Proving that R is an Equivalence Relation.

Question

Consider the relations R and S on $\Bbb N$ defined by $x\; R\; y$ iff

$2 \;$divides $x + y$ and $x \;S \;y$ iff $3$ divides $x + y.$

$\text{QN: Prove that $R$ is an equivalence Relation }$

So my question for this one is that I understand how one is able to prove that it is reflexive and symmetric. But I am struggling to figure out how to prove that it is transitive. I cannot seem to grasp that concept.

$x+y = 2a \Rightarrow x+y = 3a$

$(i)$ $xRx$ , then $2$ divides $x+x = 2x$, therefore reflexive.

$(ii)$ If $xRy$ $2$ divides $x+y$, $2$ divides $y+x$ ,$yRx$. Thus Symmetric.

Answer 1

For transitivity, like so:

$xRy$, so $2$ divides $x+y$

$yRz$, so $2$ divides $y+z$

Now add the two together, you know that $2$ has to divide the sum, so $2$ divides $x + y + y + z = x + z + 2y$.

But since $2 | 2y$ and $2 | (x + z + 2y)$ then $2$ must divide $x + z$, hence $xRz$.