Show that the relation $R$ in the set $\mathbb{Z}$ of integers given by $R = \{(a, b): 2 \text{ divides } a - b\}$ is an equivalence relation.

> [!TLDR] > The relation $R$ defined by $a - b$ being divisible by $2$ is an equivalence relation because it satisfies reflexivity ($a - a = 0$ is divisible by $2$), symmetry (if $a - b$ is divisible by $2$, then $b - a$ is also divisible by $2$), and transitivity (if $a - b$ and $b - c$ are divisible by $2$, then $a - c$ is also divisible by $2$). The core idea here is that $R$ groups integers by their parity — whether they are even or odd. Two integers are related if their difference is even, which means they have the same remainder when divided by $2$. This is a classic example of an equivalence relation, and proving it requires checking three properties: reflexivity, symmetry, and transitivity. Let’s work through each property step by step. 1. **Reflexivity**: For any integer $a \in \mathbb{Z}$, we need to show that $(a, a) \in R$. Compute $a - a = 0$. Since $0$ is divisible by $2$ (because $0 = 2 \times 0$), the condition $2 \mid (a - a)$ holds. Thus, every integer is related to itself, so $R$ is reflexive. 2. **Symmetry**: For any $a, b \in \mathbb{Z}$, if $(a, b) \in R$, then we must show $(b, a) \in R$. If $(a, b) \in R$, then $2 \mid (a - b)$. This means $a - b = 2k$ for some integer $k$. Now consider $b - a = -(a - b) = -2k = 2(-k)$. Since $-k$ is an integer, $b - a$ is also divisible by $2$. Hence, $(b, a) \in R$, proving symmetry. 3. **Transitivity**: For any $a, b, c \in \mathbb{Z}$, if $(a, b) \in R$ and $(b, c) \in R$, then we must show $(a, c) \in R$. From $(a, b) \in R$, we have $a - b = 2k$ for some integer $k$. From $(b, c) \in R$, we have $b - c = 2m$ for some integer $m$. Add these two equations: $$(a - b) + (b - c) = 2k + 2m$$ This simplifies to $a - c = 2(k + m)$. Since $k + m$ is an integer, $a - c$ is divisible by $2$. Therefore, $(a, c) \in R$, and transitivity holds. …

Show that the relation R in the set mathbbZ of int – Class 12 Mathematics

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The relation $R$ defined by $a - b$ being divisible by $2$ is an equivalence relation because it satisfies reflexivity ( $a - a = 0$ is divisible by $2$ ), symmetry (if $a - b$ is divisible by $2$ , then $b - a$ is also divisible by $2$ ), and transitivity (if $a - b$ and $b - c$ are divisible by $2$ , then $a - c$ is also divisible by $2$ ).

The core idea here is that $R$ groups integers by their parity — whether they are even or odd. Two integers are related if their difference is even, which means they have the same remainder when divided by $2$ . This is a classic example of an equivalence relation, and proving it requires checking three properties: reflexivity, symmetry, and transitivity.

Let’s work through each property step by step.

Reflexivity: For any integer $a \in \mathbb{Z}$ , we need to show that $(a, a) \in R$ .

Compute $a - a = 0$ . Since $0$ is divisible by $2$ (because $0 = 2 \times 0$ ), the condition $2 \mid (a - a)$ holds. Thus, every integer is related to itself, so $R$ is reflexive.
Symmetry: For any $a, b \in \mathbb{Z}$ , if $(a, b) \in R$ , then we must show $(b, a) \in R$ .

If $(a, b) \in R$ , then $2 \mid (a - b)$ . This means $a - b = 2k$ for some integer $k$ . Now consider $b - a = -(a - b) = -2k = 2(-k)$ . Since $-k$ is an integer, $b - a$ is also divisible by $2$ . Hence, $(b, a) \in R$ , proving symmetry.
Transitivity: For any $a, b, c \in \mathbb{Z}$ , if $(a, b) \in R$ and $(b, c) \in R$ , then we must show $(a, c) \in R$ .

From $(a, b) \in R$ , we have $a - b = 2k$ for some integer $k$ . From $(b, c) \in R$ , we have $b - c = 2m$ for some integer $m$ . Add these two equations:

(a - b) + (b - c) = 2k + 2m

This simplifies to $a - c = 2(k + m)$ . Since $k + m$ is an integer, $a - c$ is divisible by $2$ . Therefore, $(a, c) \in R$ , and transitivity holds. …

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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.Show that the relation $R$ in the set $\mathbb{Z}$ of integers given by $R = \{(a, b): 2 \text{ divides } a - b\}$ is an equivalence relation.

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

Method: Proving an equivalence relation defined by a difference/divisibility condition

Steps

Common Mistakes

Q.Show that the relation RRR in the set Z\mathbb{Z}Z of integers given by R={(a,b):2 divides a−b}R = \{(a, b): 2 \text{ divides } a - b\}R={(a,b):2 divides a−b} is an equivalence relation.

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Q.Show that the relation $R$ in the set $\mathbb{Z}$ of integers given by $R = \{(a, b): 2 \text{ divides } a - b\}$ is an equivalence relation.

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