A relation R on set $A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$ is defined as $R = \{(x, y) : x + y \text{ is an integer divisible by } 2\}$. Show that R is an equivalence relation. Also, write the equivalence class $[2]$.

> [!TLDR] > R is an equivalence relation because it is reflexive, symmetric, and transitive — it essentially groups numbers by parity (even/odd). The equivalence class $[2]$ is $\{-4, -2, 0, 2, 4\}$. The key insight here is that $x + y$ being divisible by 2 means $x$ and $y$ have the same parity — both even or both odd. Why? Because an even number plus an odd number gives an odd sum, which is not divisible by 2. So R is really just "same parity" in disguise. Once you see that, proving it's an equivalence relation becomes almost trivial, and finding $[2]$ is just listing all even numbers in A. Let's prove it formally step by step. 1. **Reflexive**: For any $x \in A$, we need $(x, x) \in R$. That means $x + x = 2x$ must be divisible by 2. Since $2x$ is always a multiple of 2, this holds for every $x$ in A. So R is reflexive. 2. **Symmetric**: If $(x, y) \in R$, then $x + y$ is divisible by 2. But $x + y = y + x$, so $y + x$ is also divisible by 2, meaning $(y, x) \in R$. Symmetry is immediate from commutativity of addition. 3. **Transitive**: This is the only step that needs a little work. Suppose $(x, y) \in R$ and $(y, z) \in R$. Then $x + y = 2a$ and $y + z = 2b$ for some integers $a, b$. We need to show $x + z$ is divisible by 2. Add the two equations: $(x + y) + (y + z) = 2a + 2b$, so $x + 2y + z = 2(a + b)$. Subtract $2y$ from both sides: $x + z = 2(a + b - y)$. Since $a + b - y$ is an integer, $x + z$ is divisible by 2. Hence $(x, z) \in R$, and transitivity holds. …

A relation R on set A = \-4, -3, -2, -1, 0, 1, 2, 3, 4\ is defined as R = \(x, y) : x + y is an integer divisi

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A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}

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R is an equivalence relation because it is reflexive, symmetric, and transitive — it essentially groups numbers by parity (even/odd). The equivalence class $[2]$ is $\{-4, -2, 0, 2, 4\}$ .

The key insight here is that $x + y$ being divisible by 2 means $x$ and $y$ have the same parity — both even or both odd. Why? Because an even number plus an odd number gives an odd sum, which is not divisible by 2. So R is really just "same parity" in disguise. Once you see that, proving it's an equivalence relation becomes almost trivial, and finding $[2]$ is just listing all even numbers in A.

Let's prove it formally step by step.

Reflexive: For any $x \in A$ , we need $(x, x) \in R$ . That means $x + x = 2x$ must be divisible by 2. Since $2x$ is always a multiple of 2, this holds for every $x$ in A. So R is reflexive.
Symmetric: If $(x, y) \in R$ , then $x + y$ is divisible by 2. But $x + y = y + x$ , so $y + x$ is also divisible by 2, meaning $(y, x) \in R$ . Symmetry is immediate from commutativity of addition.
Transitive: This is the only step that needs a little work. Suppose $(x, y) \in R$ and $(y, z) \in R$ . Then $x + y = 2a$ and $y + z = 2b$ for some integers $a, b$ . We need to show $x + z$ is divisible by 2.

Add the two equations: $(x + y) + (y + z) = 2a + 2b$ , so $x + 2y + z = 2(a + b)$ . Subtract $2y$ from both sides: $x + z = 2(a + b - y)$ . Since $a + b - y$ is an integer, $x + z$ is divisible by 2. Hence $(x, z) \in R$ , and transitivity holds. …

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

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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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Q.A relation R on set $A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$ is defined as $R = \{(x, y) : x + y \text{ is an integer divisible by } 2\}$ . Show that R is an equivalence relation. Also, write the equivalence class $[2]$ .

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

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