Proving that a relation is Reflexive, Symmetric, and Transitive

Question

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!

Answer 1

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 $x\equiv_5 z$.

Answer 2

$5\mid (x-y)$ 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 $y-x=5\times(-k)$.