Let $A$ be the set of all functions $f: \mathbb{Z} \to \mathbb{Z}$ 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:

Step 1: Check for Reflexivity

For a relation to be reflexive, every element must be related to itself. In this case, for any function $f \in A$, we need to check if $(f,f) \in R$. This means we must have $f(0) = f(1)$ and $f(1) = f(0)$. This condition is not always true for all functions $f: \mathbb{Z} \to \mathbb{Z}$. For example, if $f(x) = x$, then $f(0) = 0$ and $f(1) = 1$, so $f(0) \neq f(1)$.

Step 2: Check for Symmetry

For a relation to be symmetric, if $(f,g) \in R$, then $(g,f)$ must also be in $R$. Given $(f,g) \in R$, we have $f(0)=g(1)$ and $f(1)=g(0)$. For $(g,f) \in R$, we need $g(0)=f(1)$ and $g(1)=f(0)$. These are exactly the conditions given, just reordered. Thus, if $(f,g) \in R$, then $(g,f) \in R$.

Step 3: Check for Transitivity

For a relation to be transitive, if $(f,g) \in R$ and $(g,h) \in R$, then $(f,h)$ must also be in $R$. From $(f,g) \in R$, we have $f(0)=g(1)$ and $f(1)=g(0)$. From $(g,h) \in R$, we have $g(0)=h(1)$ and $g(1)=h(0)$. Substituting $g(1)$ from the first condition into the second, we get $f(0)=h(0)$. Substituting $g(0)$ from the first condition into the second, we get $f(1)=h(1)$. For $(f,h) \in R$, we need $f(0)=h(1)$ and $f(1)=h(0)$. Since we found $f(0)=h(0)$ and $f(1)=h(1)$, the transitivity condition $f(0)=h(1)$ and $f(1)=h(0)$ is not necessarily satisfied. For example, let $f(x)=0$, $g(x)=0$, $h(x)=1$. Then $f(0)=0, f(1)=0$. $g(0)=0, g(1)=0$. $h(0)=1, h(1)=1$. $(f,g) \in R$ since $f(0)=g(1)=0$ and $f(1)=g(0)=0$. $(g,h) \in R$ since $g(0)=h(1)=1$ (this is false, $0 \neq 1$). Let's choose functions carefully. Let $f(x) = x$, $g(x) = 1-x$, $h(x) = x$. Then $f(0)=0, f(1)=1$. $g(0)=1, g(1)=0$. $h(0)=0, h(1)=1$. $(f,g) \in R$ because $f(0)=0=g(1)$ and $f(1)=1=g(0)$. $(g,h) \in R$ because $g(0)=1=h(1)$ and $g(1)=0=h(0)$. Now check $(f,h) \in R$. This requires $f(0)=h(1)$ and $f(1)=h(0)$. We have $f(0)=0$ and $h(1)=1$, so $f(0) \neq h(1)$. Thus, the relation is not transitive.

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