1. Key Concepts and Formulas

• Relation: A relation $R$ on a set $A$ is a subset of $A \times A$. We denote $(x, y) \in R$ as $xRy$.
• Reflexive Relation: For all $x \in A$, $xRx$.
• Symmetric Relation: For all $x, y \in A$, if $xRy$, then $yRx$.
• Transitive Relation: For all $x, y, z \in A$, if $xRy$ and $yRz$, then $xRz$.
• Equivalence Relation: A relation that is reflexive, symmetric, and transitive.

2. Step-by-Step Solution

The given relation is $R = \{(x, y): x, y \in \mathbb{Z} \text{ and } x+y \text{ is even}\}$. The set is $\mathbb{Z}$, the set of integers.

• Step 1: Check for Reflexivity. We need to determine if $xRx$ for all $x \in \mathbb{Z}$. The condition for $xRx$ is that $x+x$ must be an even integer. $x+x = 2x$. Since $x$ is an integer, $2x$ is always an even integer. Thus, $xRx$ for all $x \in \mathbb{Z}$. The relation $R$ is reflexive.

• Step 2: Check for Symmetry. We need to determine if for all $x, y \in \mathbb{Z}$, if $xRy$, then $yRx$. Assume $xRy$. This means $x+y$ is an even integer. We need to check if $yRx$, which means $y+x$ is an even integer. By the commutative property of addition, $y+x = x+y$. Since $x+y$ is even, $y+x$ is also even. Thus, if $xRy$, then $yRx$. The relation $R$ is symmetric.

• Step 3: Check for Transitivity. We need to determine if for all $x, y, z \in \mathbb{Z}$, if $xRy$ and $yRz$, then $xRz$. Assume $xRy$ and $yRz$. $xRy$ implies $x+y$ is even. Let $x+y = 2k_1$ for some integer $k_1$. $yRz$ implies $y+z$ is even. Let $y+z = 2k_2$ for some integer $k_2$. We need to check if $xRz$, which means $x+z$ is even. Consider the sum $(x+y) + (y+z) = 2k_1 + 2k_2$. $x + 2y + z = 2(k_1 + k_2)$. Rearranging to find $x+z$: $x+z = 2(k_1 + k_2) - 2y$ $x+z = 2(k_1 + k_2 - y)$. Since $k_1, k_2,$ and $y$ are integers, $k_1 + k_2 - y$ is also an integer. Let $K = k_1 + k_2 - y$. Then $x+z = 2K$, which means $x+z$ is an even integer. Thus, if $xRy$ and $yRz$, then $xRz$. The relation $R$ is transitive.

• Step 4: Classify the Relation. We have shown that the relation $R$ is reflexive, symmetric, and transitive. Therefore, $R$ is an equivalence relation.

• Step 5: Analyze the Options. (A) reflexive and transitive but not symmetric - Incorrect, as R is symmetric. (B) reflexive and symmetric but not transitive - Incorrect, as R is transitive. (C) an equivalence relation - Correct, as R is reflexive, symmetric, and transitive. (D) symmetric and transitive but not reflexive - Incorrect, as R is reflexive.

3. Common Mistakes & Tips

• When checking for transitivity, ensure you use distinct variables for the elements involved ($x, y, z$).
• The condition "$x+y$ is even" is equivalent to "$x$ and $y$ have the same parity" (both even or both odd). This parity concept can quickly verify reflexivity ($x$ and $x$ have the same parity), symmetry (if $x$ and $y$ have the same parity, then $y$ and $x$ do), and transitivity (if $x,y$ have same parity and $y,z$ have same parity, then $x,y,z$ all have same parity, so $x,z$ do).
• Do not stop at finding the relation to be an equivalence relation if the options present finer details. In this case, the relation is an equivalence relation, which is one of the options.

4. Summary

The relation $R = \{(x, y): x, y \in \mathbb{Z} \text{ and } x+y \text{ is even}\}$ is analyzed for reflexivity, symmetry, and transitivity. It is found to be reflexive because the sum of an integer with itself ($2x$) is always even. It is symmetric because if $x+y$ is even, then $y+x$ is also even due to the commutative property of addition. It is transitive because if $x+y$ and $y+z$ are both even, then their sum $(x+y)+(y+z) = x+2y+z$ is even. From this, $x+z = (x+2y+z) - 2y$ is also even. Since the relation satisfies all three properties, it is an equivalence relation.

The final answer is \boxed{C}