Step 1: Understanding the Concept:
An equivalence relation is a specific type of binary relation that partitions a set into disjoint equivalence classes.
To qualify as an equivalence relation, the relation must simultaneously satisfy three properties: reflexivity, symmetry, and transitivity.
In the context of parity (even/odd), we check how the sum of numbers behaves.
Step 2: Detailed Explanation:
Reflexivity:
A relation is reflexive if every element is related to itself: \((a, a) \in R\).
Substitute \(a\) into the rule: \(a + a = 2a\).
Since \(2a\) is always a multiple of 2, the sum is even for any natural number \(a\).
Thus, R is reflexive.
Symmetry:
A relation is symmetric if \((a, b) \in R \implies (b, a) \in R\).
If \((a, b) \in R\), then \(a + b = E\) (some even number).
By the commutative property of addition, \(b + a = a + b = E\).
Since \(b + a\) is also even, \((b, a) \in R\).
Thus, R is symmetric.
Transitivity:
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, then both \(a\) and \(b\) must have the same parity (both are even or both are odd).
2. If \(b + c\) is even, then \(b\) and \(c\) must have the same parity.
Since \(b\) has the same parity as \(a\) AND \(b\) has the same parity as \(c\), it follows that \(a\) and \(c\) must have the same parity.
If \(a\) and \(c\) have the same parity, their sum \(a + c\) is always even.
Thus, R is transitive.
Step 4: Final Answer:
R is reflexive, symmetric, and transitive. Therefore, it is an Equivalence Relation.
This matches option (A).