Question

Let a relation \( R \) be defined on the set of all natural numbers \( \mathbb{N} \) by the rule: \( (a, b) \in R \) if and only if \( a + b \) is an even number. This relation \( R \) is classified as an:

• A. Equivalence Relation
• B. Symmetric but not Transitive Relation
• C. Reflexive but not Symmetric Relation
• D. Anti-symmetric Relation

Hint:

Parity relationships (checking if a sum or difference matches an even number or is divisible by an integer) are classic templates that almost always evaluate to Equivalence Relations in competitive exam modules.

Answer 1

The Correct Option is A

Concept: A relation is classified as an Equivalence Relation if and only if it satisfies three independent properties simultaneously: it must be Reflexive, Symmetric, and Transitive.

Step 1: Verify the Reflexive property.
A relation is reflexive if \( (a, a) \in R \) for every element. Substitute \( a \) into our addition rule: \[ a + a = 2a \] Since \( 2a \) is always a multiple of 2, the sum is always an even number for any natural number, making the relation Reflexive.

Step 2: Verify the Symmetric property.
A relation is symmetric if \( (a, b) \in R \) implies \( (b, a) \in R \). If \( a + b \) is even, then because addition is commutative (\( a + b = b + a \)), the expression \( b + a \) must also yield the exact same even value, proving the relation is Symmetric.

Step 3: Verify the Transitive property.
A relation is transitive if \( (a, b) \in R \) and \( (b, c) \in R \) implies \( (a, c) \in R \). 1. If \( a + b \) is even, both numbers must share the same parity (either both are even or both are odd).
2. If \( b + c \) is even, then since \( b \)'s parity is fixed, \( c \) must also share that same parity.
This matching parity ensures that \( a \) and \( c \) are either both even or both odd, meaning their sum \( a + c \) will always be an even number. Since the relation is reflexive, symmetric, and transitive, it is an Equivalence Relation.