WBCHSE Class 12 Maths Solutions For Relations

Relations And Functions – Chapter 1 Relations

In class 9 we discussed about the Cartesian product of sets. Now, we extend our ideas to relation in a set and then in next chapter we shall be taking up functions.

Relation In A Set

A relation R in a set A is a subset of A x A.

Thus, R is a relation in a set A ⇔ R ⊆ A x A.

If (a,b) ∈ R, then we say that a is related to b and write, a R b.

If (a,b) ∉ R, then we say that a is not related to b and write, a \(\not R\) b.

Example Let A = {1,2,3,4,5,6} and let R be a relation in A, given by

R = {(a, b): a – b = 2}.

Then, R = {(3,1),(4,2),(5,3),(6,4)}.

Clearly, 3 R 1, 4 R 2, 5 R 3 and 6 R 4.

But, 1 \(\not R\) 3, 2 \(\not R\) 4, 5 \(\not R\) 6, etc.

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Domain And Range Of A Relation

Let R be a relation in a set A. Then, the set of all first coordinates of elements of R is called the domain of R, written as dom (R) and the set of all second coordinates of R is called the range of R, written as range (R).

∴ dom (R) = {a:(a,b) ∈ R} and range (R) = {b:(a,b) ∈ R}.

Example Let A = {1,2,3,4,…,15,16} and let R be a relation in A, given by

R = {(a,b):b = a2}.

Then, R = {(1,1), (2,4), (3,9), (4,16)}.

∴ dom (R) = {1,2,3,4} and range (R) = {1,4,9,16}.

Some Particular Types of Relations

Empty Relation (Or Void Relation) A relation R in a set A is called an empty relation, if no element of A is related to any element of A and we denote such a relation by Φ

Thus, R = Φ ⊆ A x A.

Example Let A = {1,2,3,4,5} and let R be a relation in A, given by R = {(a,b): a – b = 6}.

Clearly, no element (a,b) ∈ A x A satisfies the property a – b = 6.

∴ R is an empty relation in A.

Universal Relation A relation R in a set A is called a universal relation, if each element of A is related to every element of A.

Thus, R = (A x A) ⊆ (A x A) is the universal relation on A.

Example Let A = {1,2,3}. Then,

R = (A x A) = {(1,1),(1,2),(1,3),(2,1)(2,2),(2,3),(3,1),(3,2),(3,3)} is the universal relation in A.

Identity Relation The relation IA = {(a, a): a ∈ A} is called the identity relation on A.

Example Let A = {1,2,3}. Then,

IA = {(1,1),(2,2),(3,3)} is the identity relation on A.

Various Types Of Relations

Let A be a nonempty set. Then, a relation R on A is said to be

(1) reflexive if (a,a) ∈ R for each a ∈ A,

i.e., if a R a for each a ∈ A.

(2) Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R for all a,b ∈ A,

i.e., if a R b ⇒ b R a for all a, b ∈ A.

(3) transitive if (a,b) ∈ R, (b,c) ∈ R ⇒ (a,c) ∈ R for all a,b,c ∈ A,

i.e., if a R b and b R c ⇒ a R c.

Equivalence Relation A relation R in a set A is said to be an equivalence relation if it is reflexive, symmetric and transitive.

WBCHSE Class 12 Maths Solutions For Relations Solved Examples

Example 1 Let A be the set of all triangles in a plane and let R be a relation in A, defined by R = {(△1, △2): △1 ≅ △2}. Show that R is an equivalence relation in A.

Solution

The given relation satisfies the following properties:

(1) Reflexivity

Let △ be an arbitrary triangle in A. Then,

△ ≅ △ ⇒ (△, △) ∈ R for all values of △ in A.

∴ R is reflexive.

(2) Symmetry

Let △1, △2 ∈ A such that (△1, △2) ∈ R. Then,

(△1, △2) ∈ R ⇒ △1 ≅ △2

⇒ △2 ≅ △1

⇒ (△2, △1) ∈ R.

(3) Transitivity

Let △1, △2, △3 ∈ A such that (△1, △2) ∈ R and (△2, △3) ∈ R.

Then, (△1, △2) ∈ R and (△2, △3) ∈ R

⇒ △1 ≅ △2 and △2 ≅ △3

⇒ △1 ≅ △3

⇒ (△1, △3) ∈ R.

∴ R is transitive

Thus, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

Example 2 Let A be the set of all lines in xy-plane and let R be a relation in A, defined by R = {(L1, L2): L1 ∥ L2}. show that R is an equivalence relation in A. Find the set of all lines related to the line y = 3x + 5.

Solution

The given relation satisfies the following properties:

(1) Reflexivity

Let L be an arbitrary line in A. Then,

L ∥ L ⇒ (L, L) ∈ R ∀ L ∈ A.

Thus, R is reflexive.

(2) Symmetry

Let L1, L2 ∈ A such that (L1, L2) ∈ R. Then,

(L1, L2) ∈ R ⇒ L1 ∥ L2

⇒ L2 ∥ L1

⇒ (L2, L1) ∈ R.

∴ R is symmetric.

(3) Transitivity

Let L1, L2, L3 ∈ A such that (L1, L2) ∈ R and (L2, L3) ∈ R.

Then, (L1, L2) ∈ R and (L2, L3) ∈ R

⇒ L1 ∥ L2 and L2 ∥ L3

⇒ L1 ∥ L3

⇒ (L1, L3) ∈ R.

∴ R is transitive

Thus R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

The family of lines parallel to the line y = 3x + 5 is given by y = 3x + k, where k is real.

Example 3 Let Z be the set of all integers and let R be a relation in Z, defined by R = {(a, b):(a-b) is even}. Show that R is an equivalence relation in Z.

Solution

R = {(a, b):(a-b) is even}

Here, R satisfies the following properties:

(1) Reflexivity

Let a be an arbitrary element of Z.

Then, (a-a) = 0, which is even.

∴ (a,a) ∈ R ∀ a ∈ Z.

So, R is reflexive.

(2) Symmetry

Let a,b ∈ Z such that (a,b) ∈ R. Then,

(a,b) ∈ R ⇒ (a-b) is even

⇒ -(a-b) is even

⇒ (b-a) is even

⇒ (b,a) ∈ R.

∴ R is symmetric.

(3) Transitivity

Let a, b, c ∈ Z such that (a,b) ∈ R and (b,c) ∈ R. Then,

(a,b) ∈ R and (b,c) ∈ R

⇒ (a-b) is even and (b-c) is even

⇒ {(a-b) + (b-c)} is even

⇒ (a-c) ∈ R.

⇒ (a,c) ∈ R.

∴ R is transitive.

Thus, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation in Z.

Example 4 Let A be the set of all lines in a plane and let R be a relation in A defined by R = {(L1, L2): L1 ⊥ L2}. Show that R is symmetric but neither reflexive nor transitive.

Solution

Given

Let A be the set of all lines in a plane and let R be a relation in A defined by R = {(L1, L2): L1 ⊥ L2}.

Clearly, any line L cannot be perpendicular to itself.

∴ (L, L) ∉ R for any L ∈ A.

So, R is not reflexive.

Again, let (L1, L2) ∈ R. Then,

(L1, L2) ∈ R ⇒ L1 ⊥ L2

⇒ L2 ⊥ L1

⇒ (L2,L3) ∈ R.

∴ R is symmetric.

Now, let L1, L2, L3 ∈ A such that L1 ⊥ L2 and L2 ⊥ L3.

Then, clearly L1 is not perpendicular to L3.

Thus, (L1,L2) ∈ R and (L2,L3) ∈ R, but (L1, L3) ∉ R.

Class 12 Maths Relations Example 4

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Example 5 Let S be the set of all real numbers and let R be a relation in S defined by R{(a, b): (1+ab) > 0}. Show that R is reflexive and symmetric but not transitive.

Solution

Given

Let S be the set of all real numbers and let R be a relation in S defined by R{(a, b): (1+ab) > 0}.

Let a be any real number. Then,

(1 + aa) = (1 + a2) > 0 shows that (a,a) ∈ R ∀ a ∈ S.

∴ R is reflexive.

Also, (a,b) ∈ R ⇒ (1 + ab) > 0

⇒ (1 + ba) > 0 [∵ ab = ba]

⇒ (b,a) ∈ R.

∴ R is symmetric.

In order to show that R is not transitive, consider (-1,0) and (0,2).

Clearly, (-1,0) ∈ R, since [1 + (-1) x 0] > 0.

And, (0,2) ∈ R, since [1 + 0 x 2] > 0.

But, (-1,2) ∉ R, since [1 + (-1) x 2] is not greater than 0.

Hence, R is reflexive and symmetric but not transitive.

Example 6 Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b}. Show that R is reflexive and transitive but not symmetric.

Solution

Given

Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b}.

Here, R satisfies the following properties:

(1) Reflexivity

Let a be an arbitrary real number.

Then, a ≤ a ⇒ (a,a) ∈ R.

Thus, (a,a) ∈ R ∀ a ∈ S.

∴ R is reflexive.

(2) Transitivity

Let a, b, c be real numbers such that (a,b) ∈ R and (b,c) ∈ R .

Then, (a,b) ∈ R and (b,c) ∈ R

⇒ a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a,c) ∈ R.

∴ R is transitive.

(3) Nonsymmetry

Clearly, (4,5) ∈ R since 4 ≤ 5.

But, (5,4) ∉ R since 5 ≤ 4 is not true.

∴ R is not symmetric.

Example 7 Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b2}. Show that R satisfies none of reflexivity, symmetry and transitivity.

Solution

Given

Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b2}.

(1) Nonreflexivity

Clearly, \(\frac{1}{2}\) is a real number and \(\frac{1}{2} \leq\left(\frac{1}{2}\right)^2\) is not true.

∴ \(\left(\frac{1}{2}, \frac{1}{2}\right) \notin R .\)

Hence, R is not reflexive.

(2) Nonsymmetry

Consider the real numbers \(\frac{1}{2}\) and 1.

Clearly, \(\frac{1}{2}\) ≤ 12 ⇒ \(\left(\frac{1}{2}, 1\right) \in R \text {. }\)

But, 1 ≤ (\(\frac{1}{2}\))2 is not true and so \(\left(1, \frac{1}{2}\right) \notin R \text {. }\)

Thus, \(\left(\frac{1}{2}, 1\right) \in R but \left(1, \frac{1}{2}\right) \notin R .\)

Hence, R is not symmetric.

(3) Nontransitivity

Consider the real numbers 2, -2 and 1.

Clearly, 2 ≤ (-2)2 and -2 ≤ (1)2 but 2 ≤ 12 is not true.

Thus, (2,-2) ∈ R and (-2,1) ∈ R, but (2,1) ∉ R.

Hence, R is not transitive.

Example 8 Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b3}. Show that R satisfies none of reflexivity, symmetry and transitivity.

Solution

Given 

Let S be the set of all real numbers and let R be a relation in S, defined by R = {(a,b): a ≤ b3}.

(1) Nonreflexivity

Clearly, \(\frac{1}{2}\) is a real number and \(\frac{1}{2}\) ≤ (\(\frac{1}{2}\))3 is not true.

∴ \(\left(\frac{1}{2}, \frac{1}{2}\right) \notin R\)

Hence, R is not reflexive.

(2) Nonsymmetry

Take the real numbers \(\frac{1}{2}\) and 1.

Clearly, \(\frac{1}{2}\) ≤ 13 is true and therefore, \(\left(\frac{1}{2}, 1\right) \in R \text {. }\)

But, \(1 \leq\left(\frac{1}{2}\right)^3 is not true and so \left(1, \frac{1}{2}\right) \notin R .\)

Hence R is not symmetric.

(3) Nontransitivity

Consider the real numbers, 3, \(\frac{3}{2}\) and \(\frac{4}{3}\).

Clearly, 3 \(\leq\left(\frac{3}{2}\right)^3 \text { and } \frac{3}{2} \leq\left(\frac{4}{3}\right)^3 \text { but } 3 \leq\left(\frac{4}{3}\right)^3\) is not true.

Thus, \(\left(3, \frac{3}{2}\right) \in R \text { and }\left(\frac{3}{2}, \frac{4}{3}\right) \in R \text {, but }\left(3, \frac{4}{3}\right) \notin R\).

Hence, R is not transitive.

Thus, R satisfies none of reflexivity, symmetry and transitivity.

Example 9 Let N be the set of all natural numbers and let R be a relation in N, defined by R = {(a, b): a is a factor of b}. Then, show that R is reflexive and transitive but not symmetric.

Solution

Given 

Let N be the set of all natural numbers and let R be a relation in N, defined by R = {(a, b): a is a factor of b}.

Here, R satisfies the following properties:

(1) Reflexivity

Let a be an arbitrary element of N.

Then, clearly, a is a factor of a.

∴ (a,a) ∈ R ∀ a ∈ N.

So, R is reflexive.

(2) Transitivity

Let a, b, c ∈ N such that (a,b) ∈ R and (b,c) ∈ R.

Now, (a,b) ∈ R and (b,c) ∈ R

⇒ (a is a factor of b) and (b is a factor of c)

⇒ b = ad and c = be for some d,e ∈ N

⇒ c = (ad)e = a(de) [by associative law]

⇒ a is a factor of c

⇒ (a,c) ∈ R.

∴ (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R.

Hence, R is transitive.

(3) Nonsymmetry

Clearly, 2 and 6 are natural numbers and 2 is a factor of 6.

∴ (2,6) ∈ R.

But, 6 is not a factor of 2.

∴ (6,2) ∉ R.

Thus, (2,6) ∈ R and (6,2) ∉ R.

Hence, R is not symmetric.

Example 10 Let N be the set of all natural numbers and let R be a relation in N, defined by R = {(a, b): a is a multiple of b}. Show that R is reflexive and transitive but not symmetric.

Solution

Given

Let N be the set of all natural numbers and let R be a relation in N, defined by R = {(a, b): a is a multiple of b}.

Here R satisfies the following properties:

(1) Reflexivity

Let a be an arbitrary element of N.

Then, a = (a x a) shows that a is a multiple of a.

∴ (a,a) ∈ R ∀ a ∈ N.

So, R is reflexive.

(2) Transitivity

Let a, b, c ∈ N such that (a,b) ∈ R and (b,c) ∈ R.

Now, (a,b) ∈ R and (b,c) ∈ R

⇒ (a is a multiple of b) and (b is a multiple of c)

⇒ a = bd and b = ce for some d ∈ N and e ∈ N

⇒ a = (ce)d

⇒ a = c(ed)

⇒ a is a multiple of c

⇒ (a,c) ∈ R.

∴ (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R.

Hence, R is transitive.

(3) Nonsymmetry

Clearly, 6 and 2 are natural numbers and 6 is a multiple of 2.

∴ (6,2) ∈ R.

But, 2 is not a multiple of 6.

∴ (2,6) ∉ R.

Thus, (6,2) ∈ R and (2,6) ∉ R.

Hence, R is not symmetric.

Example 11 Let X be a nonempty set and let S be the collection of all subsets of X. Let R be a relation in S, defined by R = {(A,B): A ⊂ B}. Show that R is transitive but neither reflexive nor symmetric.

Solution

Given

Let X be a nonempty set and let S be the collection of all subsets of X. Let R be a relation in S, defined by R = {(A,B): A ⊂ B}.

Clearly, R satisfies the following properties:

(1) Transitivity

Let A, B, C ∈ S such that (A,b) ∈ R and (B,c) ∈ R.

Now, (A,b) ∈ R and (B,C) ∈ R

⇒ A ⊂ B and B ⊂ C

⇒ A ⊂ C

⇒ (A, C) ∈ R.

∴ R is transitive.

(2) Nonreflexivity

Let A be any set in S.

Then, A ⊄ A shows that (A,A) ∉ R.

∴ R is not reflexive.

(3) Nonsymmetry

Now (A, B) ∈ R ⇒ A ⊂ B

⇒ B ⊄ A

⇒ (B, A) ∉ R.

∴ R is not symmetric.

Hence, R is transitive but neither reflexive nor symmetric.

Example 12 Give an example of a relation which is

  1. reflexive and transitive but not symmetric;
  2. symmetric and transitive but not reflexive;
  3. reflexive and symmetric but not transitive;
  4. symmetric but neither reflexive nor transitive;
  5. transitive but neither reflexive nor symmetric.

Solution

Let A = {1,2,3).

Then, it is easy to verify that the relation

(1) R1 = {(1,1), (2,2), (3,3), (1,2)} is reflexive and transitive.

R1 is not symmetric, since

(1,2) ∈ R and (2,1) ∉ R.

(2) R2 = {(1,1), (2,2), (1,2), (2,1)} is symmetric and transitive.

But, R2 is not reflexive, since (3,3) ∉ R2.

(3) R3 = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)} is reflexive and symmetric.

But, R3 is not transitive, since

(1,2) ∈ R3, (2,3) ∈ R3 but (1,3) ∉ R3.

(4) R4 = {(2,2), (3,3), (1,2), (2,1)} is symmetric.

But, R4 is not reflexive since (1,1) ∉ R4.

Also, R4 is not transitive, as

(1,2) ∈ R4 and (2,1) ∈ R4 but (1,1) ∉ R4.

(5) R5 = {(2,2), (3,3), (1,2)} is transitive.

But, R5 is not reflexive, since (1,1) ∉ R.

And, R5 is not symmetric as (1,2) ∈ R5 but (2,1) ∉ R5.

Example 13 Let N be the set of all natural numbers and let R be a relation on N x N, defined by (a,b) R (c,d) ⇔ ad = bc. Show that R is an equivalence relation.

Solution

Given

Let N be the set of all natural numbers and let R be a relation on N x N, defined by (a,b) R (c,d) ⇔ ad = bc.

Here R satisfies the following properties:

(1) Reflexivity

Let (a,b) ∈ R. Then,

(a,b) R (a,b), since ab = ba [by commutative law of multiplication on N].

Thus, (a,b) R (a,b) ∀ (a,b) ∈ R.

∴ R is reflexive.

(2) Symmetry

Let (a,b) R (c,d). Then,

(a,b) R (c,d) ⇒ ad = bc

⇒ bc = ad

⇒ cb = da

[by commutativity of multiplication on N]

⇒ (c,d) R (a,b).

∴ R is symmetric.

(3) Transitivity

Let (a,b) R (c,d) and (c,d) R (e,f). Then,

ad = bc and cf = de

⇒ adcf = bcde

⇒ (af)(cd) = (be)(cd)

⇒ af = be [by cancellation law]

⇒ (a,b) R (e,f).

∴ (a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f).

∴ R is transitive.

Thus, R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

Example 14 If R1 and R2 be two equivalence relations on a set A, prove that R1 ∩ R2 is also an equivalence relation on A.

Solution

Given

R1 and R2 be two equivalence relations on a set A

Let R1 and R2 be two equivalence relations on a set A.

Then, R1 ⊆ A x A, R2 ⊆ A x A ⇒ (R1 ∩ R2) ⊆ A x A.

So, (R1 ∩ R2) is a relation on A.

This relation on A satisfies the following properties.

(1) Reflexivity

R1 is reflexive and R2 is reflexive

⇒ (a,a) ∈ R1 and (a,a) ∈ R2 for all a ∈ A

⇒ (a,a) ∈ R1 ∩ R2 for all a ∈ A

⇒ R1 ∩ R2 is reflexive.

(2) Symmetry

Let (a,b) be an arbitrary element of R1 ∩ R2. Then,

(a,b) ∈ R1 ∩ R2

⇒ (a,b) ∈ R1 and (a,b) ∈ R2

⇒ (b,a) ∈ R1 and (b,a) ∈ R2

[∵ R1 is symmetric and R2 is symmetric]

⇒ (b,a) ∈ R1 ∩ R2.

This shows that R1 ∩ R2 is symmetric.

(3) Transitivity

(a,b) ∈ R1 ∩ R2 and (b,c) ∈ R1 ∩ R2

⇒ (a,b) ∈ R1, (a,b) ∈ R2, and (b,c) ∈ R1, (b,c) ∈ R2

⇒ {(a,b) ∈ R1, (b,c) ∈ R1}, and {(a,b) ∈ R2, (b,c) ∈ R2}

⇒ (a,c) ∈ R1 and (a,c) ∈ R2

[∵ R1 is transitive and R2 is transitive]

⇒ (a,c) ∈ R1 ∩ R2.

This shows that (R1 ∩ R2) is transitive.

Thus, R1 ∩ R2 is reflexive, symmetric and transitive.

Hence, R1 ∩ R2 is an equivalence relation.

Example 15 Give an example to show that the union of two equivalence relations on a set A need not be an equivalence relation on A.

Solution

Let R1 and R2 be two relations on a set A = {1,2,3}, given by

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

and R2 = {(1,1), (2,2), (3,3), (1,3), (3,1)}.

Then, it is easy to verify that each one of R1 and R2 is an equivalence relation.

But, R1 ∪ R2 = {(1,1), (2,2), (3,3), (,2), (2,1), (1,3), (3,1)} is not transitive, as

(3,1) ∈ R1 ∪ R2 and (1,2) ∈ R1 ∪ R2 but (3,2) ∉ R1 ∪ R2.

Hence, (R1 ∪ R2) is not an equivalence relation.

Equivalence Classes Let R be an equivalence relation in a set A and let a ∈ A. Then, the set of all those elements of A which are related to a, is called the equivalence class determined by a and it is denoted by [a].

Thus, [a] = {b ∈ A: (a,b) ∈ R}.

Two equivalence classes are either disjoint or identical.

An Important Result An equivalence relation R on a set A partitions the set into mutually disjoint equivalence classes.

Example 16 On the set Z all integers, consider the relation R = {(a,b):(a-b) is divisible by 3}. Show that R is an equivalence relation on Z. Also find the partitioning of Z into mutually disjoint equivalence classes.

Solution

Given

On the set Z all integers, consider the relation R = {(a,b):(a-b) is divisible by 3}.

The relation R on Z satisfies the following properties:

(1) Reflexivity

Let a ∈ Z.

Then, (a-a) = 0, which is divisible by 3.

∴ a R a ∀ a ∈ Z.

So, R is reflexive.

(2) Symmetry

Let a, b ∈ Z such that a R b. Then,

a R b ⇒ a – b is divisible by 3

⇒ -(a-b) is divisible by 3

⇒ (b-a) is divisible by 3

⇒ b R a.

∴ a R b ⇒ b R a ∀ a,b ∈ Z.

So, R is symmetric.

(3) Transitivity

Let a, b, c ∈ Z such that a R b and b R c. Then,

a R b, b R c ⇒ (a-b) is divisible by 3 and (b-c) is divisible by 3

⇒ [(a-b)+(b-c)] is divisible by 3

⇒ (a-c) is divisible by 3.

Thus, a R b, b R c ⇒ a R c ∀ a, b, c ∈ Z.

∴ R is an equivalence relation on Z.

Now, let us consider [0], [1] and [2].

We have:

[0] = {x ∈ Z: x R 0}

= {x ∈ Z:(x-0) is divisible by 3}

= {…, -6,-3,0,3,6,9,…}.

∴ [0] = {…, -6,-3,0,3,6,9,…}.

Similarly, [1] = {x ∈ Z: x R 1}

= {x ∈ Z: (x-1) is divisible by 3}

= {…,-5,-2,1,4,7,10,…}.

∴ [1] = {…,-5,-2,1,4,7,10,…}.

And, [2] = {x ∈ Z: x R 2}

= {x ∈ Z: (x-2) is divisible by 3}

= {…,-4,-1,2,5,8,11,…}

∴ [2] = {…,-4,-1,2,5,8,11,…}.

Clearly, [0], [1] and [2] are mutually disjoint and Z = [0] ∪ [1] ∪ [2].

Example 17 Let A = {1,2,3,4,5,6,7} and let R be a relation on A, defined by R = {(a,b): both a and b are either odd or even}. Prove that R is an equivalence relation.

Let B = {1,3,5,7} and C = {2,4,6}.

Show that

  1. all elements of B are related to each other;
  2. all elements of C are related to each other;
  3. no element of B is related to any element of C.

Solution

The given relation satisfies the following properties:

(1) Reflexivity

Let a ∈ A.

Then, it is clear that a are both odd or both even.

∴ (a,a) ∈ R ∀ a ∈ A.

So, R is reflexive.

(2) Symmetry

Let (a,b) ∈ R. Then,

(a,b) ∈ R ⇒ both a and b are either odd or even

⇒ both b and a are either odd or even

⇒ (b,a) ∈ R.

∴ R is symmetric.

(3) Transitivity

Let (a,b) ∈ R and (b,c) ∈ R. Then,

(a,b) ∈ R and (b,c) ∈ R

⇒ {both a and b are either odd or even} and {both b and c are either odd or even}

⇒ both a and c are either odd or even

⇒ (a,c) ∈ R.

∴ R is transitive.

Hence, R is an equivalence relation.

  1. If we pick up any two elements of B, then both being odd, they are related to each other.
  2. If we pick up any two elements of C, then both being even, they are related to each other.
  3. If we pick up one element of B and one element of C, then one is even while the other is odd.

So, they are not related to each other.

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