Given the binary relation $xRy$ defined in $\mathbb{Z}$ such that $\exists \space k \in \mathbb{Z}(y+x=2k)$.
To prove that the relation $R$ is an equivalence relation, I must prove that $R$ is symmetric, reflexive and transitive.
-Symmetric
$\forall x,y\space(x+y=2k \wedge y+x=2k)$. So one have two equations that are equal to $2k$, I can write $x+y=y+x$. That will result in $0=0$, that is true in $\mathbb{Z}$.
-Reflexive
Let be $x \in \mathbb{Z}$. Then $\space \forall x \space (x+x=2k)$. So I write $2x=2k \Leftrightarrow x=k$. If $k$ is a integer then $x$ will be too.
-Transitive
$\forall x,y,z \space (x+y=2k \wedge y+z=2k \Rightarrow x+z=2k)$. Before all, one can see that $2k$ will allways be an integer pair number. Then I can sum $x+y=2k$ with $y+z=2k$ and get $x+2y+z=4k$. After, one can write $x+z=2(2k-y)$, that is an integer pair number too.
Are my arguments correct?