Relation Defined by \( |a – b| \leq 5 \) on \( \mathbb{Z} \)
📺 Video Explanation
📝 Question
Let \( R_2 \) be a relation on \( \mathbb{Z} \) (set of integers) defined by:
\[ (a, b) \in R_2 \iff |a – b| \leq 5 \]
Test whether \( R_2 \) is reflexive, symmetric, and transitive.
✅ Solution
🔹 Step 1: Reflexive
A relation is reflexive if: \[ (a, a) \in R_2 \quad \forall a \in \mathbb{Z} \]
\[ |a – a| = 0 \leq 5 \]
✔ True for all integers.
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
A relation is symmetric if: \[ (a, b) \in R_2 \Rightarrow (b, a) \in R_2 \]
\[ |a – b| = |b – a| \]
So if \( |a – b| \leq 5 \), then \( |b – a| \leq 5 \).
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
A relation is transitive if: \[ (a, b) \in R_2 \text{ and } (b, c) \in R_2 \Rightarrow (a, c) \in R_2 \]
Given: \[ |a – b| \leq 5 \quad \text{and} \quad |b – c| \leq 5 \]
Using triangle inequality: \[ |a – c| \leq |a – b| + |b – c| \leq 5 + 5 = 10 \]
But this does NOT guarantee: \[ |a – c| \leq 5 \]
Counterexample: \[ a = 1,\ b = 6,\ c = 11 \]
\[ |1 – 6| = 5,\quad |6 – 11| = 5 \Rightarrow \text{both in } R_2 \]
\[ |1 – 11| = 10 > 5 \Rightarrow (a,c) \notin R_2 \]
❌ Therefore, the relation is Not Transitive.
🎯 Final Answer
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: No
\[ \therefore R_2 \text{ is reflexive and symmetric but not transitive} \]
🚀 Exam Insight
- Modulus relations are usually reflexive and symmetric.
- Always test transitivity using a counterexample.
- Triangle inequality helps but may not satisfy the required condition.