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Solution


Q.

Let A be the set of all functions f : Z → Z and R be a relation on A such that R = {(f, g) : f(0) = g(1) and f(1) = g(0)}. Then R is :          [2025]
1 Transitive but neither reflexive nor symmetric.  
2 Symmetric and transitive but not reflexive.  
3 Symmetric but neither reflexive nor transitive.  
4 Reflexive but neither symmetric nor transitive.  

Ans.

(3)

We have, R = {(f, g) : f(0) = g(1) and f(1) = g(0)}

For reflexive: (f, f) ∈ R

⇒  f(0) = f(1) must hold ∀f

But f(0) ≠ f(1)

Therefore, R is not reflexive.

For symmetric: If (f, g) ∈ R ⇒ (g, f) ∈ R

Now, g(0) = f(1) and g(1) = f(0), which is true ∀f

∴  R is symmetric

For transitive : IF (f, g) ∈ R and (g, h) ∈ R ⇒ (f, h) ∈ R

Now, (f, g) ∈ R ⇒ f(0) = g(1) and f(1) = g(0)

          (g, h) ∈ R ⇒ g(0) = h(1) and g(1) = h(0)

For (f, h) ∈ R, we need f(0) = h(1) and f(1) = h(0)

But, f(0) = g(1) = h(0) ≠ h(1) and f(1) = g(0) = h(1) ≠ h(0)

Hence, R is not transitive.