Relation Defined by \( a = \frac{1}{b} \) on \( Q_0 \)
📺 Video Explanation
📝 Question
Let \( R_1 \) be a relation on \( Q_0 \) (set of non-zero rational numbers) defined by:
\[ (a, b) \in R_1 \iff a = \frac{1}{b} \]
Test whether \( R_1 \) is reflexive, symmetric, and transitive.
✅ Solution
🔹 Step 1: Reflexive
A relation is reflexive if: \[ (a, a) \in R_1 \quad \forall a \in Q_0 \]
This requires: \[ a = \frac{1}{a} \Rightarrow a^2 = 1 \Rightarrow a = \pm 1 \]
This is true only for \( a = 1 \) or \( -1 \), not for all elements.
❌ Therefore, the relation is Not Reflexive.
🔹 Step 2: Symmetric
If \( (a, b) \in R_1 \), then: \[ a = \frac{1}{b} \]
Taking reciprocal: \[ b = \frac{1}{a} \]
So, \( (b, a) \in R_1 \).
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
If: \[ (a, b) \in R_1 \Rightarrow a = \frac{1}{b} \] \[ (b, c) \in R_1 \Rightarrow b = \frac{1}{c} \]
Substitute: \[ a = \frac{1}{b} = \frac{1}{(1/c)} = c \]
So, \[ a = c \]
For transitivity, we need: \[ a = \frac{1}{c} \]
But generally \( a = c \neq \frac{1}{c} \).
❌ Therefore, the relation is Not Transitive.
🎯 Final Answer
✔ Reflexive: No
✔ Symmetric: Yes
✔ Transitive: No
\[ \therefore R_1 \text{ is symmetric only} \]
🚀 Exam Insight
- Relations involving reciprocals are usually symmetric.
- Check reflexive by substituting same variables.
- Always verify transitivity using substitution carefully.