Question

Let $ A $ be the set of all functions $ f: \mathbb{Z} \to \mathbb{Z} $ and $ R $ be a relation on $ A $ such that $$ R = \{ (f, g) : f(0) = g(1) \text{ and } f(1) = g(0) \} $$ Then $ R $ is:

• A. Symmetric and transitive but not reflective
• B. Symmetric but neither reflective nor transitive
• C. Reflexive but neither symmetric nor transitive
• D. Transitive but neither reflexive nor symmetric

Hint:

Always verify reflexivity, symmetry, and transitivity with definitions and counterexamples.

Answer 1

The Correct Option is B

- Reflexive For \( f \in A \), \( fRf \Rightarrow f(0) = f(1) \). But not true for all \( f \). 
So not reflexive. 
- Symmetric If \( fRg \Rightarrow f(0) = g(1), f(1) = g(0) \), then clearly \( gRf \). 
So symmetric. 
- Transitive 
Suppose \( fRg \) and \( gRh \Rightarrow f(0) = g(1), g(0) = f(1), g(0) = h(1), g(1) = h(0) \). 
This implies \( f(0) = h(0) \) and \( f(1) = h(1) \), which does not satisfy \( fRh \Rightarrow \) Not transitive.