Relation: \( xy \) is a Perfect Square on \( \mathbb{N} \)
📺 Video Explanation
📝 Question
Let relation \( R \) on \( \mathbb{N} \) be defined as:
\[ (x, y) \in R \iff xy \text{ is a perfect square} \]
Check whether \( R \) is reflexive, symmetric, and transitive.
✅ Solution
🔹 Step 1: Reflexive
A relation is reflexive if: \[ (x, x) \in R \quad \forall x \in \mathbb{N} \]
Check: \[ x \cdot x = x^2 \]
Since \( x^2 \) is always a perfect square,
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
A relation is symmetric if: \[ (x, y) \in R \Rightarrow (y, x) \in R \]
Since: \[ xy = yx \]
If \( xy \) is a perfect square, then \( yx \) is also a perfect square.
✔ 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: \[ xy = a^2,\quad yz = b^2 \]
Multiply: \[ (xy)(yz) = a^2 b^2 \Rightarrow x y^2 z = (ab)^2 \]
But this does not guarantee: \[ xz \text{ is a perfect square} \]
Counterexample:
Take: \[ x = 2,\ y = 8,\ z = 18 \]
\[ xy = 16 = 4^2,\quad yz = 144 = 12^2 \]
But: \[ xz = 36 = 6^2 \ (\text{this works, try better}) \]
Try: \[ x=2,\ y=2,\ z=8 \]
\[ xy = 4,\ yz = 16 \ (\text{both squares}) \]
But: \[ xz = 16 \ (\text{still square}) → try different: \]
Take: \[ x=2,\ y=18,\ z=8 \]
\[ xy=36,\ yz=144 \ (\text{both squares}) \]
\[ xz=16 \ (\text{still square}) \]
👉 Actually, relation is Transitive (based on prime factorization argument)
✔ Therefore, the relation is Transitive.
🎯 Final Answer
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🚀 Exam Insight
- Perfect square → even powers in prime factorization
- Square × square = square → helps in transitivity
- This is a standard equivalence relation question