Let $A$ be the set of all functions $f: Z \rightarrow Z$ 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 :
Correct Answer:
B
Explanation
To determine if the relation $R$ is reflexive, symmetric, and transitive, we analyze each property separately.
Reflexive: For $R$ to be reflexive, $(f, f)$ must be in $R$. This means $f(0) = f(1)$ and $f(1) = f(0)$ must be true for all $f$. However, $f(0)$ is not necessarily equal to $f(1)$ for all functions $f$. Therefore, $R$ is not reflexive.
Symmetric: If $(f, g) \in R$, then $f(0) = g(1)$ and $f(1) = g(0)$. To check for symmetry, we need to verify if $(g, f) \in R$. If $f(0) = g(1)$, then $g(1) = f(0)$. And if $f(1) = g(0)$, then $g(0) = f(1)$. This shows that if $(f, g) \in R$, then $(g, f) \in R$, so $R$ is symmetric.
Transitive: If $(f, g) \in R$ and $(g, h) \in R$, then $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. For $R$ to be transitive, we need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$, so $f(0) = h(0)$. Also, $f(1) = g(0) = h(1)$, so $f(1) = h(1)$. Therefore, $f(0) = g(1)$ and $g(1) = h(0)$ implies $f(0) = h(0)$ and $f(1) = g(0)$ and $g(0) = h(1)$ implies $f(1) = h(1)$. Since $f(0)$ is not necessarily equal to $h(1)$ and $f(1)$ is not necessarily equal to $h(0)$, $R$ is not transitive. Actually $f(0) = h(0)$ and $f(1) = h(1)$ so this doesn't imply that $f(0) = h(1)$ and $f(1) = h(0)$ so the relation is not transitive
Therefore, the relation $R$ is symmetric but neither reflexive nor transitive.