Determine whether each of the following relations in the set A of human beings in a city at a particular time are reflexive, symmetric and transitive R = { ( x , y ) : x , y ∈ A , x and y are hardworking } .What are the advantages of students being honest, obedient and hardworking? (Value Based)

Given that, A be the set of all students at a particular time. R = { ( x , y ) : x , y ∈ A , x and y are hardworking } Symmetric: ∀ x , y ∈ A x and y are hardworking. y and x are hardworking. ⇒ ( y , x ) ∈ R ⇒ R is symmetric. Transitive: ∀ x , y , z ∈ A Now, ( x , y ) ∈ R ⇒ x and y are hardworking. And ( y , z ) ∈ R ⇒ y and z are hardworking. ∴ x and z are hardworking. ⇒ ( x , z ) ∈ R ⇒ R is transitive. Thus, R is symmetric and transitive. If the students are honest, obedient and hardworking then they will definitely rise in their life and will become ideal citizen of the country.Hence the values promoted are honesty, obedient and hardworking.

Let R be the relation on W (set of whole numbers) defined by R = { ( x , y ) : x , y ∈ W , 3 x + 2 y = 12 } .Check the relation R for reflexivity, symmetricity and transitivity.

Given that, R be the relation on W (set of whole numbers) defined by R = { ( x , y ) : x , y ∈ W , 3 x + 2 y = 12 } . Since 2 y = 12 - 3 x ⇒ y = 12 - 3 x 2 When x = 0 , = 12 - 0 2 = 6 ∈ W ∴ ( 0 , 6 ) ∈ R When x = 1 , y = 12 - 3 2 = 9 2 ∉ W When x = 2 , y = 12 - 6 2 = 3 ∈ W ∴ ( 2 , 3 ) ∈ R When x = 3 , y = 12 - 9 2 = 3 2 ∉ W When x = 4 , y = 12 - 12 2 = 0 ∈ W ∴ ( 4 , 0 ) ∈ R For other values of x we don't get the values of y that belongs to set of whole numbers. ∴ R = { ( 0 , 6 ) , ( 2 , 3 ) , ( 4 , 0 ) } Clearly ( 1 , 1 ) ∉ R ∴ R is not reflexive on W . Further ( 0 , 6 ) ∈ R but ( 6 , 0 ) ∈ R ∴ R is not symmetric on W . Clearly ( 4 , 0 ) , ( 0 , 6 ) ∈ R but ( 4 , 6 ) ∉ R R is not transitive on W . Hence, R is neither reflexive, symmetric nor transitive.

Let R be the relation defined on Q 0 (set of non-zero rational numbers) defined by R = ( a , b ) : a , b ∈ Q 0 , b = 1 a . Check the relation R for reflexivity, symmetricity and transitivity.

Given that, R be the relation defined on Q 0 (set of non-zero rational numbers) defined by R = ( a , b ) : a , b ∈ Q 0 , b = 1 a . Clearly ( 1 , 1 ) ∈ R as 1 = 1 1 But ( 2 , 2 ) ∈ R as 2 ≠ 1 2 ∴ R is not reflexive on Q 0 ∀ a ∈ Q 0 Now, ( a , a ) ∈ R ⇒ a = 1 a ⇒ a 2 = 1 ⇒ a - ± 1 For any other ordered pair reflexivity does not hold. Symmetry: ∀ a , b ∈ Q 0 , ( a , b ) ∈ R ⇒ b = 1 a ⇒ a = 1 b ⇒ ( b , a ) ∈ R ∴ R is symmetric on Q 0 . Transitivity : ∀ a , b , c ∈ Q 0 , ( a , b ) ∈ R and ( b , c ) ∈ R Since ( a , b ) ∈ R ⇒ b = 1 a And ( b , c ) ∈ R ⇒ c = 1 b ∴ c = 1 1 / a = a c ≠ 1 a ⇒ ( a , c ) ∈ R R is not transitive on Q 0 . Clearly 2 , 1 2 · 1 2 , 2 ∈ R but ( 2 , 2 ) ∉ R Thus R is neither reflexive nor transitive but symmetric.

Let R be the relation on R (set of real numbers) defined by R = { ( a , b ) : a , b ∈ R , a 2 + b 2 = 1 . Show that R is symmetric but neither reflexive nor transitive.

Given that, R is the relation defined by R = ( a , b ) : a , b ⊂ R , a 2 + b 2 = 1 Reflexivity: ( a , a ) ∈ R ∀ a ∈ R ⇒ a 2 + a 2 = 1 ⇒ a 2 = 1 2 ⇒ a = ± 1 2 1 2 , 1 2 : - 1 2 , - 1 2 ∈ R Also 1 ∈ R but 1 2 + 1 2 ≠ 1 ∴ ( 1 , 1 ) ∉ R ∴ R is not reflexive. Symmetry: ∀ a , b ∈ R ( a , b ) ∈ R ⇒ a 2 + b 2 = 1 ⇒ b 2 + a 2 = 1 Commutative law holds under addition in R ⇒ ( b , a ) ∈ R ∴ R is symmetric on real number system Transitivity: ∀ a , b , c ∈ R ( a , b ) , ( b , c ) ∈ R Then ( a , c ) ∈ R Now, ( 1 , 0 ) , ( 0 , 1 ) ∈ R But ( 1 , 1 ) ∉ R Since 1 2 + 1 2 = 2 ≠ 1 Thus R is not transitive on real number system. Hence, R is symmetric but neither reflexive nor transitive.

Let T be the set of all triangles drawn in a plane with R as a relation in T given by R = T 1 , T 2 : T 1 ≡ T 2 . Show that R is an equivalence relation.

Let R be relation on T defined by a R b If a is congruent to b , ∀ a , b ∈ T where T be the set of all triangles in the Euclidean plane. ∀ a ∈ T , we know that every triangle is congruent to itself. ∴ a is congruent to a . ∴ a Ra ∴ R is reflexive on T . Let ( a , b ) ∈ R , ∀ a , b ∈ T ∴ a is congruent to b . Thus triangle b is congruent to triangle a . ( b , a ) ∈ R b R a Hence R is symmetric on T . Let ( a , b ) , ( b , c ) ∈ R ∀ a , b , c ∈ T Since ( a , b ) ∈ R ⇒ Triangle a is congruent to triangle b . And ( b , c ) ∈ R ⇒ Triangle b is congruent to triangle c . ∴ Triangle a is congruent to triangle c . ∴ R is transitive on T . Hence, R is reflexive, symmetric and transitive on T . Therefore R is equivalence relation on set T .

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

MEDIUM

12th ICSE

IMPORTANT

Earn 100

Determine whether following relation in the set $A$ of human beings in a city at a particular time are reflexive, symmetric and transitive $R = {(x, y) : x, y \in A$ , $x$ and $y$ are obedient $}$ .

Important Questions on Relations and Functions

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

Determine whether each of the following relations in the set $A$ of human beings in a city at a particular time are reflexive, symmetric and transitive $R = {(x, y) : x, y \in A$ , $x$ and $y$ are hardworking $}$ .What are the advantages of students being honest, obedient and hardworking? (Value Based)

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

Show that the relation $R$ in the set $A = {1, 2, 3}$ given by $R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}$ is reflexive but neither symmetric nor transitive.

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

Let $R$ be the relation on $W$ (set of whole numbers) defined by $R = {(x, y) : x, y \in W, 3 x + 2 y = 12}$ .Check the relation $R$ for reflexivity, symmetricity and transitivity.

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

Let $R$ be the relation defined on $Q_{0}$ (set of non-zero rational numbers) defined by $R = \{(a, b) : a, b \in Q_{0}, b = \frac{1}{a}\}$ . Check the relation $R$ for reflexivity, symmetricity and transitivity.

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 1 - Relations and Functions>EXERCISE 1.1

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