Relation \( a + b \) is even on \( \mathbb{Z} \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{Z} \) be defined as:
\[ (a, b) \in R \iff a + b \text{ is even} \]
Show that \( R \) is an equivalence relation.
✅ Solution
🔹 Step 1: Reflexive
For reflexive, we need: \[ (a,a) \in R \quad \forall a \in \mathbb{Z} \]
\[ a + a = 2a \]
Since \( 2a \) is always even, \[ (a,a) \in R \]
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
Assume: \[ (a,b) \in R \Rightarrow a + b \text{ is even} \]
\[ b + a = a + b \]
So, \[ (b,a) \in R \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
Assume: \[ (a,b) \in R,\ (b,c) \in R \]
\[ a + b = 2m,\quad b + c = 2n \]
Add: \[ (a + b) + (b + c) = a + c + 2b = 2m + 2n \]
\[ a + c = 2(m+n-b) \]
Thus, \( a + c \) is even.
\[ (a,c) \in R \]
✔ Therefore, the relation is Transitive.
🎯 Final Answer
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🚀 Exam Insight
- \( a + b \) even ⇒ both are either even or both are odd
- This splits integers into two classes: even and odd
- Such relations always form equivalence classes