TN 11th, Maths Chapter 1, Exercise 1.2
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Class : XI
Subject : Maths
Chapter : 1
Exercise : 1.2
Que 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”.
(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”.
(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”.
(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”.
(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
(i) Solution :
S = {set of all positive integers}
(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive
(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
It is not symmetric
(c) mRn ⇒ nRr as n divides r
It is transitive.
(ii) 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) Solution :
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) 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 a 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 sister of c
It is not transitive.
(v) Solution :
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set
(a) xRx ⇒ x + 2x = 1 ⇒ x = 13 ∉ N.
It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that
y + 2x = 1 as y = 1−x2 It is not symmetric.
(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.
Que 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.
Que 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: 
(i) (c, c)
(ii) (c, a)
(iii) Nothing
(iv) (c, c) and (c, a)
Que 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.
Que 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 = (30−2a)/3
(b) (3, 8) ∈ R(8, 3) ∉ R
⇒ R is not symmetric
(c) (a, b) (b, c) ∉ R ⇒ R is transitive
∴ It is not equivalence relation.
Que 6 :
Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.
Solution :
(a) S = aRa
(i.e. ) a person can be a friend to himself or herself.
So it is reflextive.
(b) aRb ⇒ bRa so it is symmetric
(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation.
Que 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
Que 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:
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3
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.
Que 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
(c)
mRn = (m – n) is divisible by 7
= (m-n)/7
= k/7
nRr = (n - r) is divisible by 7
= (n-r)/7
= l/7
mRr = m-r= [(k/7)+n]-[n-(l/7)]
m-r= k/7+n-n+1/7
m-r =1/7 (k+1) is divisible by 7
It is transitive
mRn if m – n is divisible by 7
∴ R is an equivalence relation.
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