Show that the relation R on R is defined as R = { ( a , b ) : a ≤ b } , is reflexive and transitive but not symmetric.

Given, R = a , b : a ≤ b , a , b ∈ R For the relation R to be reflexive: If the relation R is reflexive, then a , a ∈ R i.e. a ≤ a , which is always true for all the values of a , a ∈ R So, the given relation R is reflexive. For the relation R to be symmetric: If a , b ∈ R then b , a ∈ R i.e. if a ≤ b then b ≤ a , which is not possible for all the values of a , b ∈ R . For example: if a = 2 and b = 5 Here, a < b but b > a So, the given relation R is not symmetric. For the relation R to be transitive: If a , b ∈ R and b , c ∈ R then a , c ∈ R i.e.,if a ≤ b and b ≤ c then a ≤ c , which is true for all the values of a , b , c ∈ R . For example: if a = 2 , b = 3 and c = 4 Here, a < b < c So, the given relation is transitive. Hence, the given relation R is reflexive and transitive but not symmetric.

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 be the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N . Show that R is an equivalence relation.

Given that, N is the set of natural numbers and R be the relation on N × N defined by ( a , b ) R ( c , d ) iff a d = b c for all a , b , c , d ∈ N We know that, a relation R that is reflexive, symmetric and transitive is called an equivalence relation. Reflexive: For any ( a , b ) ∈ N × N a · b = b · a ∴ ( a , b ) R ( a , b ) Thus, R is reflexive Symmetric: For any ( a , b ) , ( c , d ) ∈ N × N ( a , b ) R ( c , d ) ⇒ a · d = b · c ⇒ c · b = d · a ⇒ ( c , d ) R ( a , b ) Thus, R is symmetric Transitive: For any ( a , b ) , ( c , d ) , ( e , f ) , ∈ N × N ( 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 ) Thus, R is transitive Therefore, R is an equivalence Relation.

Let n be a fixed positive integer. Define a relation R on Z as follows: ( a , b ) ∈ R ⇔ a − b is divisible by n . Show that R is an equivalence relation on Z .

We have, n a fixed positive integer and relation R on Z as ( a , b ) ∈ R ⇔ a − b is divisible by n Let a ∈ R , t hen a − a = 0 = 0 × n ⇒ a − a is divisible by n ⇒ ( a , a ) ∈ R for all a ∈ Z So , R is reflexive on Z Let ( a , b ) ∈ R Here , a − b is divisible by n ⇒ a − b = n p for some p ∈ Z ⇒ b − a = n ( − p ) ⇒ b − a is divisible by n [ p ∈ Z ⇒ − p ∈ Z ] ⇒ ( b , a ) ∈ R So , R is symmetric on Z Let ( a , b ) and ( b , c ) ∈ R Here , a − b is divisible by n and b − c is divisible by n . ⇒ a − b = n p for some p ∈ Z and b − c = n q for some q ∈ Z Adding both, we get a − b + b − c = n p + n q ⇒ a − c = n ( p + q ) ⇒ ( a , c ) ∈ R for all a , c ∈ Z So , R is transitive on Z Hence, R is an equivalence relation on Z .

Defines a relation on N : x y is square of an integer, x , y ∈ N Determine the above relation is reflexive, symmetric and transitive.

We have, R = { ( x , y ) : x y is square of an integer, x , y ∈ N } As , x · x = x 2 , which is a square of an integer x ⇒ ( x , x ) ∈ R So , R is a reflexive relation. Let ( x , y ) ∈ R ⇒ x y is square of an integer ⇒ y x is also a square of an integer ⇒ ( y , x ) ∈ R So , R is a symmetric relation. Let ( x , y ) ∈ R and ( y , z ) ∈ R ⇒ x y is square of an integer and y z is also a square of an integer ⇒ x z must be a square of an integer ⇒ ( x , z ) ∈ R So , R is a transitive relation

Let Z be the set of all integers and Z 0 be the set of all non-zero integers. Let a relation R on Z × Z 0 be defined as ( a , b ) R ( c , d ) ⇔ a d = b c , for all ( a , b ) , ( c , d ) ∈ Z × Z 0 , Prove that R is an equivalence relation on Z × Z 0 .

Let ( a , b ) be part of Z × Z 0 . Then , ( a , b ) ∈ Z × Z 0 ⇒ a b = b a ⇒ ( a , b ) R a , b So , R is reflexive on Z × Z 0 . Let ( a , b ) , ( c , d ) ∈ Z × Z 0 Then , ( a , b ) R ( c , d ) ⇒ a d = b c ⇒ c b = d a ⇒ ( c , d ) R ( a , b ) Thus , ( a , b ) R ( c , d ) ⇒ ( c , d ) R ( a , b ) So, R is symmetrical Z × Z 0 Let ( a , b ) , ( c , d ) , ( e , f ) ∈ Z × Z 0 . Then, ( a , b ) R ( c , d ) ⇒ a d = b c ( c , d ) R ( e , f ) ⇒ c f = d e Multiplying both, we get ( a d ) ( c f ) = ( b c ) ( d e ) ⇒ a f = b e ⇒ ( a , b ) R ( e , f ) So, R is transitive on Z × Z 0 So, R is an equivalence relation on Z × Z 0 .

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Important Questions on Integrals

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Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

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Sulphur sol is a type of

MEDIUM

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

R

be a relation defined on

ℕ

as a

R

b is

2 a + 3 b

is a multiple of

5, a, b \in ℕ

. Then

R

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

A relation

R

in a set

A

is called _____, if

(a_{1}, a_{2}) \in R

implies

(a_{2}, a_{1}) \in R

, for all

a_{1}, a_{2} \in A

MEDIUM

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Show that the relation

R

R

is defined as

R = {(a, b) : a \leq b}

, is reflexive and transitive but not symmetric.

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Which of the following statement is NOT correct about solution?

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

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.

HARD

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

N

be the set of natural numbers and

R

be the relation on

N \times N

defined by

(a, b) R (c, d) iff a d = b c

for all

a, b, c, d \in N

. Show that

R

is an equivalence relation.

HARD

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

A = {x \in Z : 0 \leq x \leq 12}

. Show that

R = {(a, b) : a, b \in A, | a - b | is divisible by 4}

is an equivalence relation. Find the set of all elements related to

1

. Also write the equivalence class

[2]

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

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

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let A be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive:

R = {(x, y) : x

is father of

y}

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Test whether the following relation $R_{2}$ is reflexive, symmetric and transitive :

$R_{2}$ on $Z$ defined by $(a, b) \in R_{2} \Leftrightarrow | a - b | \leq 5$

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let $A$ be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive :

R = {(x, y) : x

and $y$ work at the same place

}

MEDIUM

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

n

be a fixed positive integer. Define a relation

R

Z

as follows:

(a, b) \in R \Leftrightarrow a - b

is divisible by

n

. Show that

R

is an equivalence relation on

Z

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

A

be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive:

R = {(x, y) : x

is wife of

y}

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

A

be the set of all human beings in a town at a particular time. Determine whether of the following relation is reflexive, symmetric and transitive :

R =

{(x, y) : x

and

y

live in the same locality

}

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Defines a relation on $N :$ $x y$ is square of an integer, $x, y \in N$

Determine the above relation is reflexive, symmetric and transitive.

EASY

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

A = {a, b, c}

and the relation $R$ be defined on $A$ as follows:

R = \{(a, a), (b, c), (a, b)\}

. Then, write minimum number of ordered pairs to be added in $R$ to make it reflexive and transitive.

MEDIUM

Mathematics>Integral Calculus>Integrals>Integration by Partial Fractions

Let

Z

be the set of all integers and

Z_{0}

be the set of all non-zero integers. Let a relation

R

Z \times Z_{0}

be defined as

(a, b) R (c, d) \Leftrightarrow a d = b c

, for all

(a, b), (c, d) \in Z \times Z_{0}

, Prove that

R

is an equivalence relation on

Z \times Z_{0}

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Important Questions on Integrals

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