Q.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 $\{2, 4\}$.

Concept: Relation Properties — To prove equivalence, we check reflexivity, symmetry, and transitivity.

Step 1 — Reflexive: For any $a \in A$, $|a - a| = 0$, which is even. So $(a, a) \in R$ for all $a$.

Step 2 — Symmetric: If $(a, b) \in R$, then $|a - b|$ is even. Since $|b - a| = |a - b|$, it is also even, so $(b, a) \in R$.

Step 3 — Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $|a - b|$ and $|b - c|$ are even. Their sum $|a - b| + |b - c|$ is even, and $|a - c|$ differs from this sum by an even number (parity argument), so $|a - c|$ is even. Hence $(a, c) \in R$.

Thus $R$ is an equivalence relation. …

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.

1. Reflexive: For any $a \in A$, $|a-a| = 0$, which is even. So $(a,a) \in R$ for every $a$. Reflexivity holds.

2. 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.

3. 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. …

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

Use this for relations like "$|a-b|$ is even" or "$a-b$ is divisible by $n$", and then identifying the equivalence classes.

Steps

Step 1: Recast the rule as "same remainder"

"$|a-b|$ even" means $a$ and $b$ have the same parity; "$a-b$ divisible by $n$" means $a\equiv b \pmod n$. This reframing makes all three proofs immediate.

Step 2: Reflexive

$a-a=0$ is even (and divisible by anything), so $(a,a)\in R$.

Step 3: Symmetric

$|a-b|=|b-a|$, so if one is even the other is; hence $(b,a)\in R$.

Step 4: Transitive …

Common Mistakes

Mistake 1: Reading "$|a-b|$ even" as "$a$ and $b$ are even"

Why it's wrong: two odd numbers also have an even difference; the rule means same parity, not "both even". Correct approach: split into the odd class and the even class.

Mistake 2: Proving transitivity with the triangle inequality alone …