Relation \( x + y = 10 \) on \( \mathbb{N} \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{N} \) be defined as:
\[ (x, y) \in R \iff x + y = 10 \]
Check whether \( R \) is reflexive, symmetric, and transitive.
✅ Solution
🔹 Step 1: Reflexive
A relation is reflexive if: \[ (x, x) \in R \quad \forall x \in \mathbb{N} \]
This requires: \[ x + x = 10 \Rightarrow 2x = 10 \Rightarrow x = 5 \]
True only for \( x = 5 \), not for all natural numbers.
❌ Therefore, the relation is Not Reflexive.
🔹 Step 2: Symmetric
A relation is symmetric if: \[ (x, y) \in R \Rightarrow (y, x) \in R \]
Since: \[ x + y = 10 \Rightarrow y + x = 10 \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
A relation is transitive if: \[ (x, y) \in R \text{ and } (y, z) \in R \Rightarrow (x, z) \in R \]
Given: \[ x + y = 10 \quad \text{and} \quad y + z = 10 \]
Subtract: \[ x = z \]
For transitivity, need: \[ x + z = 10 \]
But: \[ x + z = x + x = 2x \neq 10 \text{ (in general)} \]
Example: \[ (3,7),(7,3) \in R \]
But: \[ (3,3) \notin R \ (\text{since } 6 \neq 10) \]
❌ Therefore, the relation is Not Transitive.
🎯 Final Answer
✔ Reflexive: No
✔ Symmetric: Yes
✔ Transitive: No
\[ \therefore R \text{ is symmetric only} \]
🚀 Exam Insight
- Equations like \( x + y = \text{constant} \) are always symmetric
- Check reflexive by putting \( x = y \)
- Use simple numbers (like 3,7) to test transitivity