Question

The relation \( R = \{(x, y) : x, y \in \mathbb{Z} \text{ and } x + y \text{ is even} \} \) is: 

• A. reflexive and transitive but not symmetric
• B. reflexive and symmetric but not transitive
• C. symmetric and transitive but not reflexive
• D. an equivalence relation

Hint:

When dealing with relations, always verify the properties of reflexivity, symmetry, and transitivity to determine equivalence.

Answer 1

The Correct Option is D

To determine whether the relation is an equivalence relation, we check if it is reflexive, symmetric, and transitive. 

Step 1: Check if the relation is reflexive by checking if \( x + x \) is even for all integers \( x \). 

Step 2: Check if the relation is symmetric by ensuring if \( x + y \) is even, then \( y + x \) is also even. 

Step 3: Check if the relation is transitive by verifying that if \( x + y \) and \( y + z \) are even, then \( x + z \) is also even. 

Final Conclusion: The relation is an equivalence relation, which is Option 4.

Answer 2

Step 1: Understand the given relation.
The relation is defined as:
\[ R = \{(x, y) : x, y \in \mathbb{Z} \text{ and } x + y \text{ is even} \}. \] We need to determine whether this relation is reflexive, symmetric, and transitive — that is, whether it is an equivalence relation.

Step 2: Check reflexivity.
For any integer \( x \in \mathbb{Z} \),
\[ x + x = 2x, \] which is always even. Hence, \( (x, x) \in R \) for all \( x \in \mathbb{Z} \).
Therefore, the relation is reflexive.

Step 3: Check symmetry.
If \( (x, y) \in R \), then \( x + y \) is even.
Since addition is commutative (\( x + y = y + x \)), it follows that \( y + x \) is also even.
Thus, \( (y, x) \in R \).
Hence, the relation is symmetric.

Step 4: Check transitivity.
If \( (x, y) \in R \) and \( (y, z) \in R \), then both \( x + y \) and \( y + z \) are even.
Let \( x + y = 2m \) and \( y + z = 2n \), where \( m, n \in \mathbb{Z} \).
Subtracting these equations:
\[ (x + y) - (y + z) = 2m - 2n \Rightarrow x - z = 2(m - n). \] Since \( x - z \) is even, \( x + z \) will also be even (sum of two even or two odd integers is even).
Thus, \( (x, z) \in R \).
Therefore, the relation is transitive.

Step 5: Conclusion.
Since the relation is reflexive, symmetric, and transitive, it satisfies all the conditions of an equivalence relation.

Final Answer:
\[ \boxed{\text{R is an equivalence relation.}} \]