Key Concepts and Formulas

• Reflexive Relation: A relation R on a set A is reflexive if for every element $a \in A$, $(a, a) \in R$.
• Symmetric Relation: A relation R on a set A is symmetric if for any two elements $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$.
• Transitive Relation: A relation R on a set A is transitive if for any three elements $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Step-by-Step Solution

Let A be the set of all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$. The relation R is defined on A as $R = \{(f, g) \in A \times A \mid f(0) = g(1) \text{ and } f(1) = g(0)\}$.

Step 1: Check for Reflexivity To check if R is reflexive, we need to determine if for every function $f \in A$, $(f, f) \in R$. According to the definition of R, $(f, f) \in R$ if $f(0) = f(1)$ and $f(1) = f(0)$. This condition simplifies to $f(0) = f(1)$. However, the set A contains all functions from $\mathbf{Z}$ to $\mathbf{Z}$. We can easily find functions where $f(0) \neq f(1)$. For example, consider the function $f(x) = x$. Then $f(0) = 0$ and $f(1) = 1$. Since $f(0) \neq f(1)$, $(f, f) \notin R$. Therefore, R is not reflexive.

Step 2: Check for Symmetry To check if R is symmetric, we need to determine if for any two functions $f, g \in A$, if $(f, g) \in R$, then $(g, f) \in R$. Assume $(f, g) \in R$. This means that $f(0) = g(1)$ and $f(1) = g(0)$. Now we need to check if $(g, f) \in R$. According to the definition of R, $(g, f) \in R$ if $g(0) = f(1)$ and $g(1) = f(0)$. From our assumption, we have $f(0) = g(1)$ and $f(1) = g(0)$. These conditions are exactly the same as the conditions required for $(g, f) \in R$. Therefore, if $(f, g) \in R$, then $(g, f) \in R$. R is symmetric.

Step 3: Check for Transitivity To check if R is transitive, we need to determine if for any three functions $f, g, h \in A$, if $(f, g) \in R$ and $(g, h) \in R$, then $(f, h) \in R$. Assume $(f, g) \in R$ and $(g, h) \in R$. From $(f, g) \in R$, we have: (1) $f(0) = g(1)$ (2) $f(1) = g(0)$

From $(g, h) \in R$, we have: (3) $g(0) = h(1)$ (4) $g(1) = h(0)$

Now we need to check if $(f, h) \in R$. This requires checking if $f(0) = h(1)$ and $f(1) = h(0)$. Let's use the given equations: From (1) and (4), we have $f(0) = g(1)$ and $g(1) = h(0)$. By the transitivity of equality, $f(0) = h(0)$. From (2) and (3), we have $f(1) = g(0)$ and $g(0) = h(1)$. By the transitivity of equality, $f(1) = h(1)$.

So, we have $f(0) = h(0)$ and $f(1) = h(1)$. However, the condition for $(f, h) \in R$ is $f(0) = h(1)$ and $f(1) = h(0)$. Our derived conditions are $f(0) = h(0)$ and $f(1) = h(1)$. These are not necessarily the same. For example, let's try to construct a counterexample.

Let $f(x) = x$. Then $f(0) = 0$ and $f(1) = 1$. Let $g(x) = 1-x$. Then $g(0) = 1$ and $g(1) = 0$. Check if $(f, g) \in R$: $f(0) = 0$, $g(1) = 0$. So $f(0) = g(1)$. $f(1) = 1$, $g(0) = 1$. So $f(1) = g(0)$. Thus, $(f, g) \in R$.

Now let's find a function $h$ such that $(g, h) \in R$, but $(f, h) \notin R$. Let $h(x) = x+2$. Then $h(0) = 2$ and $h(1) = 3$. Check if $(g, h) \in R$: $g(0) = 1$, $h(1) = 3$. So $g(0) \neq h(1)$. This choice of $h$ does not satisfy $(g, h) \in R$.

Let's try again to find $h$ such that $(g, h) \in R$. We need $g(0) = h(1)$ and $g(1) = h(0)$. We have $g(0) = 1$ and $g(1) = 0$. So we need $h(1) = 1$ and $h(0) = 0$. Let $h(x) = x$. Then $h(0) = 0$ and $h(1) = 1$. Check if $(g, h) \in R$: $g(0) = 1$, $h(1) = 1$. So $g(0) = h(1)$. $g(1) = 0$, $h(0) = 0$. So $g(1) = h(0)$. Thus, $(g, h) \in R$.

Now we check if $(f, h) \in R$ for $f(x) = x$ and $h(x) = x$. We need $f(0) = h(1)$ and $f(1) = h(0)$. $f(0) = 0$, $h(1) = 1$. So $f(0) \neq h(1)$. Therefore, $(f, h) \notin R$.

In this case, $(f, g) \in R$ and $(g, h) \in R$, but $(f, h) \notin R$. This demonstrates that R is not transitive.

Let's re-examine the conditions for transitivity: We have $f(0) = g(1)$, $f(1) = g(0)$ for $(f, g) \in R$. We have $g(0) = h(1)$, $g(1) = h(0)$ for $(g, h) \in R$. For $(f, h) \in R$, we need $f(0) = h(1)$ and $f(1) = h(0)$.

From $f(0) = g(1)$ and $g(1) = h(0)$, we get $f(0) = h(0)$. From $f(1) = g(0)$ and $g(0) = h(1)$, we get $f(1) = h(1)$.

So, if $(f, g) \in R$ and $(g, h) \in R$, then $f(0) = h(0)$ and $f(1) = h(1)$. The condition for $(f, h) \in R$ is $f(0) = h(1)$ and $f(1) = h(0)$. For transitivity to hold, we would need $f(0) = h(0)$ and $f(1) = h(1)$ to imply $f(0) = h(1)$ and $f(1) = h(0)$. This is not generally true. For example, if $f(0) = 5$ and $f(1) = 10$, and $h(0) = 5$ and $h(1) = 10$, then $f(0)=h(0)$ and $f(1)=h(1)$. But for $(f,h) \in R$, we would need $f(0)=h(1)$ (i.e., $5=10$, false) and $f(1)=h(0)$ (i.e., $10=5$, false).

Thus, R is not transitive.

Common Mistakes & Tips

• Confusing function properties with relation properties: Remember that reflexivity, symmetry, and transitivity are properties of the relation, not the individual functions themselves.
• Careful with universal quantifiers: For reflexivity and transitivity, you need to prove it for all elements. For symmetry, you need to prove it for all pairs. If you find even one counterexample, the property does not hold.
• Constructing counterexamples: When checking for non-reflexivity or non-transitivity, try to construct specific functions that violate the conditions. This is often easier than trying to prove it for all functions.

Summary

We analyzed the relation R on the set of functions A. We found that R is not reflexive because not all functions satisfy $f(0) = f(1)$. We proved that R is symmetric by showing that if $f(0) = g(1)$ and $f(1) = g(0)$, then the conditions for $(g, f) \in R$ are also met. Finally, we demonstrated that R is not transitive by constructing a counterexample where $(f, g) \in R$ and $(g, h) \in R$, but $(f, h) \notin R$. Therefore, R is symmetric but neither reflexive nor transitive.

The final answer is \boxed{A}.