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(…

Question

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

Accepted Answer

For $R$ to be reflexive, $(f, f)$ must be in $R$.

The means $f(0)=f(1)$ and $f(1)=f(0)$ must be true for all $f$.

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

Therefore, $R$ is not reflexive

If $(f, g) \in R$, then $f(0)=g(1)$ and $f(1)=g(0)$

$$\begin{aligned}
& \because \quad f(0)=g(1) \Rightarrow g(1)=f(0) \
& \text { and } f(1)=g(0) \Rightarrow g(0)=f(1)
\end{aligned}$$

$R$ is symmetric

If $(f, g) \in R$ and $(g, h) \in \mathrm{R}$, then $f(0)=g(1)$, $f(1)=g(0), g(0)=n(1) \& g(1)=h(0)$

Since, $f(0)=g(1)$ and $g(1)=h(0)$, then $f(0)$ is not necessarily equal to $h(0)$.

Therefore, $R$ is not transitive.

$\therefore \quad$ The relation $R$ is symmetric but not reflexive or transitive.

Solutions · 1