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]

Question

Let A be the set of all functions f : Z $\to$ 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]

Smart Achievers · Accepted Answer

(3)

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

For reflexive: (f, f) $\in$ R

$\Rightarrow$ f(0) = f(1) must hold $\forall$ f

But f(0) $\neq$ f(1)

Therefore, R is not reflexive.

For symmetric: If (f, g) $\in$ R $\Rightarrow$ (g, f) $\in$ R

Now, g(0) = f(1) and g(1) = f(0), which is true $\forall$ f

$∴$ R is symmetric

For transitive : IF (f, g) $\in$ R and (g, h) $\in$ R $\Rightarrow$ (f, h) $\in$ R

Now, (f, g) $\in$ R $\Rightarrow$ f(0) = g(1) and f(1) = g(0)

(g, h) $\in$ R $\Rightarrow$ g(0) = h(1) and g(1) = h(0)

For (f, h) $\in$ R, we need f(0) = h(1) and f(1) = h(0)

But, f(0) = g(1) = h(0) $\neq$ h(1) and f(1) = g(0) = h(1) $\neq$ h(0)

Hence, R is not transitive.

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