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: Relation Properties — we check reflexivity, symmetry, and transitivity.

1. Reflexive: For any $a \in \mathbb{Z}$, $a - a = 0$, and $2$ divides $0$. So $(a, a) \in R$.

2. Symmetric: If $(a, b) \in R$, then $2 \mid (a - b)$. Hence $2 \mid -(a - b) = b - a$, so $(b, a) \in R$. …

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:

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. …

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

Use this for relations of the form $aRb \iff n \text{ divides } (a-b)$ (or any "same remainder / same value of some quantity" rule).

Steps

Step 1: Reflexivity

Show $a-a=0$ satisfies the condition. Since $n$ divides $0$ for any $n$, every element relates to itself.

Step 2: Symmetry

Assume $n\mid(a-b)$, i.e. $a-b=nk$. Then $b-a=-nk=n(-k)$, which is also a multiple of $n$, so $(b,a)\in R$.

Step 3: Transitivity …

Common Mistakes

Mistake 1: Thinking $0$ is not divisible by $2$, so reflexivity fails

Why it's wrong: $a-a=0$ and $0=2\times0$, so $2$ divides $0$; every element relates to itself. Correct approach: remember $0$ is a multiple of every integer — reflexivity holds.

Mistake 2: Proving symmetry/transitivity only with specific numbers …