## 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.

**Read and Learn More Class 12 Math Solutions **

**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 = a^{2}}.

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 I_{A} = {(a, a): a ∈ A} is called the identity relation on A.

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

I_{A} = {(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 = {(L _{1}, L_{2}): L_{1} ∥ L_{2}}. 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 L_{1}, L_{2} ∈ A such that (L_{1}, L_{2}) ∈ R. Then,

(L_{1}, L_{2}) ∈ R ⇒ L_{1} ∥ L_{2}

⇒ L_{2} ∥ L_{1}

⇒ (L_{2}, L_{1}) ∈ R.

∴ R is symmetric.

(3) Transitivity

Let L_{1}, L_{2}, L_{3} ∈ A such that (L_{1}, L_{2}) ∈ R and (L_{2}, L_{3}) ∈ R.

Then, (L_{1}, L_{2}) ∈ R and (L_{2}, L_{3}) ∈ R

⇒ L_{1} ∥ L_{2} and L_{2} ∥ L_{3}

⇒ L_{1} ∥ L_{3}

⇒ (L_{1}, L_{3}) ∈ 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 = {(L _{1}, L_{2}): L_{1} ⊥ L_{2}}. 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 = {(L_{1}, L_{2}): L_{1} ⊥ L_{2}}.

Clearly, any line L cannot be perpendicular to itself.

∴ (L, L) ∉ R for any L ∈ A.

So, R is not reflexive.

Again, let (L_{1}, L_{2}) ∈ R. Then,

(L_{1}, L_{2}) ∈ R ⇒ L_{1} ⊥ L_{2}

⇒ L_{2} ⊥ L_{1}

⇒ (L_{2},L_{3}) ∈ R.

∴ R is symmetric.

Now, let L_{1}, L_{2}, L_{3} ∈ A such that L_{1} ⊥ L_{2} and L_{2} ⊥ L_{3}.

Then, clearly L_{1} is not perpendicular to L_{3}.

Thus, (L_{1},L_{2}) ∈ R and (L_{2},L_{3}) ∈ R, but (L_{1}, L_{3}) ∉ R.

∴ 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 + a^{2}) > 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 ≤ b ^{2}}. 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 ≤ b^{2}}.

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

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 ≤ 1^{2} 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 ≤ b ^{3}}. 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 ≤ b^{3}}.

(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, \(\frac{1}{2}\) ≤ 1^{3 }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**

**reflexive and transitive but not symmetric;****symmetric and transitive but not reflexive;****reflexive and symmetric but not transitive;****symmetric but neither reflexive nor transitive;****transitive but neither reflexive nor symmetric.**

**Solution**

Let A = {1,2,3).

Then, it is easy to verify that the relation

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

R_{1} is not symmetric, since

(1,2) ∈ R and (2,1) ∉ R.

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

But, R_{2} is not reflexive, since (3,3) ∉ R_{2}.

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

But, R_{3} is not transitive, since

(1,2) ∈ R_{3}, (2,3) ∈ R_{3} but (1,3) ∉ R_{3}.

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

But, R_{4} is not reflexive since (1,1) ∉ R_{4}.

Also, R_{4} is not transitive, as

(1,2) ∈ R_{4} and (2,1) ∈ R_{4} but (1,1) ∉ R_{4}.

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

But, R_{5} is not reflexive, since (1,1) ∉ R.

And, R_{5} is not symmetric as (1,2) ∈ R_{5} but (2,1) ∉ R_{5}.

**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 R _{1} and R_{2} be two equivalence relations on a set A, prove that R_{1 }∩ R_{2} is also an equivalence relation on A.**

**Solution**

**Given**

R_{1} and R_{2} be two equivalence relations on a set A

Let R_{1} and R_{2} be two equivalence relations on a set A.

Then, R_{1} ⊆ A x A, R_{2} ⊆ A x A ⇒ (R_{1} ∩ R_{2}) ⊆ A x A.

So, (R_{1} ∩ R_{2}) is a relation on A.

This relation on A satisfies the following properties.

(1) Reflexivity

R_{1} is reflexive and R_{2} is reflexive

⇒ (a,a) ∈ R_{1} and (a,a) ∈ R_{2} for all a ∈ A

⇒ (a,a) ∈ R_{1} ∩ R_{2} for all a ∈ A

⇒ R_{1} ∩ R_{2} is reflexive.

(2) Symmetry

Let (a,b) be an arbitrary element of R_{1} ∩ R_{2}. Then,

(a,b) ∈ R_{1} ∩ R_{2}

⇒ (a,b) ∈ R_{1} and (a,b) ∈ R_{2}

⇒ (b,a) ∈ R_{1} and (b,a) ∈ R_{2}

[∵ R_{1} is symmetric and R_{2} is symmetric]

⇒ (b,a) ∈ R_{1} ∩ R_{2}.

This shows that R_{1} ∩ R_{2} is symmetric.

(3) Transitivity

(a,b) ∈ R_{1} ∩ R_{2} and (b,c) ∈ R_{1} ∩ R_{2}

⇒ (a,b) ∈ R_{1}, (a,b) ∈ R_{2}, and (b,c) ∈ R_{1}, (b,c) ∈ R_{2}

⇒ {(a,b) ∈ R_{1}, (b,c) ∈ R_{1}}, and {(a,b) ∈ R_{2}, (b,c) ∈ R_{2}}

⇒ (a,c) ∈ R_{1} and (a,c) ∈ R_{2}

[∵ R_{1} is transitive and R_{2} is transitive]

⇒ (a,c) ∈ R_{1} ∩ R_{2}.

This shows that (R_{1} ∩ R_{2}) is transitive.

Thus, R_{1} ∩ R_{2} is reflexive, symmetric and transitive.

Hence, R_{1} ∩ R_{2} 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 R_{1} and R_{2} be two relations on a set A = {1,2,3}, given by

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

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

Then, it is easy to verify that each one of R_{1} and R_{2} is an equivalence relation.

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

(3,1) ∈ R_{1} ∪ R_{2} and (1,2) ∈ R_{1} ∪ R_{2} but (3,2) ∉ R_{1} ∪ R_{2}.

Hence, (R_{1} ∪ R_{2}) 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**

**all elements of B are related to each other;****all elements of C are related to each other;****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.

- If we pick up any two elements of B, then both being odd, they are related to each other.
- If we pick up any two elements of C, then both being even, they are related to each other.
- 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.