Relation \( a^2 + b^2 = 1 \) on \( \mathbb{R} \)
📺 Video Explanation
📝 Question
Let relation \( S \) on \( \mathbb{R} \) be defined as:
\[ (a, b) \in S \iff a^2 + b^2 = 1 \]
Show that \( S \) is not an equivalence relation.
✅ Concept Used
A relation is an equivalence relation if it is:
- Reflexive
- Symmetric
- Transitive
(All three must hold) :contentReference[oaicite:0]{index=0}
❌ Check Reflexive Property
For reflexive, we need: \[ (a,a) \in S \quad \forall a \in \mathbb{R} \]
Check: \[ a^2 + a^2 = 2a^2 \]
For this to be in \( S \), \[ 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2} \]
This is NOT true for all real numbers.
❌ Therefore, relation is Not Reflexive.
🔹 Check Symmetric Property
If: \[ (a,b) \in S \Rightarrow a^2 + b^2 = 1 \]
Then: \[ b^2 + a^2 = 1 \]
So, \[ (b,a) \in S \]
✔ Relation is Symmetric.
❌ Check Transitive Property
Assume: \[ (a,b) \in S,\ (b,c) \in S \]
\[ a^2 + b^2 = 1,\quad b^2 + c^2 = 1 \]
Take example: \[ a = 1,\ b = 0,\ c = 1 \]
Check: \[ 1^2 + 0^2 = 1 ✔ \] \[ 0^2 + 1^2 = 1 ✔ \]
But: \[ 1^2 + 1^2 = 2 \ne 1 \]
So, \[ (a,c) \notin S \]
❌ Therefore, relation is Not Transitive.
🎯 Final Conclusion
✔ Reflexive: No
✔ Symmetric: Yes
✔ Transitive: No
\[ \therefore S \text{ is NOT an equivalence relation} \]
🚀 Exam Insight
- Fails reflexive ⇒ automatically not equivalence
- Always try counterexample for transitivity
- Equation-based relations often fail reflexivity