ABHAY MAHAJANB.Tech. IIT Roorkee➔ 11+ years teaching experience➔ Mentored and taught around 20000+ students➔ Produced 400+ IIT selections ➔ 10+ INMO selections ➔ 1 EGMO Selection

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RelationRelation and Function

● Description of Relation

● Number of Relation

● Inverse of Relation

● Types of Relation

Relation

Description of RelationRelation

CARTESIAN PRODUCT

Defining A X B

A = { 1, 2, 3}B = {a, b}

A X B =

Relation

Any subset of A × B is defined as a relation from set A to set B, where A and B are non-empty sets

Remark:

Relation from A to A is also called relation on A.

For Ex:A = {1, 2, 3}B = {a, b}

R = { (1, a), (2, b)}

Description of relation

There are three ways of writing a relation Let us do it by examples

Let A = {1, 2, 3} & B = {1, 3, 5, 7}

(1) R = {(1,1), (2,3), (3,5)}

a ∈ A b ∈ B

(2) R = {(a,b)| b = 2a –1, and a ∈ A (or) b ∈ B }

(3) a R b ⇔ b = 2a –1, and

or

Q. Write relation on A defined as: x R y ⇔ y = x2 whereA = {1, –1, 2, 3, 4}

Solution

R = {(1, 1), (-1, 1), (2, 4)}

Q. Write relation on A defined as: x R y ⇔ y = x2 whereA = {1, –1, 2, 3, 4}

Domain and RangeRelation

Domain and Range

Let R be a relation from a set A to a set B.

Then, Dom (R) = {a : (a, b) ∈ R} Range (R) = {b : (a, b) ∈ R}

It is evident from the definition that the domain of a relation from A to B is a subset of A and its range isa subset of B.

Q. Let A = {2, 3, 4, 5, 6, 7, 8, 9}. Let R be the relation on A defined by {(x, y) : x ∈ A, y ∈ A & x2 = y or x = y2}. Find domain and range of R.

Q. Let A = {2, 3, 4, 5, 6, 7, 8, 9}. Let R be the relation on A defined by {(x, y) : x ∈ A, y ∈ A & x2 = y or x = y2}. Find domain and range of R.

Solution

The relation R isR = {(2, 4), (3, 9), (4, 2), (9, 3)}Domain of R = {2, 3, 4, 9}Range of R = {2, 3, 4, 9}

Number of RelationRelation

Result

If n (A) = p & n (B) = q then number of relations that can be defined from A to B is 2pq.

Q. A relation on A = {-2, -1, 0, 1, 2} is defined asR = {(x, y) : y = |x|, x ≠ -1}. Then find number of elements in power set of R.

Hint:

No. of elements in power set of a set = 2no. of elements in that

set

R = {(–2, 2),(-1, 1), (0 ,0), (1, 1), (2, 2)}

Solution

Number of elements in power set of R = 25 = 32.

Q. A relation on A = {-2, -1, 0, 1, 2} is defined asR = {(x, y) : y = |x|, x ≠ -1}. Then find number of elements in power set of R.

Inverse of RelationRelation

Inverse of a relation

If R is a relation on set A, then the relation R−1 on A, defined by

R−1 = {(b, a) : (a, b) ∈ R} is called an inverse relation to A.Clearly,

Domain (R−1) = Range of (R)

Range of (R−1) = domain of (R).

Types of RelationRelation

Identity relation

If every element of A is related to itself only. Symbolically, the relation IA = {(a, a) : a ∈ A} is called an Identity relation on A.

For example, let A = {1, 2, 3} then relation R1 on A, defined as R1 = {(1, 1), (2, 2), (3, 3)} is an identity relation.

Note R2 = {(1, 1), (2, 2)} and R3 = {(1, 1), (2, 2), (3, 3), (1, 2)} are not identity relations on A.

Reflexive relation

(a, a) ∈ R, ∀ a ∈ A.

Reflexive relation

A Relation R on a set A is said to be reflexive, if every element of A is related to itself.

Thus if R is reflexive, then (a, a) ∈ R, ∀ a ∈ A.For example, let A = {1, 2, 3} then R1 = {(x, y) | y ≥ x} is a reflexive relation on A.

Note Every Identity relation is reflexive but every reflexive relation is not Identity.

Symmetric relation

(a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A

Symmetric relation

A relation R on a set A is said to be a symmetric relation iff

(a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A

i.e. a R b ⇒ b R a, ∀ a, b ∈ A

For example, let A = {1, 2, 3} then

R1 = {(1, 2), (1, 3), (2, 2), (2, 1), (3, 1)} is a symmetric relation on A.

Q. Let A be the set of all lines in a plane. Relation R defined on A as L1 R L2 ⇔ L1 ⊥ L2, Show that R is a symmetric relation.

Q. Let A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 4), (2, 3)}. At least how many elements need to be added in R so that it becomes reflexive and symmetric.

Q. Let A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 4), (2, 3)}. At least how many elements need to be added in R so that it becomes reflexive and symmetric.

Solution

Let A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 4), (2, 3)}

To make it reflexive,

We have to add (3, 3), (4, 4)And to make it symmetric

(4, 3), (3, 2)

So in total 4 elements need to be added.

Transitive relation

(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A

Transitive relation

Relation R on set A is transitive iff (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A

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

1 R1 = {1, 2), (2,3), (1,3)}

It is transitive relation.

2 R2 = {(1, 2), (2, 3), (1, 3), (4, 5), (5, 1)}

It is not transitive relation.

3 R3 = {(1, 3), (2, 4)} is transitive.

Q. Prove that relation R = {(x, y) : x is factor of y} is a transitive relation of integers.

Q. Prove that relation R = {(x, y) : x is factor of y} is a transitive relation of integers.

Solution

Hence it is transitive relation

Let (a, b) ∈ R & (b, c) ∈ R∴ b = ak1 & c = bk2 for some k1, k2 ∈ Z

⇒ c = ak1k2

∴ a is a factor of c.

∴ (a, c) ∈ R.

Q. Consider a relation of whole numbers defined as xRy ⇔ x ≤ y2. Prove that it is reflexive but neither symmetric nor transitive.

Q. Consider a relation of whole numbers defined as xRy ⇔ x ≤ y2. Prove that it is reflexive but neither symmetric nor transitive.

SolutionClearly, it is reflexive as

(a, a) ∈ R as a ≤ a2

But, if (1, 2) ∈ RThen (2, 1) ∉ R So it is not symmetric

(5, 3) ∈ R, (3, 2) ∈ R

Now, let us consider

But

Thus R is not transitive.

(5, 2) ∉ R,

Equivalence relation

Relation R defined on any set A is equivalence relation ifR is reflexive

R is symmetric

R is transitive

For example, let us Consider A = {1, 2, 3, 4, 5}

Defined as R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5), (5, 1)} is equivalence relation.

Q. A relation R is defined on Set A of triangles in a plane as R = {(T1, T2)|T1 is similar to T2}. OBSERVE THAT it is equivalence relation.

Solution

Reflexivity:

So, R is reflexive relation

Symmetric:

Let T1, T2 ∈ A

Such that (T1, T2) ∈ R⇒ T1 is similar to T2

So, R is Symmetric relation.

⇒ (T2, T1) ∈ R

(T1, T1) ∈ R ∀ T1 ∈ AAs T1 is similar to T1

⇒ T2 is similar to T1

Transitivity:

Let T1, T2, T3 ∈ A

So that (T1, T2) ∈ R and (T2, T3) ∈ R

⇒ T1 is similar to T2 and T2 is similar to T3

So, R is transitive relation.

⇒ T1 is similar to T3 and (T1, T3) ∈ R

So, R is an equivalence relation.

Q. Show that relation R defined on the set A = {0, 1, 2, …., 12} given by R = {(a, b)|a–b is multiple of 4} is an equivalence relation. Also find elements related to 3.

Solution

A = {0, 1, 2, …., 12}

R = {(a, b)|a–b is multiple of 4}

Reflexivity:

For any a ∈ A, (a, a) ∈ R

As, a – a = 0 is multiple of 4

So, R is reflexive

Symmetric:

⇒ a–b = 4λfor some λ ∈ Z

(b – a) = 4(–λ)

So, (b–a) is multiple of 4

Let a, b ∈ ASuch that (a,b) ∈ Ra – b is multiple 4

(b–a) ∈ R

It is symmetric

Transitivity: Let a, b, c ∈ A

Such that (a, b) ∈ R and (b, c) ∈ R

⸫ a − b is multiple of 4 and (b − c) is multiple of 4

(a –b ) = 4λ & (b–c) = 4μ for λ, μ ∈ N

Add these equations

a – b + b – c = 4λ + 4μ

a – c = 4 (λ + μ)

So, (a–c) is multiple of 4

Thus, It is transitive

So, R is an equivalence relation.

Elements related to 3

Let x be the element of A such that (x, 3) ∈ R

Then (x − 3) is multiple of 4

∴ (x - 3) = 0, 4, 8, 12, ….

⇒ x = 3, 7, 11, 15, …

set A = {0, 1, 2, …. , 12}

x = 3, 7, 11

Q. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on A = {3, 6, 9, 12}. Relation R is

Reflexive and transitive only A

Reflexive only B

An equivalence relationC

Reflexive and symmetric onlyD

IIT 2005

Q. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on A = {3, 6, 9, 12}. Relation R is

Reflexive and transitive only A

Reflexive only B

An equivalence relationC

Reflexive and symmetric onlyD

IIT 2005

Q. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on A = {3, 6, 9, 12}. Relation R is

IIT 2005

SolutionGiven A = {3, 6, 9, 12}

Since (3, 3), (6, 6), (9, 9), (12, 12) ∈ R

Thus its reflexive relation

Here, (6, 12) ∈ R, but

Thus R is not symmetric

Here, (3, 6) ∈ R and (6, 12) ∈ R, also (3, 12) ∈ RSo relation is transitive also

(12, 6) ∉ R

Q. Let W denote the words in the English dictionary. Define the relation R on W by : R = {(x, y) : the words x and y have at least one letter in common}, then R is

Symmetric, transitive but not reflexive. A

Reflexive, symmetric but not transitive B

EquivalenceC

Reflexive, transitive but not symmetricD

IIT 2006

Symmetric, transitive but not reflexive. A

Reflexive, symmetric but not transitive B

EquivalenceC

Reflexive, transitive but not symmetricD

Q. Let W denote the words in the English dictionary. Define the relation R on W by : R = {(x, y) : the words x and y have at least one letter in common}, then R is

IIT 2006

Solution

Thus, R is reflexive and symmetric only but not transitive.

Here if ‘x’ is some word, then (x, x) ∈ R, as they have all letters in common.Let (x, y) ∈ R,∴ x and y has one letter in common

⇒ y and x have one letter in common

⇒ (y, x) ∈ R

∴ R is symmetric

Now, if (TOM, MAD) ∈ R and (MAD, DEN) ∈ R, but (TOM, DEN) ∉ R

Q. Consider the following two binary relations on the set A = {a, b, c} R1 = {c, a), (b, b), (a, c), (c,c), (b,c), (a,a)} and R2 = {a,b), (b,a), (c,c), (c,a), (a,a), (b,b), (a,c)}. Then:

R2 is symmetric but it not transitive A

both R1 and R2 are not symmetric B

both R1 and R2 are transitiveC

R R2 is not symmetric but it is transitive D

JEE Main 2018

Q. Consider the following two binary relations on the set A = {a, b, c} R1 = {c, a), (b, b), (a, c), (c,c), (b,c), (a,a)} and R2 = {a,b), (b,a), (c,c), (c,a), (a,a), (b,b), (a,c)}. Then:

R2 is symmetric but it not transitive A

both R1 and R2 are not symmetric B

both R1 and R2 are transitiveC

R R2 is not symmetric but it is transitive D

JEE Main 2018

Q. Consider the following two binary relations on the set A = {a, b, c} R1 = {c, a), (b, b), (a, c), (c,c), (b,c), (a,a)} and R2 = {a,b), (b,a), (c,c), (c,a), (a,a), (b,b), (a,c)}. Then:

Solution

JEE Main 2018

Q. Let R be the set of real numbers.Statement – 1: A ={(x,y) ∈ R × R: y –x is an integer} is an equivalence relation on R.Statement – 2 B = {(x, y) ∈ R × R: x = ⍺y for some rational number ⍺} is an equivalence relation on R.

S1 is true, S2 is true; S2 is correct explanation for S1.A

State S1 is true, S2 is true; S2 is not a correct explanation for S1B

S1 is true, S2 is false.C

S1 is false, S2 is true D

AIEEE 2011

Q. Let R be the set of real numbers.Statement – 1: A ={(x,y) ∈ R × R: y –x is an integer} is an equivalence relation on R.Statement – 2 B = {(x, y) ∈ R × R: x = ⍺y for some rational number ⍺} is an equivalence relation on R.

S1 is true, S2 is true; S2 is correct explanation for S1.A

State S1 is true, S2 is true; S2 is not a correct explanation for S1B

S1 is true, S2 is false.C

S1 is false, S2 is true D

AIEEE 2011

Solution:

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8th Jan 2020-(Shift 1)