Q.Let $R$ be the relation defined in the set $A = \{1, 2, 3, 4, 5, 6, 7\}$ by $R = \{(a, b): \text{both } a \text{ and } b \text{ are either odd or even}\}$. Show that $R$ is an equivalence relation. Further, show that all the elements of the subset $\{1, 3, 5, 7\}$ are related to each other and all the elements of the subset $\{2, 4, 6\}$ are related to each other, but no element of the subset $\{1, 3, 5, 7\}$ is related to any element of the subset $\{2, 4, 6\}$.

Concept: Relation Properties — We check reflexivity, symmetry, and transitivity to prove equivalence. The relation groups numbers by parity (odd/even).

Step 1 — Reflexive: For any $a \in A$, $a$ is either odd or even, so $(a,a) \in R$. Hence $R$ is reflexive.

Step 2 — Symmetric: If $(a,b) \in R$, then $a$ and $b$ share the same parity. So $b$ and $a$ also share that parity, giving $(b,a) \in R$. Hence $R$ is symmetric.

Step 3 — Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $a$ and $b$ have the same parity, and $b$ and $c$ have the same parity. Thus $a$ and $c$ have the same parity, so $(a,c) \in R$. Hence $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Step 4 — Partition into equivalence classes: …

Why parity? The core idea

A relation is an equivalence relation if it is reflexive, symmetric, and transitive. Here, the rule is simple: two numbers are related when they have the same "type" — both odd or both even. That’s exactly what parity means.

This kind of relation always creates neat, non-overlapping groups (called equivalence classes). Every odd number is related to every other odd, every even to every even, and no cross-relations exist. Let’s verify each property.

Step-by-step verification

1. Reflexive: Is every element related to itself?

Take any $a \in A$. Is $a$ either odd or even? Yes — every integer is one or the other. So $a$ and $a$ are both odd or both even. Hence $(a, a) \in R$ for all $a \in A$. Reflexivity holds.

2. Symmetric: If $a$ is related to $b$, is $b$ related to $a$?

Suppose $(a, b) \in R$. That means $a$ and $b$ are both odd or both even. But that statement is symmetric: if $a$ and $b$ share parity, then $b$ and $a$ share parity too. So $(b, a) \in R$. Symmetry holds.

3. Transitive: If $a$ is related to $b$ and $b$ to $c$, is $a$ related to $c$?

Assume $(a, b) \in R$ and $(b, c) \in R$.

• From $(a, b)$: $a$ and $b$ have the same parity.
• From $(b, c)$: $b$ and $c$ have the same parity.

If $a$ and $b$ share parity, and $b$ and $c$ share parity, then $a$ and $c$ must share that same parity. So $(a, c) \in R$. Transitivity holds.

Since all three properties are satisfied, $R$ is an equivalence relation.

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Method: Proving an equivalence relation and finding its equivalence classes

Use this two-part method when a relation groups elements by a shared attribute (same parity, same remainder, same size) and you must both prove it is an equivalence relation and describe the classes.

Steps

Step 1: Prove reflexive, symmetric, transitive

Reflexive: an element shares its attribute with itself. Symmetric: if $a$ shares the attribute with $b$, then $b$ shares it with $a$. Transitive: if $a$ matches $b$ and $b$ matches $c$, then $a$ matches $c$ — because "having the same value of the attribute" is automatically consistent along a chain.

Step 2: Form the equivalence classes by the attribute …

Common Mistakes

Mistake 1: Thinking transitivity needs the three elements to be equal or consecutive

Why it's wrong: transitivity only needs the attribute (parity) to stay consistent along the chain, not the numbers themselves; $1,3,5$ are all related because all are odd. Correct approach: argue "same parity" is preserved: if $a,b$ share parity and $b,c$ share parity, then $a,c$ share parity.

Mistake 2: Describing the classes wrongly or claiming cross-relations exist …