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

Question

Aicharya · Accepted Answer

For reflexive $(x, x) \in \mathbb{R}, x \in \mathbb{Z}$
$\Rightarrow x+x=2 x \rightarrow \text { even }$
For symmetric of $(x, y) \in \mathbb{R}$ then $(y, x) \in \mathbb{R}$ when $x, y \in \mathbb{Z}$
$x+y \rightarrow$ even
$\Rightarrow y+x \rightarrow \text { even }$
for transitive if $(x, y) \in \mathbb{R} \Rightarrow x+y \rightarrow$ even
$(y, z) \in \mathbb{R} \Rightarrow y+z \rightarrow \text { even }$
$x+2 y+z \rightarrow$ even
$\Rightarrow x+z$ is even
$\Rightarrow(x, z) \in \mathbb{R}$
$\Rightarrow \mathbb{R}$ is an equivalence relation. .

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

Solution