OP Malhotra Class 12 Maths Solutions Chapter 1 Relations Ex 1

Practicing OP Malhotra Class 12 Solutions Chapter 1 Relations Ex 1 is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 12 ICSE Maths Solutions Chapter 1 Relations Ex 1

Question 1.
Consider the following properties of relations : Symmetric (S), Transitive (T), Reflexive (R), Equivalence (E), None of these (TV).
State which property/properties are satisfied by each of the following relations: (Give the answers in terms of S, T, R, E and N).
(a) “is greater than” for the set of real numbers.
(b) “is the cube of’ for the set of all real numbers.
(c) “is the sister of” for a set of children.
(d) “is similar to” for the set of triangles.
(e) “is perpendicular to” for a set of co-planar lines.
Solution:
(a) Given S be the given relation “ is greater than” for the set of real numbers.
∴ (a, b) ∈ S ⇔ a > b ∀ a, b ∈ R
Reflexive :
since a ≯ a∀ a ∈ R thus, (a, a) ∉ S
∴ S is not reflexive.
Symmetric :
∀ a, b ∈ R s.t (a, b) ∈ S
⇒ a > b ⇒ b < a i.e. b ≱ a i.e (b, a) ∉ S Thus S is not symmetric.

Transitive : Let (a, b) ∈ S & (b, c) ∈ S ∀ a, b, c ∈ R ⇒ a > b and b > c
⇒ a > c ⇒ (a, c) ∈ S
Thus S is transitive.

(b) Given relation S be “ is the cube of’ for the set of all real number.
since 1 is the cube of 1 ∴ (1, 1) ∈ S
since 23 = 8 x 2 ∴ (2, 2) ∉ S
∴ S is not reflexive
since (8, 2) ∈ S as 8 = 23
but (2, 8) ∉ S since 2 ≠ 8³
Thus S is not symmetric
As (a, b) ∈ S and (b, c) ∈ S need not
imply that (a, c) ∈ S
since (512, 8) ∈ S since 512 = 8³
& (8, 2) ∈ S since 8 = 2³
but (512, 2) ∉ S since 512 ≠ 2³
Thus S is not transitive.

(c) Given S = {(a, b) : a is the sister of b}
Reflexive:
since a can’t be the sister of itself
∴ (a, a) ∉ S
∴ S is not reflexive

Symmetric:
Let (a, b) ∈ S i.e. a is a sister of b
Then b may not be a sister of a in case b be a boy. Thus (b, a) may or may not be related to a i.e. (b, a) ∉ R
∴ S is not symmetric

Transitive:
Let (a, b) ∈ S
⇒ a be a sister of b & (b, c) ∈ S
⇒ b be a sister of c
⇒ a be a sister of c
⇒ (a, c) ∈ S
Thus, S is transitive

(d) S = { (a, b) : a is similar to b}
since every triangle is similar to itself
∴ (a, a) ∈ S
Thus S is symmetirc
If (a, b) ∈ S ⇒ a is similar to triangle b
Then triangle b is similar to triangle a
⇒ (b, a) ∈ S, Thus S is symmetric.
Let (a, b) ∈ S, (b, c) ∈ S
i.e. triangle a is similar to triangle b and triangle b is similar to triangle c. Then triangle a is similar to triangle c
∴ (a, c) ∈ S
Thus S is transitive on set of triangles
∴ R, S, T.

(e) Let L be the set of coplanar lines & relation S = {{a, b) : a ⊥ b, a, b ∈ L}
Reflexive : since a line can’t be ⊥ to itself.
∴ (a, a) ∉ S ⇒ S is not reflexive

Symmetric :
Let (a, b) ∈ S ∀ a, b ∈ L
⇒ a ⊥ b ⇒ b ⊥ a ⇒ (b, a) ∈ S
∴ S is symmetric.

Transitive :
Let (a, b) ∈ S, (b, c) ∈ S ∀ a, b, c ∈ L
Now (a, b) ∈ S ⇒ a ⊥ b
& (b, c) ∈ S ⇒ b ⊥ c ,
Then a is not ⊥ to c so a is parallel to c
Thus, (a, c) ∉ S ∴ S is not transitive
∴ S

Question 2.
Write down a relation which is
(a) Only transitive
(b) Only symmetric
(c) Only reflexive and transitive
(d) Only symmetric and reflexive.
Solution:
(a) Let S = {(a, b) : a > b, a, b ∈ R}
since a ≯ a ∴ a (a, a) ∉ S or S is not reflexive.
Let (a, b) ∈ S ∀ a, b ∈ R
⇒ a > b ⇒ b ≯ a ⇒ (b, a) ∉ S
∴ S is not symmetric
Let (a, b) ∈ S, (b, c) ∈ S ∀ a, b, c ∈ R
since (a, b) ∈ S ⇒ a > b … (1)
(b, c) ∈ S ⇒ b > c … (2)
∴ a > c [using (1) & (2)]
⇒ (a, c) ∈ S ∴ S is transitive.

(b) Let S = {(a, b); a is ⊥ to b, a, b ∈ L}
where L be the set of coplaner lines.
since a line can’t be ⊥ to itself
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ S ∀ a, b ∈ L
⇒ a ⊥ b ⇒ b ⊥ a
∴ (b, a) ∈ S
Thus S is symmetric.
Let (a, b) ∈ S, (b, c) ∈ S ∀ a, b, c ∈ L
since (a, b) ∈ S ⇒ a ⊥ b
& (b, c) ∈ S ⇒ b ⊥ c
Thus a is parallel to c. Thus (a, c) ∉ S
Thus S is symmetric only.

(c) Let S = {(a, b); a multiple of b, a, b ∈ R}
Clearly a is a multiple of Ia as a = a.I
∴ (a, a) ∈ S ⇒ S is reflexive.
Let (a, b) ∈ S ⇒ a is a multiple b
⇒ a = λb where λ ∈ I
Clearly b is a factor of Ia i.e. not a multiple of a
∴ (b, a) ∉ S
Thus S is not symmetric.
Let (a, b) ∈ S ⇒ (b, c) ∈ S
Since (a, b) ∈ S ⇒ a is a multiple of b
⇒ a = λb
& (b, c) ∈ S
⇒ b is a multiple of c where λ ∈ I
⇒ b = μc where μ ∈ I
⇒ a = λμc ⇒ a = λ’c where λ’ = λμ ∈ I
Thus a be a multiple of c ⇒ (a, c) ∈ S
Therefore S is transitive.

(d) R = {(a, b); a is a friend of b}
since a must be a friend of itself
∴ (a, a) ∈ R
∴ R is reflexive
Let (a, b) ∈ R ⇒ a is a friend of b
⇒ b is a friend of a ⇒ (b, a) ∈ R
Thus R is symmetric
Let (a, b) ∈ R ⇒ a is a friend of b
(b, c) ∈ R ⇒ b is a friend of c
Then a may not be a friend of c
⇒ (a, c) ∉ R (always)
Hence, R is not transitive.
Aliter : Let A ={ 1, 2, 3} & relation R on
A given by {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3) (3, 1)}
Clearly (1, 1), (2, 2), (3, 3) ∈ R
⇒ R is relflexive
(1, 3), (3, 1), (2, 1), (1, 2)eR
⇒ R is symmetric.
i.e. if (a, b) ∈ R then (b, a) ∈ R
Now (2, 1), (1, 3) ∈ R but (2, 3) ∈ R
i.e. (a, b), (b, c) ∈ R but (a, c) g R
⇒ R is not transitive.

Question 3.
Prove that if A is the set of the members of a family and R means “is brother of” then it is a transitive relation.
Solution:
Given A be the set of members of a family & R means “is brother of’
∀ a ∈ A i.e. a can’t be a brother of itself
⇒ (a, a) ∉ R ⇒ R is not reflexive.
∀ a, b ∈ A s.t. (a, b) e R ⇒ a is a brother of b. Then b may not be a brother of a (in case when b is female).
⇒ (b, a) ∉ R
⇒ R is not symmetric on A.
∀ s.t. (a b) ∈ A, (b, c) ∈ A
Now (a, b) ∈ A ⇒ a is a brother of b
(b, c) ∈ A ⇒ b is a brother of c
Thus a is a brother of c ⇒ (a, c) ∈ R
Therefore, R is transitive on A.

Question 4.
Show that the relation R in the set (1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Solution:
Let A = {1, 2, 3} and relation R on a = {(1, 2), (2, 1)}
Clearly (1, 1), (2, 2), (3, 3) ∉ R
∴ R is not reflexive
Now (1, 2) ∈ R(∀a,b ∈ A) and (2, 1) ∈ R
i.e., (a, b) ∈ R ∀ a, b ∈ A
Then (b, a) ∈ R. ∴ R is symmetric on A.
Clearly (1, 2), (2, 1) ∈ R but (1, 1) ∉ R
∴ R is transitive on A.

Question 5.
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.
Choose the correct answer :
(а) R is reflexive and symmetric but not transitive.
(b) R is reflexive and transitive but not symmetric.
(c) R is symmetric and transitive but not reflexive.
(d) R is an equivalence relation.
Solution:
Let A = {1, 2, 3, 4} and R be a relation on A is given by
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
Clearly (1, 1), (2, 2), (3, 3), (4, 4) ∉ R
⇒ R is reflexive on A.
Since (1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric on A.
Clearly,
(1, 2), (2, 2) ∈ R ⇒ (1, 2) ∈ R
(1, 1), (1, 2) ∈ R ⇒ (1, 2) ∈ R
(1, 1), (1, 3) ∈ R ⇒ (1, 3) ∈ R
(1, 3), (3, 3) ∈ R ⇒ (1, 3) ∈ R
(l, 3), (3, 2) ∈ R ⇒ (1, 2) ∈ R
(3, 2), (2, 2) ∈ R ⇒ (3, 2) ∈ R
∀ a, b, c ∈ A s.t. (a, b) ∈ R, (b, c) ∈ R
Then (a, c) ∈ R
Then R is transitive on A.
Hence, R is reflexive and transitive but not symmetric.
∴ R is reflexive and transitive but not symmetric.

Question 6.
Show that the relation R in the set A of real numbers defined as R = {(a, b) : a ≤ b) is reflexive and transitive but not symmetric
Solution:
Let A be the set of real numbers & R = {{a, b) : a ≤ b)
Reflexive : ∀ a ∈ R, a ≤ a
⇒ (a, a) ∈ R
⇒ R is reflexive.

Symmetric:
Let (a, b) ∈ R ∀ a, b ∈ A
⇒ a ≤ b ⇒ b ≰ a⇒(b, a) ∈ R
∴ R is not symmetric on A.

Transitive:
∀ a, b, c ∈ A s.t. (a, b) ∈ R, (b, c) ∈ R
since (a, b) ∈ R ⇒ a ≤ b … (1)
and (b, c) ∈ R ⇒ b ≤ c … (2)
⇒ a ≤ c [using (1) and (2)]
⇒ (a, c) ∈ R
Thus, R is transitive on A.
Hence, R is reflexive and transitive but not symmetric.

Question 7.
Show that the relation R in the set A of real numbers defined as R = {(a, b): a < b²} is neither reflexive nor symmetric nor transitive.
Solution:
Given relation R on A set of real numbers is given by R = {(a, b) : a < b²}
since \(\frac { 1 }{ 2 }\) ∈ A but \(\frac { 1 }{ 2 }\) ≮ (\(\frac { 1 }{ 2 }\))² = \(\frac { 1 }{ 4 }\)
since \(\frac { 1 }{ 4 }\) < \(\frac { 1 }{ 2 }\) ∴ (\(\frac { 1 }{ 2 }\),\(\frac { 1 }{ 2 }\)) ∉ R
∴ R is not reflexive.
since (a, a) ∉ R ∀ a ∈ A

Symmetric:
Let (a, b) ∈ R ∀ a < b ∈ A it need not
imply that (b, a) ∈ R
e.g. (1, 2) ∈ R since 1 ≤ 2² = 4
but (2, 1) ∉ R since 2 < 1² = 1
Thus, R is not symmetric on A.
Transitive : (a, b), (b, c) ∈ R it does not imply (a, c) ∈ R
e.g. (40, 7) ∈ R since 40 ≤ 7² = 49
(7, 3) ∈ R since 7 ≤ 3² = 9
but (40, 3) ∉ R since 40 ≮ 32 = 9
Thus, R is not transitive on A.
Hence R is neither reflexive nor symmetric and transitive on A.

Question 8.
In the set of all triangles in a plane, show that the relation of similarity is an equivalence relation.
Solution:
Let S = set cf all triangles in a plane & relation R on S is given by R = {{a, b), a is similar to b}
Reflexive :
∀a ∈ S since every triangle is similar to itself
∴ (a, a) ∈ R
⇒ a is reflexive on S.
Symmetric :
∀ a, b ∈ S s.t. (a, b) ∈ S.
⇒ a is similar to triangle b
Then b is similar to triangle a
⇒ (a, b) ∈ R
Thus, R is symmetric.
Transitive:
∀ a, b, c ∈ S s.t. (a, b), (b, c) ∈ R
Now (a, b) ∈ R ⇒ a is similar to b
& (b, c) ∈ R ⇒ b is similar to c.
Thus a is similar to c ⇒ (a, c) ∈ R
∴ R is transitive on S.
Hence, R is reflexive, symmetric and transitive on S. Thus R be an equivalence relation on S.

Question 9.
Is the relation ‘is the square of’ for the set of natural numbers N an equivalence relation?
Solution:
Let R = {(a, b); a is the square of b, a, b ∈ N}
Reflexive : Clearly 2 is not the square of 2
∴ (2, 2) ∉ R as 2 ∈ N
Thus (a, a) ∉ R ∀a ∈ N
∴ R is not reflexive on N.
Hence R is not an equivalence relation on N.

Question 10.
Consider the following properties of relations : Symmetric (S), Transitive (T), Reflexive (R) and Equivalence (E). A relation may have all, some or none of these properties. For each part, state whether the relation has some or all of the properties mentioned by writing the letters S, T, R, E in the space provided. Write N, if none of the properties satisfy.
(a) ‘is smaller than’,
(b) ‘is the father of’,
(c) ‘is parallel to’ for set of straight lines.
(d) ‘is a multiple of for a set of positive integers’
(e) ‘is congruent to’.
Solution:
(a) Let R₁ = {{a, b) : a < b)
Clearly a ≮ a ⇒ (a, a) ∉ R,
∴ R₁ is not reflexive
Let (a, b) ∈ R₁ ⇒ a < b
⇒ b ≮ (as b < a
⇒ (b, a) ∉ R₁
e.g. 1 < 2 ⇏ 2 < 1
∴ R₁ is not symmetric.
Let (a, b) , (b, c) ∈ R₁
⇒ a < b & b < c
⇒ a < c
⇒ (a, c) ∈ R₁
∴ R₁ is transitive only
More over R₁ is not an equivalence relation since R₁ is not reflexive & symmetric.

(b) Let R₁ = {{a, b) : a is the father of b}
since a cannot be the father of itself
∴ (a, a) ∉ R₁ ∀ a
Thus R₁ is not reflexive
Let (a, b) ∈ R₁ ⇒ a is a father of b
Then b be the son/daughter of a
i.e. b is not the father of a
⇒ (b, a) ∈ R₁ ⇒ R₁ is not symmetric.
Let (a, b) ∈ R₁ ⇒ a is a father of b.
and (b, c) ∈ R₁ ⇒ b is a father of c.
Then a is a grand father of c
⇒ (a, c) ∉ R₁
∴ R₁ is not transitive.
Thus, R₁ is neither reflexive, nor symmetric and transitive and hence not an equivalence relation.

(c) Let R₁ = {(a, b) ; a is parallel to b; a, b ∈ L}
Where L be the set of all straight lines.
Since every line is parallel to itself
∴ (a, a) ∈ R
Thus R is reflexive on L.
Symmetric :
∀ a, b ∈ L s.t. (a, b) ∈ R₁
⇒ a is parallel to b
⇒ line b is parallel to line a
⇒ (b, a) ∈ R₁
∴ R₁ is symmetric on L.

Transitive:
∀ a, b, c ∈ L s.t. (a, b) ∈ R₁, (b, c) ∈ R₁
Now (a, b) ∈ R₁ ⇒ a is parallel to b
⇒ (b, c) ∈ R₁ ⇒ b is parallel to c
⇒ a is parallel to c
⇒ (a, c) ∈ R₁
Thus R₁ is transitive on L
Hence R₁ is reflexive symmetric and transitive on L
Thus, R₁ be an equivalence relation on L.

(d) Let R₁ = {(a, b) : a is a multiple of b : a, b ∈ Z}
Clearly a is a multiple of a as a = a. 1
∴ (a, a) ∈ S ⇒ S is reflexive.
Let (a, b) ∈ S ⇒ a is a multiple b
⇒ a = λb where λ ∈ I
Clearly A is a factor of a i.e. not a
multiple of a ∴ (b, a) ∉ S
Thus S is not symmetric.
Let (a, b) ∈ S ⇒ (b, c) ∈ S
Since (a, b) ∈ S ⇒ a is a multiple of b
⇒ a = λb
& (b, c) ∈ S
⇒ b is a multiple of c where λ ∈ I
⇒ b = µc where µ ∈ I
⇒ a = λµc = λ’c where λ’ = λµ ∈ I
Thus a be a multiple of c ⇒ (a, c) ∈ S
Therefore S is transitive.

(e) Let R₁ = {(a, b) : a is congruent to b}
Since every triangle is congruent to itself
∴ (a, a) ∈ R₁
Thus, R₁ is reflexive.
Symmetric : Let (a, b) ∈ R₁
⇒ a is congruent to b
Then triangle b is congruent to triangle a
⇒ R₁ is symmetric.
Transitive : Let (a, b) ∈ R₁
⇒ a is congruent to b
& (b, c) ∈ R₁ ⇒ b is congruent to c
Thus a is congruent to c ⇒ (a, c) ∈ R₁
∴ R₁ is transitive.
Hence, R₁ is reflexive, symmetric and transitive and thus R₁ be an equivalence relation set of all triangles in a plane.

Question 11.
Answer true or false:
The relation “is congruent to” in a set of triangles is a transitive relation.
Solution:
Let R₁ = {(a, b) : a is congruent to b}
Since eveiy triangle is congruent to itself
∴ (a, a) ∈ R₁
Thus, R₁ is reflexive.
Symmetric : Let (a, b) ∈ R₁
⇒ a is congruent to b
Then triangle b is congruent to triangle a
⇒ R₁ is symmetric.
Transitive : Let (a, b) ∈ R₁
⇒ a is congruent to b
& (b, c) ∈ R₁ ⇒ b is congruent to c
Thus a is congruent to c ⇒ (a, c) ∈ R₁
∴ R₁ is transitive.
Hence, R₁ is reflexive, symmetric and transitive and thus R₁ be an equivalence relation set of all triangles in a plane.

Question 12.
If is a relation in N x N defined by (a, b) R (c, d) if and only if a + d = b + c, show that R is an equivalence relation.
Solution:
Given R be relation in N x N defined by
(a, b) R (c, d) iff a + d = b + c
Reffexive : ∀ (a, b) ∈ N x N
s.t. (a, b) R (a, b)
⇒ a + b = b + a which is true
[since commutative law holds under addition for natural numbers]
Thus R is reflexive on N x N.

Symmetric :
Let (a, b) R (c, d) ∀ a, b, c, d ∈ N
⇒ a + d = b + c
⇒ d + a = c + b
⇒ c + b = d + a ⇒ (c, d) R (a, b)
∴ R is symmetric on N x N.

Transitive :
Let a, b, c, d, e, f ∈ N
s.t. (a, b) R (c, d) & (c, d) R(e, f)
since (a, b) R(c, d)
⇒ a + d = b + c … (1)
and (c, d) R(e, f)
⇒ c + f = d + e … (2)
on adding (1) & (2); we have
a + d + c + f = b + c + d + e
[Since commutative law under addtition holds on set f natural numbers]
⇒ (a, b) R (e, f)
⇒ a + f = b + e
Thus R is transitive on N x N.
Hence, R be an equivalence relation on N x N.

Question 13.
Prove that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): | a – b | is even} is an equivalence relation.
Solution:
Given A = { 1, 2, 3, 4, 5}
& relation R on A be given by R = {{a, b) : |a – b| is even}
Reflexive : Now (a, a) ∈ R
⇒ |a – a| is even ⇒ 0 is even, which is true
∴ R is reflexive on A.

Symmetric:
Let (a, b) ∈ R ∀ a, b ∈ A
⇒ |a – b| is even
⇒ |a – b| = 2 m , m ∈ Z
⇒ |-(b – a) | = 2 m
⇒ |b – a| = 2 m ⇒ |b – a| be even
⇒ (b, a) ∈ R
Thus, R is symmetric on A.

Transitive:
∀ a, b, c ∈ A s.t. (a, b), (b, c) ∈ R
Now (a, b) ∈ R ⇒ |a – b| is even
⇒ a – b = ± 2m … (1)
& (b, c) ∈ R ⇒ |b – c| is even
⇒ b – c = ± 2n … (2)
On adding (1) & (2), we have
a – c = ± 2(m + n) = ± 2m’
where m + n = m’ ∈ Z
⇒ |a – c| – 2m’ ⇒ (a, c) ∈ R
Thus, R is transitive on A.
Hence, R is reflexive, symmetric and transitive on A.
Thus, R be on equivalence relation on A.

Question 14.
Let I be the set of all integers and R be the relation on I defined by a R b iff (a + b) is an even integer for all a, b ∈ I. Prove that R is an
equivalence relation.
Solution:
Given I be the set of integers and R be the relation on I defined by a R b iff a + b is an even integer ∀ a, b ∈ I
Reflexive : Now a R a ⇒ a + a = 2a,
which is clearly an even integer ∀ as I.

Symmetric :
∀ a, b ∈ I s.t. (a, b) ∈ R
⇒ a + b is an even integer
⇒ b + a is an even integer
⇒ (b, a) ∈ R
∴ R is symmetric on I.

Transitive:
∀ a, b, c ∈ I s.t. (a, b), (b, c) ∈ R
Now (a, b) ∈ R
⇒ a + b is an even integer
⇒ a + b = 2m, where m ∈ I
& (b, c) ∈ R
⇒ b + c is an even integer
⇒ b + c = 2n, where n ∈ I
∴ a + b + b + c = 2(m + n)
⇒ a + c = 2(m + n – b) = 2m’
where m’ = m + n – b
since m, n & b ∈ I
Thus, a + c is an even integer
⇒ (a, c) ∈ R
Hence, R be an transitive relation on I.
Thus, R is reflexive, symmetric and transitive relation on I. Hence R be an equivalence relation on I.

Question 15.
Let I be the set of all integers and R be the relation on I defined by R – {(x, y): x, y ∈ I, x – y is divisible by 11}. Prove that R is an equivalence relation.
Solution:
Let I be the set of all integers and R be the relation on I defined by
R = {(x, y) : x, y ∈ I, x – y is divisible by 11}
Reflexive :
∀ x ∈ I, x – x is divisible by 11
i.e. 0 is divisible by 11, which is true
⇒ (x, x) ∈ R
Thus, R is reflexive on I.

Symmetric :
∀ x, y ∈ I s.t. (x, y) ∈ R
⇒ x – y is divisible by 11
⇒ x – y = 11K where K ∈ I
⇒ y – x = – 11K = 11K’
where K’ = – K ∈ I
⇒ y – x is divisible by 11.
⇒ (y, x) ∈ R
∴ R is symmetric on I.

Transitive :
∀ x, y, z ∈ I s.t. (x, y), (y, z) ∈ R
Now (x, y) ∈ R ⇒ x – y is divisible by 11.
⇒ x – y = 11K₁, where K₁ ∈ I … (1)
& (y, z) ∈ R ⇒ y – z is divisible by 11
⇒ y – z = 11K₂, where K₂ ∈ I … (2)
On adding (1) & (2); we have
x – y + y – z = 11 (K₁ + K₂)
⇒ x – z = 11K’, where K’ = K₁ + K₂ ∈ I
⇒ x – z is divisible by 11.
∴ (x, z) ∈ I
Thus, R is transitive on I.
Hence R is reflexive, symmetric and transitive relation on I.
Thus, R be an equivalence relation on I.