Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions 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 “m R n if m divides n”.
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “ l R m if l is perpendicular to m”.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by “a R b if a is not a sister of b”.
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by “a R b if a is not a sister of b”.
(v) On the set of natural numbers the relation R defined by “x R y if x + 2y = 1” (i) The relation R defined on the set of all positive integers by “m R n if m divides n”.
S = {set of all positive integers}

(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive

(b) mRn ⇒ m divides n but
nRm ⇒ n does not divide m
(i.e.,) mRn ≠ nRm
It is not symmetric

(c) mRn ⇒ nRr as n divides r
It is transitive (ii) Let P denote the set of all straight lines in a plane. The relation R defined by“ l R m if l is perpendicular to m”.
Let P denote the set of all straight lines in a plane. The relation R is defined by l R m if l is perpendicular to m.
R = {(l, m): l is perpendicular to m}

(a) Reflexive:
Let l be any line in the plane P. Then line l is not perpendicular to itself.
{1, 1) ∉ R
∴ R is not reflexive.

(b) Symmetric:
Let (1, m) ∉ R ⇒ l is perpendicular to m
∴ m is perpendicular to l.
Hence (m, l) ∈ R
∴ R is symmetric.

(c) Transitive;
Let (l, m), (m, n) ∈ R
⇒ l is perpendicular to m.
∴ l is parallel to n. (l , n) ∉ R
Hence R is not transitive. (iii) Let A be the set consisting of all the members of a family. The relation R defined by “a R b if a is not a sister of b”.
A = {set of all members of the family}
aRb is a is not a sister of b

(a) aRa ⇒ a is not a sister of a It is reflexive

(b) aRb ⇒ a is not a sister of b.
bRa ⇒ b is not a sister of a.
It is symmetric

(c) aRb ⇒ a is not a sister of b.
bRc ⇒ b is not a sister of c.
⇒ aRc ⇒ a can be a sister of c
It is not transitive. (iv) Let A be the set consisting of all the female members of a family. The relation R defined by “a R b if a is not a sister of b”.
Given A is the set containing female members of the family.
Let M = Mother
H = Female child
A = { M, H }
The relation R on A is defined by aRb if a is not a sister of b.
R = {(M, M), (M, H), (H, H), (H, M)}

(a) Reflexive:
Clearly (M, M) and (H, H) ∈ R.
∴ R is reflexive.

(b) Symmetric:
For (M, H) ∈ R, we have (H, M ) ∈ R
∴ R is symmetric.

(c) Transitive:
For (M,M),(M,H) ∈ R ⇒ (M, H) ∈ R
(M,H),(H,H) ∈ R ⇒ (M, H) ∈ R
(H,H),(H,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,H) ∈ R (H, H) ∈ R
∴ R is transitive. (v) On the set of natural numbers the relation R defined by “x R y if x + 2y = 1”
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set

(a) xRx ⇒ x + 2x = 1 ⇒ x = $$\frac{1}{3}$$ ∉ N. It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = $$\frac{1-x}{2}$$ It is not symmetric.

(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.

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
Given X = { a, b, c, d }
R = { (a, a), (b, b), (a, c) }
(i) The minimum ordered pairs to be included to R in order to make R to be reflexive is (c, c) and (d, d)
(ii) The minimum ordered pairs to be included to R in order to make R to be symmetric is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After adding the ordered pairs (c, c),(d, d), (c, a) the new relation becomes
R, = {(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)}
The new relation satisfies, reflexive, symmetric and transitive property.
∴ R1 is an 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
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 a R b if a is similar to b. Prove that R is an equivalence relation.
Given P = the set of all triangles in a plane.
R is the relation defined by a R b if a is similar to b.
R = {(a, b) : a is similar to b for a, b ∈ p }

(a) Reflexive:
(a, a) ⇒ a is similar to a for all a ∈ P
∴ R is reflexive.

(b) Symmetric: .
Let (a,b) ∈ R ⇒ a is similar to b
⇒ b is similar to a
∴ (b, a) ∈ R
Hence R is symmetric.

c) Transitive:
Let (a, b) and ( b, c) ∈ R
(a, b) ∈ R ⇒ a is similar to b
(b, c) ∈ R ⇒ b is similar to c
∴ a is similar to c.
Hence R is transitive.
∴ R is an equivalence relation on P. Question 5.
On the set of natural numbers let R be the relation defined by a R b if 2a + 3b = 30. Write down the relation by listing all the pairs. Cheek whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Given N = set of natural numbers
R is the relation defined by a R b if 2a + 3b = 30  When a > 15, b negative and does not belong to N.
∴ R = { (3,8),(6,6), (9,4), (12,2)}.
(i) R is not reflexive since (a, a) ∉ R for all a ∈ N.
(ii) R is not symmetric since for (3, 8) ∈ R, (8, 3) ∉ R
(iii) Clearly R is transitive since we cannot find elements (a, b), (b, c) in R such that (a, c) ∉ R
∴ R 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.
(a) S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflective.

(b) aRb ⇒ bRa so it is symmetric

(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation

Question 7.
On the set of natural numbers let R be the relation defined by a R b 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.
N = the set of natural numbers.
R is the relation defined on N by
a R b if a + b ≤ 6
R = { (a, b), a, b ∈ N / a + b ≤ 6}
a + b ≤ 6 ⇒ b ≤ 6 – a

a = 1,
b ≤ 6 – 1 = 5
b is 1, 2, 3, 4, 5
∴ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) ∈ R

a = 2,
b ≤ 6 – 2 = 4
b is 1, 2, 3, 4
∴ (2, 1), (2, 2),(2, 3), (2, 4) ∈ R

a = 3,
b < 6 – 3 = 3
b is 1, 2, 3
∴ (3, 1), (3, 2), (3, 3) ∈ R

a = 4 ,
b < 6 – 4 = 2
b is 1, 2
∴ (4, 1), (4, 2) ∈ R

a = 5,
b < 6 – 5 = 1
b is 1
∴ (5, 1) ∈ R
∴ 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) Reflexive:
R is not reflexive since (4, 4), (5, 5) ∈ R

(ii) Symmetric:
Cleary R is symmetric forever (x, y) ∈ R, we have (y, x) ∈ R.

(iii) Transitive:
(3, 1), (1, 5) ∈ R ⇒ (3,5) ∉ R
∴ R is not transitive.

(iv) R 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?
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3

(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}
n(R) = 9 is the largest cardinality of R to make it equivalence. Question 9.
In the set Z of integers, define m Rn if m – n is divisible by 7. Prove that R is an equivalence relation.
Z = set of all integers
Relation R is defined on Z by m R n if m – n is divisible by 7.
R = {(m, n), m, n ∈ Z/m – n divisible by 7}
m – n divisible by 7
∴ m – n = 7k where k is an integer.

a) Reflexive:
m – m = 0 = 0 × 7
m – m is divisible by 7
∴ (m, m ) ∈ R for all m ∈ Z
Hence R is reflexive.

b) Symmetric:
Let (m, n ) ∈ R ⇒ m – n is divisible by 7
m – n = 7k
n – m = – 7k
n – m = (-k)7
∴ n – m is divisible by 7
∴ (n, m) ∈ R. c) Transitive:
Let (m, n) and (n , r) ∈ R
m – n is divisible by 7
m – n = 7k ——— (1)
n – r is divisible by 7
n – r = 7k1 ——— (2)
(m – n) + (n – r) = 7k + 7k1
m – r = ( k + k1) 7
m – r is divisible by 7.
∴ (m, r) ∈ R
Hence R is transitive.
R is an equivalence relation.