Proving that a relation is Reflexive, Symmetric, and Transitive 
The relation $\equiv_5$ on $\mathbb Z$(integers) defined as follows:
  $x\equiv_5 y$ if and only if $5 \mid (x−y)$.
  Prove that it is reflexive, symmetric, and transitive.

Hey guys, stuck on this one because I cant quite figure out what is meant by some of the symbols here. Particularly confused by "$5 \mid (x-y)$". Is that supposed to be divisible? Or is the point to prove the nonsensical? 
Thanks for any help!
 A: $5\mid (x-y)$ means "$5$ is a factor of $(x-y)$". 
So, for example, to show $\equiv_5$ is symmetric, you need to show that if $5$ is a factor of $x-y$ then it is a factor of $y-x$ (which translates to "if $x\equiv_5 y$ then $y\equiv_5 x$"). This is true because if $5$ is a factor of $x-y$ we can write $x-y=5k$ for some integer $k$, and then $y-x=5\times(-k)$.
A: For a relation to be an equivalence relation we need that it is reflexive, symmetric and transitive. So let us check these if
 $
\equiv_5
$
is an equivalence relation.
Reflexivity of a relation means that if you have a relation $R$ on a set $X$ then for all $x\in X$ the following is true

$$
xRx
$$

that is all the elements are in relation with themselves. So is $x\equiv_5 x$? According to your definition we have 
$$
x\equiv_5 x \iff 5\mid x-x=0
$$
which is true. So, yes $\equiv_5$ is reflexive.
For symmetry we need that for all $x,y\in X$

$$
xRy\iff yRx
$$

using thegiven definition of $\equiv_5$ we have to show that
$$
5\mid (x-y) \iff 5\mid (y-x)
$$
these translates to
$$
\exists k\in\mathbb{Z},\quad x-y=5k
$$
multiplying bot sides with $-1$ we get 
$$
y-x=5(-k)
$$
so yes, the relation is symmetric.
Now for transitivity we need to show that 

$$
xRy\ \text{and}\ yRz\Rightarrow xRz. 
$$

So assume $x\equiv_5 y$ that is $5\mid x-y$ and that $y\equiv_5 z$ that is $5\mid y-z$. Writing in equation form
$$
x-y=5k,\quad y-z=5l, \ \text{for some} \ k,l\in\mathbb{Z}
$$
that is
$$
x-y=5k, \quad y=5l+z \Rightarrow x-5l-z=5k
$$
that is
$$
x-z=5(k+l)
$$
which concludes transitivity and we are done since this shows that $x\equiv_5 z$.
