Check for an equivalence relation on the integers Given the set of integer $\mathbb Z$, define $\sim$ as $x\sim y$ precisely when $2$ divides $x-y$.
I'm having a hard time showing that the three properties of an equivalence relation hold.
Any help is appreciated.

(added from comments)
I have  $$x-y=2n $$for all n in the set of integer $$y-x=2m $$for all m in the set of integer... Intuitively, there is a connection involving, $$(x-y), (y-x) \text{ and } 2$$  $$y-x= (something) - (x-y)$$ I'm not sure how to proceed forward
 A: To start with, reflexivity is easy since for any $x$, $x-x=0$ and we know $2\mid 0$ since $0 = 2 \cdot 0$.
For symmetry, how can you use the fact that $x-y=-(y-x)$? Hint: suppose $2\mid y-x$. Then $y-x=2k$ for some integer $k$ (by the definition). Can you find another integer $\ell$ such that $x-y=2\ell$? What does this tell you?
For transitivity, you assume $2\mid y-x$ and $2\mid z-y$. Write equations corresponding to those inequalities and try to reach the conclusion that $2\mid z-x$. Hint: something will cancel when you add the two equations. For instance, the equations become $y-x=2k$ and $z-y=2\ell$ for some integers $k,\ell$. You can add these equations to get a new relationship.
EDIT: Let's work through the definition of the relation. To say that $2\mid y-x$ means that there exists an integer $n$ such that $y-x=2n$. This is the definition. Now, if you're proving symmetry, you first start by assuming $x\sim y$, which means $$y-x=2n$$ for an integer $n$ (this is exactly the definition above. Multiplying the above by $-1$ gives $$x-y = -2n.$$ You're absolutely right that $-2n\neq 2n$ - but we don't need this to happen to have $y\sim x$ which is what we're trying to prove. For $y\sim x$ to be true, we must have $x-y=2m$ for some integer $m$. But we know that $x-y=-2n$ which can be written as $$x-y=2(-n).$$ But $-n$ is also an integer, so we can pick $m = -n$. Plugging this in we get $$x-y=2m$$ and so $y\sim x$.
A: It is easy that $2|x-x$ and if $2|x-y$ then $2|y-x$;
let $2|x-y$ and $2|y-z$ we should have $2|x-z$:
By assumption $x-y=2n$ and $y-z =2m$. So $x-z=(x-y)+(y-z)=2(n+m)$
