Relation \( a – b \) divisible by 3 on \( \mathbb{Z} \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{Z} \) be defined as:
\[ (a, b) \in R \iff a – b \text{ is divisible by } 3 \]
Show that \( R \) is an equivalence relation.
✅ 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 3, \[ (a,a) \in R \]
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
Assume: \[ (a,b) \in R \Rightarrow a – b = 3k \]
Then: \[ b – a = -3k = 3(-k) \]
So, \[ (b,a) \in R \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
Assume: \[ (a,b) \in R,\ (b,c) \in R \]
\[ a – b = 3m,\quad b – c = 3n \]
Adding: \[ a – c = (a – b) + (b – c) = 3m + 3n = 3(m+n) \]
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
- Think in terms of modulo: \( a \equiv b \ (\text{mod } n) \)
- These relations form equivalence classes