Show that the relation R in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b) : |a - b|$ is even$\}$, is an equivalence relation. Show that all the elements of $\{1, 3, 5\}$ are related to each other and all the elements of $\{2, 4\}$ are related to each other. But no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Show that the relation R in the set A = \1, 2, 3, – Class 12 Mathematics

Exercise 1.1 · Q8

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

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The relation $R$ groups numbers by parity: two numbers are related if their difference is even, which means they share the same parity. This makes $R$ an equivalence relation, partitioning $A$ into the odd set $\{1,3,5\}$ and the even set $\{2,4\}$ .

We need to show three things: that $R$ is reflexive, symmetric, and transitive (the definition of an equivalence relation); then that within each parity class all elements are related; and finally that no element from one class relates to any from the other.

The core idea is simple: $|a-b|$ is even exactly when $a$ and $b$ have the same parity — both odd or both even. Why? Because the difference of two odd numbers is even, the difference of two even numbers is even, but the difference of an odd and an even is odd. So the relation is really "same parity."

Let's verify the three properties.

Reflexive: For any $a \in A$ , $|a-a| = 0$ , which is even. So $(a,a) \in R$ for every $a$ . Reflexivity holds.
Symmetric: If $(a,b) \in R$ , then $|a-b|$ is even. But $|b-a| = |a-b|$ , so it's the same number — also even. Hence $(b,a) \in R$ . Symmetry holds.
Transitive: Suppose $(a,b) \in R$ and $(b,c) \in R$ . Then $|a-b|$ and $|b-c|$ are both even. This means $a$ and $b$ have the same parity, and $b$ and $c$ have the same parity. So $a$ and $c$ must share that same parity. Therefore $|a-c|$ is even, giving $(a,c) \in R$ . Transitivity holds.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/sentence: Define equivalence relation.$
›Reveal solutionSolution

Equivalence relation = reflexive + symmetric + transitive.

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CBSE 2025Set ANNUAL1 mark
$R = \{(1,-1),(2,-2),(3,-1)\}$
›Reveal solutionSolution
State the three defining properties of an equivalence relation.

A relation $R$ on a non-empty set $A$ is called an equivalence relation if it satisfies all three of the following:

Reflexive: $(a,a)\in R$ for every $a\in A$ .

Symmetric: if $(a,b)\in R$ then $(b,a)\in R$ .

Transitive: if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$ . …
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CBSE 2024Set A1 mark
$R$
›Reveal solutionSolution

A relation that is reflexive, symmetric and transitive is called an equivalence relation.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let R be the relation in the set Z of all integers defined as, R = {(x, y) : x - y is an integer}, then R is$
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence relation
›Reveal solutionSolution

R is reflexive, symmetric, and transitive, hence an equivalence relation.

For any $x \in \mathbb{Z}$ , $x - x = 0$ , which is an integer, so $(x,x) \in R$ for every $x$ . Hence R is reflexive.

If $(x,y) \in R$ , then $x-y$ is an integer. Then $y - x = -(x-y)$ is also an integer, so $(y,x) \in R$ . Hence R is symmetric.

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CBSE 2023Set ANNUAL1 mark
$R$
›Reveal solutionSolution

For an equivalence relation R on A, both the domain and the range of R equal A itself.

Since R is an equivalence relation on A, it is reflexive: for every $a \in A$ , $(a,a) \in R$ . This means every element of $A$ appears as a first coordinate (so it is in the domain) and also as a second coordinate (so it is in the range) of some pair in R. H …

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CBSE 2023Set ANNUAL1 mark
$R$
›Reveal solutionSolution
An equivalence relation is one that is simultaneously reflexive, symmetric and transitive; the alternative just lists the images of the allowed $x$ -values.

A relation $R$ on a set $A$ is called an equivalence relation when it satisfies all three of the following properties:

Reflexive: $(a,a)\in R$ for every $a\in A$ .

Symmetric: if $(a,b)\in R$ then $(b,a)\in R$ .

Transitive: if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$ . …
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CBSE 2023Set ANNUAL1 mark
$Q. Define an equivalence relation.$
›Reveal solutionSolution
An equivalence relation is a relation satisfying all three properties: reflexivity, symmetry, and transitivity.

A relation $R$ defined on a set $A$ is called an equivalence relation if it satisfies all of the following three conditions:

Reflexive: $(a,a) \in R$ for every $a \in A$ .

Symmetric: If $(a,b) \in R$ , then $(b,a) \in R$ , for all $a,b \in A$ .

Transitive: If $(a,b)\in R$ and $(b,c)\in R$ , then $(a,c)\in R$ , for all $a,b,c \in A$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$B=\{b_1,b_2,b_3\}$
(a) $2^{6}$
(b) $2^{5}$
(c) $0$
(d) $2^{3}$
›Reveal solutionSolution

The number of relations from a set of size $m$ to a set of size $n$ is $2^{mn}$ ; here $m=3,n=2$ .

$B=\{b_1,b_2,b_3\}$ has $|B|=3$ elements; $G=\{g_1,g_2\}$ has $|G|=2$ elements.

A relation from $B$ to $G$ is any subset of $B\times G$ . The number of elements in $B\times G$ is $|B|\times|G|=3\times2=6$ .

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Which of the following relations on A = {1, 2, 3} is an equivalence relation?$
(a) {(1, 1), (2, 2), (3, 3)}
(b) {(1, 1), (2, 2), (3, 3), (1, 2)}
(c) {(1, 1), (3, 3), (1, 3), (3, 1)}
(d) None of these
›Reveal solutionSolution

Only option (a) is reflexive, symmetric, and transitive at once - the other two options fail reflexivity or symmetry.

An equivalence relation on $A=\{1,2,3\}$ must be reflexive (contain $(1,1),(2,2),(3,3)$ ), symmetric (if $(x,y)$ is in it so is $(y,x)$ ), and transitive.

(a) $\{(1,1),(2,2),(3,3)\}$ - contains all three diagonal pairs (reflexive), has no off-diagonal pair to break symmetry, and is trivially transitive. This IS an equivalence relation (it is the identity relation on $A$ ).

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CBSE 2022Set ANNUAL1 markMCQ
$y=x-4$
(a) Equivalence
(b) Only reflexive
(c) Not reflexive
(d) symmetric but not transitive
›Reveal solutionSolution

The relation 'is parallel to' on a set of lines is reflexive, symmetric and transitive, hence an equivalence relation.

$R=\{(L_1,L_2): L_1\parallel L_2\}$ on the set $L$ of all lines.

Reflexive: every line is parallel to itself, so $(L,L)\in R$ for all $L$ . (True)

Symmetric: if $L_1\parallel L_2$ then $L_2\parallel L_1$ , so $(L_1,L_2)\in R\Rightarrow(L_2,L_1)\in R$ . (True)

…

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CBSE 2020Set ANNUAL1 mark
$Q. What is meant by an equivalence relation?$
›Reveal solutionSolution

definition recall

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CBSE 2019Set ANNUAL1 markMCQ
$R\{(a, b) : a$
(a) No relation
(b) Trivial
(c) Equivalence relation
(d) Not symmetric
›Reveal solutionSolution

“Same parity” partitions A into evens and odds, so R is reflexive, symmetric and transitive — an equivalence relation.

R = {(a, b) : a and b are both even or both odd} on A = {1,2,3,4,5,6,7}.

Step 1 (Reflexive): Any a has the same parity as itself, so (a, a) ∈ R. ✓

Step 2 (Symmetric): If a and b have the same parity, so do b and a; (a,b) ∈ R ⇒ (b,a) ∈ R. ✓

…

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Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

Method: Proving an Equivalence Relation and Finding Its Classes (Parity / Modular Type)

Steps

Common Mistakes

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