Relation \( a – b \) divisible by \( n \) on \( \mathbb{Z} \)
📺 Video Explanation
📝 Question
Let \( n \) be a fixed positive integer. Define a relation \( R \) on \( \mathbb{Z} \) as:
\[ (a, b) \in R \iff a – b \text{ is divisible by } n \]
Show that \( R \) is an equivalence relation on \( \mathbb{Z} \).
✅ Solution
🔹 Step 1: Reflexive
For reflexive, we need: \[ (a,a) \in R \quad \forall a \in \mathbb{Z} \]
\[ a – a = 0 \]
Since 0 is divisible by \( n \), \[ (a,a) \in R \]
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
Assume: \[ (a,b) \in R \Rightarrow a – b = nk \]
\[ b – a = -nk = n(-k) \]
So, \[ (b,a) \in R \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
Assume: \[ (a,b) \in R,\ (b,c) \in R \]
\[ a – b = nm,\quad b – c = np \]
Adding: \[ a – c = (a – b) + (b – c) = nm + np = n(m+p) \]
So, \[ (a,c) \in R \]
✔ Therefore, the relation is Transitive.
🎯 Final Answer
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🚀 Exam Insight
- \( a – b \) divisible by \( n \) → always an equivalence relation
- This represents congruence: \( a \equiv b \ (\text{mod } n) \)
- Integers are divided into \( n \) equivalence classes