reflexive is (c, c).

(ii) The minimum ordered pair to be included to R in order to make it symmetrical is (c, a).

(iii) R is transitive. We need not add any pair.

(iv) After including the ordered pairs (c, c),(c, a) to R the new relation becomes

R_{1} = { (a, a), (b, b), (c, c) , (a, c) , (c, a) }

R_{1} is reflexive symmetric and transitive.

∴ R_{1} is an equivalence relation.

Question 4.

Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is

similar to b. Prove that R is an equivalence relation.

Solution:

P = {set of all triangles in a plane}

aRb ⇒ a similar to b

(a) aRa ⇒ every triangle is similar to itself

∴ aRa is reflexive

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a.

⇒ It is symmetric

(c) aRb ⇒ bRc ⇒ aRc

a is similar to b and b is similar to c

⇒ a is similar to a

⇒ It is transitive

∴ R is an equivalence relation

Question 5.

On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

N = {set of natural numbers}

R ={(3, 8), (6, 6), (9, 4), (12, 2)}

(a) (3, 3) ∉ R ⇒ R is not reflexive

2a + 3b = 30

3b = 30 – 2a

b = \(\frac{30-2 a}{3}\)

(b) (3, 8) ∈ R(8, 3) ∉ R

⇒ R is not symmetric

(c) (a, b) (b, c) ∉ R ⇒ R is transitive

∴ It is not an equivalence relation.

Question 6.

Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.

Solution:

If a is a friend of b and b is a friend of c, then a need not be a friend of c.

a R b and b R c does not imply a R c.

∴ R is not transitive.

∴ The relation is not an equivalence relation.

Question 7.

On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the relation by listing all the pairs. Check whether it is

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

Set of all natural numbers aRb if a + b ≤ 6

R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}

(i) (5, 1) ∈ R but(5, 5) ∉ R

It is not reflexive

(ii) aRb ⇒ bRa ⇒ It is symmetric

(iii) (4, 2), (2, 3) ∈ R ⇒ (4, 3) ∉ R

∴ It is not transitive

(iv) ∴ It is not an equivalence relation

Question 8.

Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?

Solution:

A = { a, b, c }

Let R_{1} = { (a, a),(b, b),(c, c) }

Clearly, R_{1} is reflexive, symmetric, and transitive.

Thus R_{1} is the equivalence relation on A of smallest cardinality, n (R_{1}) = 3

Let R_{2} = { (a, a), (b, b), (c, c), (a, b), (b, a), (b, c), (c, b), (c, a), (a, c)}

(i) Reflexive:

(a, a) , (b, b) , (c, c) ∈ R

∴ R_{2} is reflexive.

(ii) Symmetric:

(a , b) ∈ R_{2} we have (b, a) ∈ R_{2}

(b , c) ∈ R_{2} we have (c, b) ∈ R_{2}

(c , a) ∈ R_{2} we have (a, c) ∈ R_{2}

∴ R_{2} is symmetric.

(iii) Transitive:

(a, b), (b, c) ∈ R_{2} ⇒ (a, c) ∈ R_{2}

(b, c), (c, a) ∈ R_{2} ⇒ (b, a) ∈ R_{2}

(c, a) , (a, b) ∈ R_{2} ⇒ (c, b) ∈ R_{2}

∴ R_{2} is transitive and R_{2} is an equivalence relation of largest cardinality.

n (R_{2}) = 9

Question 9.

In the set Z of integers, define mRn if m – n is divisible by 7. Prove that R is an equivalence relation.

Solution:

mRn if m – n is divisible by 7

(a) mRm = m – m = 0

0 is divisible by 7

∴ It is reflexive

(b) mRn = {m – n) is divisible by 7

nRm = (n – m) = – {m – n) is also divisible by 7

It is symmetric

It is transitive

mRn if m – n is divisible by 7

∴ R is an equivalence relation.

### Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2 Additional Questions

Question 1.

Find the range of the function.

f = {(1, x), (1, y), (2, x), (2, y), (3, z)}

Solution:

The range of the function is {x, y, z}.

Question 2.

For n, m ∈ N, It means that it is a factor of n & m. Then find whether the given relation is an equivalence relation.

Solution:

Since n is a factor of n. So the relation is reflexive.

When n is a factor of m (where m ≠ n) then m cannot be a factor of n.

So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.

Question 3.

Verify whether the relation “is greater than” is an equivalence relation.

Solution:

You can do it yourself.