Let A be the set of all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$ and R be a relation on A such that…

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

Solution

$\begin{aligned} & \mathrm{R}=\{(\mathrm{f}, \mathrm{g}): \mathrm{f}(0)=\mathrm{g}(1) \text { and } \mathrm{f}(1)=\mathrm{g}(0)\} \\ & \text { Reflexive: }(\mathrm{f}, \mathrm{f}) \in \mathrm{R} \\ & =\mathrm{f}(0)=\mathrm{f}(1) \text { and } \mathrm{f}(1)=\mathrm{f}(0) \rightarrow \text { must hold } \\ & \Rightarrow \text { but this is not true for all function }\end{aligned}$
so not reflexive
Symmetric: If $(\mathrm{f}, \mathrm{g}) \in \mathrm{R} \Rightarrow(\mathrm{g}, \mathrm{f}) \in \mathrm{R}$
Now, $g(0)=f(1)$ and $g(1)=f(0) \rightarrow$ true
$\therefore$ symmetric
Transitive : $\operatorname{If}(\mathrm{f}, \mathrm{g}) \in \mathrm{R}$ and $(\mathrm{g}, \mathrm{h}) \in \mathrm{R}$
$\Rightarrow(\mathrm{f}, \mathrm{h}) \in \mathrm{R}$
Now $(f, g) \in R \Rightarrow f(0)=g(1)$ and $f(1)=g(0)$
$(\mathrm{g}, \mathrm{h}) \in \mathrm{R} \Rightarrow \mathrm{g}(0)=\mathrm{h}(1)$ and $\mathrm{g}(1)=\mathrm{h}(0)$
For $(f, h) \in R$ we need $f(0)=h(1)$ and $f(1)=h(0)$
Now $f(0)=g(1)=h(0)$ and $f(1)=g(0)=h(1)$
Hence not transitive ^

Asked in: JEE Main 2025 (02 Apr Shift 1)

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