87. Let N be the set of all natural numbers. Consider the relation R on N given by
R = {(m, n) : m − n is divisible by 2}.
Then
(A) R is symmetric and transitive
(B) R is symmetric but NOT transitive
(C) R is reflexive but NOT symmetric
(D) R is reflexive and transitive
Relation R = {(m, n) : m − n is Divisible by 2}
Understanding the Problem
This question belongs to the topic of Relations and Set Theory, one of the most important areas of discrete mathematics. The given relation states that two natural numbers are related whenever their difference is divisible by 2. In simple words, two numbers are related if they have the same parity; that is, they are either both even or both odd.
To determine the correct option, we must verify whether the relation satisfies the three fundamental properties of a relation: reflexive, symmetric, and transitive. Each property should be checked separately using the mathematical definition instead of relying on intuition.
Step 1: Check Whether the Relation is Reflexive
A relation R is reflexive if every element is related to itself.
For every natural number m,
m − m = 0.
Since 0 is divisible by 2, we have
(m, m) ∈ R for every natural number.
Therefore, the relation is reflexive.
Step 2: Check Whether the Relation is Symmetric
A relation is symmetric if whenever (m, n) ∈ R, then (n, m) ∈ R.
Suppose
m − n is divisible by 2.
Then
n − m = −(m − n).
The negative of an even integer is also an even integer. Therefore, n − m is also divisible by 2.
Hence,
(n, m) ∈ R.
Therefore, the relation is symmetric.
Step 3: Check Whether the Relation is Transitive
A relation is transitive if whenever
(m, n) ∈ R
and
(n, p) ∈ R,
then
(m, p) ∈ R.
Assume
m − n is divisible by 2
and
n − p is divisible by 2.
Adding these two expressions gives
(m − n) + (n − p) = m − p.
The sum of two even integers is always even. Hence, m − p is also divisible by 2.
Therefore,
(m, p) ∈ R.
Thus, the relation is transitive.
Final Conclusion
The given relation satisfies all three important properties:
- Reflexive ✔
- Symmetric ✔
- Transitive ✔
Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.
Final Answer
Option (A): R is symmetric and transitive.
Although the relation is also reflexive, Option (A) is the most appropriate choice because it correctly states that the relation is both symmetric and transitive.
Why This Method Works
The defining condition of the relation depends only on the divisibility of the difference of two numbers by 2. Numbers having the same parity (both even or both odd) automatically satisfy this condition. Equality ensures reflexivity, reversing the order preserves divisibility and proves symmetry, while adding two even differences establishes transitivity. This systematic verification is the standard approach used in discrete mathematics and competitive examinations.
Key Concepts Covered
Reflexive Relation
A relation R on a set A is reflexive if every element is related to itself.
(a, a) ∈ R for every a ∈ A.
Symmetric Relation
A relation is symmetric if
(a, b) ∈ R ⇒ (b, a) ∈ R.
Transitive Relation
A relation is transitive if
(a, b) ∈ R and (b, c) ∈ R together imply (a, c) ∈ R.
Equivalence Relation
A relation that is reflexive, symmetric, and transitive is called an equivalence relation. Such relations partition a set into equivalence classes. In this problem, the natural numbers are divided into two equivalence classes: even numbers and odd numbers.
Correct Answer: Option (A)
