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. Also find the equivalence class \( [0] \).
✅ Solution
🔹 Step 1: Reflexive
\[ a – a = 0 \]
Since 2 divides 0, \[ (a,a) \in R \]
✔ Reflexive
🔹 Step 2: Symmetric
If: \[ (a,b) \in R \Rightarrow a – b = 2k \]
Then: \[ b – a = -2k = 2(-k) \]
So, \[ (b,a) \in R \]
✔ Symmetric
🔹 Step 3: Transitive
If: \[ (a,b) \in R,\ (b,c) \in R \]
\[ a – b = 2m,\quad b – c = 2n \]
Add: \[ a – c = 2(m+n) \]
So, \[ (a,c) \in R \]
✔ Transitive
🎯 Conclusion
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🔹 Equivalence Class of 0
\[ [0] = \{a \in \mathbb{Z} : (a,0) \in R\} \]
\[ 2 \mid (a – 0) \Rightarrow 2 \mid a \]
So, \( a \) must be even.
\[ [0] = \{\ldots, -4, -2, 0, 2, 4, \ldots\} \]
🎯 Final Answer
\[ [0] = \{ \text{all even integers} \} \]
🚀 Exam Insight
- This is modulo 2 relation
- Classes: even numbers and odd numbers
- [0] always represents even class