equivalence relation (even number) I have two problems with this task. First I can´t do the  correct mathematic shape. And also when is an equivalence it´s supposed to be reflexive, symmetrical and transitive. 
The first two are okay. $x+x= 2x$ and $x+y= 2k$ and $y+x =2k$. How to solve the transitive part? Could you please show me the soolution in correct shape and with small steps? Thank you so much
$x, y \in \mathbb{Z}$       the relation is defined:
$R: = \{(x,y) \mid x + y \text{ is an even number}\}$
Please show that it´s an equivalence relation!
 A: For $R$ is transitive: suppose $(x,y)\in R$ and $(y,z)\in R$. We are about to show $(x,z)\in R$. 
So, by hypothesis we know that $x+y$ and $y+z$ are even. Then, we have
$$x+z = (x+y)+(y+z)-2y$$
is a sum of $3$ even numbers, so it is even.

An alternative way is to observe that, for integers, $x+y$ is even iff $x-y$ is even (because their difference is $2y$, even). So, we have that
$$R=\{(x,y) : 2|\,x-y\}$$
And this, in general ($2$ replaced to any integer $m$) is an equivalence relation, called congruence modulo $m$ on integers.
A: Suppose that $\langle x,y\rangle,\langle y,z\rangle\in R$; to show that $R$ is transitive, we must show that $\langle x,z\rangle\in R$, which means that we must show that $x+z$ is even. We know that $x+y$ and $y+z$ are even, and the sum of two even numbers is even, so $(x+y)+(y+z)=x+2y+z$ is even. Finally, the difference of two even numbers is even, and $2y$ is even, so $(x+2y+z)-2y=x+z$ is even. This is exactly what we needed to show in order to conclude that $R$ is transitive.
A: Hint:  If $(x,y)$ and $(y,z)$ belong to $R$, then $x + y = 2k$ and $y + z = 2l$ are both even.  Solve the two equations for $x$ and $z$, respectively, and then add them together.  After a little bit of rearrangement, you should be able to factor out a $2$.
