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

## 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 = k1n
r = k1(km) = (k1k) 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
R1 = { (a, a), (b, b), (c, c) , (a, c) , (c, a) }
R1 is reflexive symmetric and transitive.
∴ R1 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 R1 = { (a, a),(b, b),(c, c) }
Clearly, R1 is reflexive, symmetric, and transitive.
Thus R1 is the equivalence relation on A of smallest cardinality, n (R1) = 3
Let R2 = { (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
∴ R2 is reflexive.

(ii) Symmetric:
(a , b) ∈ R2 we have (b, a) ∈ R2
(b , c) ∈ R2 we have (c, b) ∈ R2
(c , a) ∈ R2 we have (a, c) ∈ R2
∴ R2 is symmetric.

(iii) Transitive:
(a, b), (b, c) ∈ R2 ⇒ (a, c) ∈ R2
(b, c), (c, a) ∈ R2 ⇒ (b, a) ∈ R2
(c, a) , (a, b) ∈ R2 ⇒ (c, b) ∈ R2
∴ R2 is transitive and R2 is an equivalence relation of largest cardinality.
n (R2) = 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.