Let Z denote the set of all integers. If a relation R is defined on Z as follows: x , y ∈ R if and only if x is multiple of y , then R is

Reflexive, transitive but not symmetric

Reflexive, symmetric but not transitive

Symmetric, transitive but not reflexive

Neither reflexive nor transitive but symmetric

Let Z denote the set of all integers. If a relation R is defined on Z as follows: x , y ∈ R if and only if x is multiple of y , then R is

Reflexive, transitive but not symmetric

Let R be the real line. Consider the following subsets of the plane R × R S = { ( x , y ) ∣ y = x + 1 , 0 < x < 2 } T = { ( x , y ) ∣ x - y is an integer } . Which one of the following is true?

T is an equivalence relation on R but S is not

Neither S nor T is an equivalence relation on R

Both S and T are equivalence relation on R

S is an equivalence relation on R and T is not

Let R be a relation in N defined by a R b if 5 divides a − b , ( a , b ∈ N ) . Prove that a R b ⇒ a 2 R b 2 .

It is given that a R b if 5 divides a - b , a , b ∈ N . Now, a R b ⇒ a - b = 5 k , for some k . Multiplying both sides by a + b , we get: a - b a + b = 5 k a + b ⇒ a - b a + b = 5 k a + b ⇒ a 2 - b 2 = 5 k a + b ⇒ a 2 - b 2 is divisible by 5 . Hence, a 2 R b 2 . Thus, a R b ⇒ a 2 R b 2 Hence proved.

Let R be the real line. Consider the following subsets of the plane R × R S = { x , y : y = x + 1 and 0 < x < 2 } T = { x , y : x - y is an integer} Which of the following is true?

T is an equivalence relation on R but S is not

Neither S nor T is an equivalence relation on R

Both S and T are equivalence relations on R

S is an equivalence relations on R and T is not

Let the relation $\rho$ be defined on R as a ρ b if 1 + a b > 0 . Then

ρ is reflexive and symmetric but not transitive.

ρ is reflexive and transitive but not symmetric

Let R 1 and R 2 be two relations defined as follows : R 1 = a , b ∈ R 2 : a 2 + b 2 ∈ Q and R 2 = a , b ∈ R 2 : a 2 + b 2 ∉ Q , where Q is the set of all rational numbers, then

Neither R 1 nor R 2 is transitive.

R 1 is transitive but R 2 is not transitive.

R 2 is transitive but R 1 is not transitive.

R 1 and R 2 are both transitive.

Let A = { 1 , 2 , 3 , 4 , 5 } and R be a relation defined by R = { ( x , y ) : x , y ∈ A , x + y = 5 } . Then, R is

neither reflexive nor transitive

reflexive and symmetric but not transitive

neither reflexive nor symmetric but transitive

Let Z be the set of integers. Show that the relation R = { ( a , b ) : a , b ∈ Z and a + b is even } is an equivalence relation.

Given that, ∀ a , b ∈ Z , R = { ( a , b ) : ( a + b ) is even } R is reflexive if ( a , a ) ∈ R ∀ a ∈ Z For any a ∈ A , we have a + a = 2 a , which is even. ⇒ ( a , a ) ∈ R Thus, R is reflexive. R is symmetric if ( a , b ) ∈ R ⇒ ( b , a ) ∈ R ∀ a , b ∈ Z ( a , b ) ∈ R ⇒ a + b is even. ⇒ b + a is even. ⇒ ( b , a ) ∈ R Thus, R is symmetric . R is transitive if ( a , b ) ∈ R and ( b , c ) ∈ R ⇒ ( a , c ) ∈ R ∀ a , b , c ∈ Z Let ( a , b ) ∈ R and ( b , c ) ∈ R ∀ a , b , c ∈ Z ⇒ a + b = 2 P and b + c = 2 Q where P , Q are any two integers Adding both, we get a + c + 2 b = 2 ( P + Q ) ⇒ a + c = 2 ( P + Q ) - 2 b ⇒ a + c is an even number ⇒ ( a , c ) ∈ R Thus, R is transitive on Z . Since R is reflexive, symmetric and transitive it is an equivalence relation on Z . Hence proved

Show that the relation R on N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c , ∀ ( a , b ) , ( c , d ) ∈ N × N is an equivalence relation.

Given that, R be the relation in N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c for a , b , c , d in N × N R is reflexive if ( a , b ) R ( a , b ) for a , b in N × N Let ( a , b ) R ( a , b ) ⇒ a + b = b + a which is true since addition is commutative on N . ⇒ ( a , b ) R ( a , b ) ⇒ R is reflexive. R is symmetric if ( a , b ) R ( c , d ) ⇒ ( c , d ) R ( a , b ) for a , b , c , d in N × N Let ( a , b ) R ( c , d ) ⇒ a + d = b + c ⇒ b + c = a + d ⇒ c + b = d + a [since addition is commutative on N ] ⇒ c , d R ( a , b ) ⇒ ( c , d ) R ( a , b ) ⇒ R is symmetric. R is transitive if a , b R c , d and c , d R e , f ⇔ ( a , b ) R e , f for ( a , b ) , ( c , d ) , e , f in N × N Let a , b R c , d and c , d R e , f ⇒ a + d = b + c and c + f = d + e ⇒ ( a + d ) + ( c + f ) = ( b + c ) + ( d + e ) ⇒ a + f = b + e ⇒ ( a , b ) R e , f ⇒ R is transitive. Since, R is reflexive, symmetric and transitive and hence it is an equivalence relation. Hence, proved

Check whether the relation R defined in the set 1 , 2 , 3 , 4 , 5 , 6 as R = a , b : b = a + 1 is reflexive, symmetric or transitive.

Suppose A = 1 , 2 , 3 , 4 , 5 , 6 . A relation R is defined on set A as: R = { ( a , b ) : b = a + 1 } ∴ R = { ( 1, 2 ) , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.

Let N denote the set of all natural numbers. Define two binary relations on N as R 1 = x , y ∈ N × N : 2 x + y = 10 and R 2 = x , y ∈ N × N : x + 2 y = 10 . Then

both R 1 and R 2 are transitive relations

both R 1 and R 2 are symmetric relations

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Earn 100

Let $Z$ denote the set of all integers. If a relation $R$ is defined on $Z$ as follows:

$(x, y) \in R$ if and only if $x$ is multiple of $y,$ then $R$ is

50% studentsanswered this correctly

Important Questions on Relations and Functions

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

A = \{1, 2, 3, 4\}

then which one of the following is reflexive?

HARD

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

R

and

S

be any two equivalence relations on a set

X

. Then which of the following is incorrect statement

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let $R$ be the real line. Consider the following subsets of the plane $R \times R$

$S = {(x, y) ∣ y = x + 1, 0 < x < 2}$

$T = {(x, y) ∣ x - y$ is an integer $} .$ Which one of the following is true?

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

R

be a relation in

N

defined by

a R b

5

divides

a - b, (a, b \in N)

. Prove that

a R b \Rightarrow a^{2} R b^{2}

MEDIUM

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

R

be the real line. Consider the following subsets of the plane

R \times R

S = {(x, y) : y = x + 1

and

0 < x < 2}

T = {(x, y) : x - y

is an integer}

Which of the following is true?

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

What is meant by an equivalence relation?

MEDIUM

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

A = \{2, 3, 4, 5, \dots ., 16, 17, 18\} .

Let

\approx

be the equivalence relation on

A \times A

cartesian product of

A

and

A

, defined by

(a, b) \approx (c, d)

a d = b c,

then the number of ordered pairs of the equivalence class of

(3, 2)

MEDIUM

Mathematics>Algebra>Relations and Functions>Types of Relations

Let the relation $\rho$ be defined on

R

a ρ b

1 + a b > 0 .

Then

HARD

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

R_{1}

and

R_{2}

be two relations defined as follows :

$R_{1} = \{(a, b) \in R^{2} : a^{2} + b^{2} \in Q\}$ and $R_{2} = \{(a, b) \in R^{2} : a^{2} + b^{2} \notin Q\}$ , where $Q$ is the set of all rational numbers, then

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

A = {1, 2, 3, 4, 5}

and

R

be a relation defined by

R = {(x, y) : x, y \in A, x + y = 5}

. Then,

R

MEDIUM

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

Z

be the set of integers. Show that the relation

R = {(a, b) : a, b \in Z

and

a + b

is even

}

is an equivalence relation.

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Consider the non-empty set consisting of children in a family and a relation

R

defined as

a R b

a

is brother of

b

. Then

R

HARD

Mathematics>Algebra>Relations and Functions>Types of Relations

Show that the relation $R$ on $N \times N$ defined by

$(a, b) R (c, d) \Leftrightarrow a + d = b + c, \forall (a, b), (c, d) \in N \times N$ is an equivalence relation.

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

R = {(P, Q) | P

and

Q

are at the same distance from the origin

}

be a relation, then the equivalence class of

(1, - 1)

is the set

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Consider the following two binary relations on the set

A = \{a, b, c\} : R_{1} = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}

and

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

, then :

HARD

Mathematics>Algebra>Relations and Functions>Types of Relations

Define a relation $R$ over a class of $n \times n$ real matrices $A$ and $B$ as " $A R B$ iff there exists a non-singular matrix $P$ such that $P A P^{- 1} = B$ ". Then which of the following is true ?

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

ρ_{1}

and

ρ_{2}

be two equivalence relations defined on a non-void set

S

. Then

MEDIUM

Mathematics>Algebra>Relations and Functions>Types of Relations

If a relation

R

on the set

\{1, 2, 3\}

be defined by

R = \{(1, 1)\}

, then

R

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Check whether the relation

R

defined in the set

\{1, 2, 3, 4, 5, 6\}

R = \{(a, b) : b = a + 1\}

is reflexive, symmetric or transitive.

EASY

Mathematics>Algebra>Relations and Functions>Types of Relations

Let

N

denote the set of all natural numbers. Define two binary relations on

N

R_{1} = \{(x, y) \in N \times N : 2 x + y = 10\}

and

R_{2} = \{(x, y) \in N \times N : x + 2 y = 10\}

. Then

Let Z denote the set of all integers. If a relation R is defined on Z as follows: x, y∈R if and only if x is multiple of y, then R is

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

Let $Z$ denote the set of all integers. If a relation $R$ is defined on $Z$ as follows:

$(x, y) \in R$ if and only if $x$ is multiple of $y,$ then $R$ is