How to prove $R$ is an equivalence relation where: $\forall x, y \in A$, $xRy \iff 4|(x − y)$

Question

$A = \{ 2, 4, 6, 8, 10 \}$ I'm unsure how to 'show' that $R$ is an equivalence relation. My conclusion is it is not as $x-x$ reflexive as given any number in the set will equal $0$ therefore $4$ divides $0$ Am I Wrong?

Answer 1

First of all, $4$ does divide $0$, because by definition, $a$ divides $b$ if $b=k\cdot a$ for some integer $k$, and you can clearly see that $0=0\cdot 4$.

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Second, you prove that $R$ is an equivalence relation the way you always prove a relation is an equivalence relation:

1. You show that it is symmetric. I.e., you show that if $xRy$, then $yRx$. This should be easy since $y-x=-(x-y)$.
2. You show that it is reflexive. See the first part of my answer for that.
3. You show that it is transitive. You take arbitrary values of $x,y,z$ and assume that $xRy$ and $xRz$ and you show that $xRz$. This should be very simple if you just write down the definition of divisability.

Answer 2

R is equivalence relation, if it is reflexive, symmetric and transitive.

Let's prove that holds:

R is reflexive: $\forall x\in A: xRx$

$xRx$ holds for every $x\in A$, since $4|0$.

R is symmetric: $\forall x,y\in A: xRy\implies yRx$

If $xRy$, then $4|x-y$ implies $4|y-x$, which means $yRx$.

R is transitive: $\forall x,y,z\in A: (xRy\land yRz)\implies xRz$

If $xRy$ and $yRz$, then $4|x-y$, $4|y-z$, which implies $4|(x-y)+(y-z)=x-z$, which means $xRz$.