Question ID: #435
The relation $R = \{(x, y) : x, y \in \mathbb{Z} \text{ and } x+y \text{ is even}\}$ is:
- (1) reflexive and transitive but not symmetric
- (2) reflexive and symmetric but not transitive
- (3) an equivalence relation
- (4) symmetric and transitive but not reflexive
Solution:
Given $R = \{(x, y) : x+y \text{ is even}\}$.
Reflexive:
For any $x \in \mathbb{Z}$, consider $x+x = 2x$.
Since $2x$ is always even, $(x, x) \in R$.
So, R is reflexive.
Symmetric:
Let $(x, y) \in R$. Then $x+y$ is even.
Since addition is commutative, $y+x = x+y$, which is also even.
Thus, $(y, x) \in R$.
So, R is symmetric.
Transitive:
Let $(x, y) \in R$ and $(y, z) \in R$.
$x+y = 2k_1$ (even) and $y+z = 2k_2$ (even).
Adding these two equations:
$(x+y) + (y+z) = 2k_1 + 2k_2$
$x + 2y + z = 2(k_1 + k_2)$
$x + z = 2(k_1 + k_2 – y)$
Since the RHS is a multiple of 2, $x+z$ is even.
Thus, $(x, z) \in R$.
So, R is transitive.
Since the relation is Reflexive, Symmetric, and Transitive, it is an equivalence relation.
Ans. (3)
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