Relation \( 2 \mid (a – b) \) on \( \mathbb{Z} \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{Z} \) be defined as:
\[ (a, b) \in R \iff 2 \mid (a – b) \]
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 2 divides 0, \[ (a,a) \in R \]
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
Assume: \[ (a,b) \in R \Rightarrow 2 \mid (a – b) \]
\[ b – a = -(a – b) \]
Since divisibility is preserved under negatives, \[ 2 \mid (b – a) \]
So, \[ (b,a) \in R \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
Assume: \[ (a,b) \in R,\ (b,c) \in R \]
\[ a – b = 2m,\quad b – c = 2n \]
Adding: \[ a – c = (a – b) + (b – c) = 2m + 2n = 2(m+n) \]
So, \[ 2 \mid (a – c) \]
\[ (a,c) \in R \]
✔ Therefore, the relation is Transitive.
🎯 Final Answer
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🚀 Exam Insight
- \( n \mid (a – b) \) type relations are always equivalence relations
- This represents congruence: \( a \equiv b \ (\text{mod } 2) \)
- It divides integers into even and odd classes