Relation Defined by \( 2x + y = 41 \) on \( \mathbb{N} \)
📺 Video Explanation
📝 Question
Let \( R = \{(x,y) : x,y \in \mathbb{N},\ 2x + y = 41\} \).
Find:
- Domain of \( R \)
- Range of \( R \)
- Check whether \( R \) is reflexive, symmetric, and transitive
✅ Solution
🔹 Step 1: Express Relation
\[ 2x + y = 41 \Rightarrow y = 41 – 2x \]
Since \( x,y \in \mathbb{N} \), we need: \[ 41 – 2x > 0 \Rightarrow x < 20.5 \]
So: \[ x = 1,2,3,\dots,20 \]
🔹 Step 2: Domain
\[ \text{Domain} = \{1,2,3,\dots,20\} \]
🔹 Step 3: Range
Compute:
\[ y = 41 – 2x \]
Values:
\[ \{39,37,35,\dots,1\} \]
\[ \text{Range} = \{1,3,5,\dots,39\} \]
🔹 Step 4: Reflexive
Reflexive requires: \[ (x,x) \in R \Rightarrow 2x + x = 41 \Rightarrow 3x = 41 \]
\[ x = \frac{41}{3} \notin \mathbb{N} \]
❌ Not Reflexive
🔹 Step 5: Symmetric
If \( (x,y) \in R \), then: \[ 2x + y = 41 \]
For symmetry, need: \[ 2y + x = 41 \]
Example: \[ x=1,\ y=39 \Rightarrow (1,39) \in R \]
Check reverse: \[ 2(39) + 1 = 79 \neq 41 \]
❌ Not Symmetric
🔹 Step 6: Transitive
If: \[ (x,y) \in R \Rightarrow 2x + y = 41 \] \[ (y,z) \in R \Rightarrow 2y + z = 41 \]
Then: \[ y = 41 – 2x \]
Substitute:
\[ z = 41 – 2y = 41 – 2(41 – 2x) = 4x – 41 \]
For transitivity, need: \[ 2x + z = 41 \]
\[ 2x + (4x – 41) = 6x – 41 = 41 \Rightarrow x = \frac{82}{6} \]
Not valid for all \( x \).
❌ Not Transitive
🎯 Final Answer
Domain: \( \{1,2,3,\dots,20\} \)
Range: \( \{1,3,5,\dots,39\} \)
✔ Reflexive: No
✔ Symmetric: No
✔ Transitive: No
🚀 Exam Insight
- Convert relation into function form first
- Domain from valid x-values
- Range from corresponding y-values
- Use substitution method for transitivity