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 S be the set of all real numbers and let R = f a , b : a , b ∈ S and a = ± b }. Show that R is an equivalence relation on S .

Given that, ∀ a , b ∈ S , R = ( a , b ) : a = ± b R is reflexive if a , a ∈ R ∀ a ∈ S For any a ∈ S , we have a = ± a ⇒ ( a , a ) ∈ R Thus, R is reflexive. R is symmetric if ( a , b ) ∈ R ⇒ ( b , a ) ∈ R ; a , b ∈ S Let ( a , b ) ∈ R ⇒ a = ± b ⇒ b = ± a ⇒ ( b , a ) ∈ R Thus, R is symmetric . R is transitive if ( a , b ) ∈ R and ( b , c ) ∈ R ⇒ ( a , c ) ∈ R ∀ a , b , c ∈ S Let ( a , b ) ∈ R and ( b , c ) ∈ R ∀ a , b , c ∈ S ⇒ a = ± b and b = ± c ⇒ a = ± c ⇒ ( a , c ) ∈ R Thus, R is transitive. Since, R is reflexive, symmetric and transitive, it is an equivalence relation. Hence proved

Let R = { ( a , b ) : a , b ∈ Z and a + b is even } . Show that R is an equivalence relation on Z .

Given that, ∀ a , b ∈ Z , R = { ( a , b ) : ( a + b ) is even } R is reflexive if ( a , a ) ∈ R ∀ a ∈ Z For any a ∈ Z , 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

Let S be the set of all sets and let R = A , B : A ⊂ B , i.e., A is a proper subset of B . Show that R is transitive.

Let A , B ∈ R and B , C ∈ R ∀ A , B , C ∈ S ⇒ A is a proper subset of B and B is a proper subset of C ⇒ A is a proper subset of C ⇒ A , C ∈ R For example, if B = 1 , 2 , 5 then A = 1 , 5 is a proper subset of B . And if C = 1 , 2 , 5 , 7 then B = 1 , 2 , 5 is a proper subset of C . We observe that A = 1 , 5 is a proper subset of C also. ⇒ R is transitive. Thus, R is transitive but not reflexive and not symmetric.

On the set S of all real numbers, define a relation R = a , b : a ≤ b . Show that R is reflexive

Let R = a , b : a ≤ b be a relation defined on S . We observe that any element x ∈ S equal to itself. It satisfies the condition x ≤ x ⇒ x , x ∈ R ∀ x ∈ S ⇒ R is reflexive. Hence proved

Let A = 1 , 2 , 3 , 4 , 5 , 6 and let R = { a , b : a , b ∈ A and b = a + 1 } . Show that R is not reflexive.

Given that, A = { 1 , 2 , 3 , 4 , 5 , 6 ) and R = { a , b : a , b ∈ A and b = a + 1 } . ∴ R = { ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 4 , 5 ) , ( 5 , 6 ) } R is reflexive if ( a , a ) ∈ R ∀ a ∈ A Since, ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) ∉ R Thus, R is not reflexive . Hence proved

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

Given that, R be the relation in N × N defined by ( a , b ) R ( c , d ) if 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 . ⇒ 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 ) ⇒ 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 ) - ( d + e ) = ( b + c ) - ( c + f ) ⇒ a - e = b - f ⇒ a + f = b + e ⇒ ( a , b ) R e , f ⇒ R is transitive. Since, R is reflexive, symmetric and transitive, it is an equivalence relation. Hence proved

EASY

Earn 100

Explain a Universal Relation.

Important Questions on Relations and Functions

EASY

Mathematics>Differential Calculus>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.

MEDIUM

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $S$ be the set of all real numbers and let $R = f (a, b) : a, b \in S$ and $a = \pm b$ }. Show that $R$ is an equivalence relation on $S$ .

EASY

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

The relation "is subset of" on the power set

P (A)

of a set

A

MEDIUM

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $R = \{(a, b) : a = b^{2}\}$ for all $a, b \in N$ . Show that $R$ satisfies none of reflexivity, symmetry and transitivity.

EASY

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $A = \{1, 2, 3\}$ and $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ . Show that $R$ is reflexive but neither symmetric nor transitive.

HARD

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $A$ be the set of all triangles in a plane. Show that the relation $R = \{(Δ_{1}, Δ_{2}) : Δ_{1} ~ Δ_{2}\}$ is an equivalence relation on $A$ .

EASY

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Give an example of a relation. Which is symmetric but neither reflexive nor transitive.

HARD

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $A$ be the set of all points in a plane and let $O$ be the origin. Show that the relation $R = {(P, Q) : P, Q \in A$ and $O P = O Q}$ ) is an equivalence relation.

HARD

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $R = {(a, b) : a, b \in Z$ and $(a + b)$ is even $}$ . Show that $R$ is an equivalence relation on $Z$ .

MEDIUM

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $S$ be the set of all real numbers. Show that the relation $R = \{(a, b) : a^{2} + b^{2} = 1\}$ is symmetric but neither reflexive nor transitive.

MEDIUM

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $S$ be the set of all points in a plane and let $R$ be a relation in $S$ defined by $R = {(A, B) : d (A, B) < 2$ units}, where $d (A, B)$ is the distance between the points $A$ and $B$ . Show that $R$ is reflexive and symmetric but not transitive.

EASY

Mathematics>Differential Calculus>Relations and Functions>Types of Relations

Let $A = \{1, 2, 3, 4\}$ and $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)\}$ . Show that $R$ is reflexive and transitive but not symmetric.