Let $n$ be a fixed positive integer. Define a relation $R$ in $\mathbb{Z}$ as follows: $\forall\, a, b \in \mathbb{Z}$, $aRb$ if and only if $a - b$ is divisible by $n$. Show that $R$ is an equivalence relation.

> [!TLDR] > The relation $R$ defined by $aRb \iff n \mid (a-b)$ is an equivalence relation because it satisfies reflexivity ($a-a=0$ is divisible by $n$), symmetry (if $n\mid(a-b)$ then $n\mid(b-a)$), and transitivity (if $n\mid(a-b)$ and $n\mid(b-c)$ then $n\mid(a-c)$). This is the standard **congruence modulo $n$** relation. The core idea here is that divisibility by a fixed integer $n$ creates a natural way to group integers into classes based on their remainder when divided by $n$. The relation $R$ simply says two integers are related if their difference is a multiple of $n$ — which is exactly the definition of being congruent modulo $n$, written as $a \equiv b \pmod{n}$. To prove any relation is an equivalence relation, we must check three properties: reflexivity, symmetry, and transitivity. Each one follows directly from basic facts about divisibility. --- 1. **Reflexivity**: For any $a \in \mathbb{Z}$, we need $aRa$. Since $a - a = 0$, and $0$ is divisible by every positive integer $n$ (because $0 = n \cdot 0$), we have $n \mid (a-a)$. Hence $aRa$ holds for all $a$. 2. **Symmetry**: If $aRb$, then $n \mid (a-b)$. We need to show $bRa$, i.e., $n \mid (b-a)$. If $a-b = nk$ for some integer $k$, then $b-a = -(a-b) = n(-k)$, which is also a multiple of $n$. So $n \mid (b-a)$, and $bRa$ follows. 3. **Transitivity**: If $aRb$ and $bRc$, then $n \mid (a-b)$ and $n \mid (b-c)$. We need $n \mid (a-c)$. Write $a-b = nk$ and $b-c = n\ell$ for integers $k, \ell$. Adding these gives: $$(a-b) + (b-c) = a-c = nk + n\ell = n(k+\ell)$$ Since $k+\ell$ is an integer, $a-c$ is a multiple of $n$. Thus $n \mid (a-c)$, so $aRc$. …

Let n be a fixed positive integer. Define a relation R in mathbbZ as follows: ∀ a, b ∈ mathbbZ, aRb if and onl

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Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

Note

Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.

The three checks

Reflexive — show $(a, a) \in R$ for every $a \in A$ .
Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$ .
Transitive — assume $(a, b) \in R$ and $(b, c) \in R$ , and deduce $(a, c) \in R$ .

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$ .

Reflexive: $a - a = 0$ , and $0$ is divisible by $5$ , so $a\,R\,a$ for every integer $a$ . ✓

Symmetric: if $a\,R\,b$ , then $a - b = 5k$ for some integer $k$ . Then $b - a = -5k = 5(-k)$ , also a multiple of $5$ , so $b\,R\,a$ . ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$ , then $a - b = 5k$ and $b - c = 5m$ . Adding, $a - c = 5(k + m)$ , a multiple of $5$ , so $a\,R\,c$ . ✓

All three hold, so $R$ is an equivalence relation. …

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Q.Let nnn be a fixed positive integer. Define a relation RRR in Z\mathbb{Z}Z as follows: ∀ a,b∈Z\forall\, a, b \in \mathbb{Z}∀a,b∈Z, aRbaRbaRb if and only if a−ba - ba−b is divisible by nnn. Show that RRR is an equivalence relation.

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

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Q.Let $n$ be a fixed positive integer. Define a relation $R$ in $\mathbb{Z}$ as follows: $\forall\, a, b \in \mathbb{Z}$ , $aRb$ if and only if $a - b$ is divisible by $n$ . Show that $R$ is an equivalence relation.

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