Prove that x ~ y if |x-y| $\leq$ 3 is an equivalence relation for x,y elements of the natural numbers

Question

Problem:

Prove the following relation is an equivalence relation on the natural numbers.

x~y if |x-y| $\leq$ 3

I believe I am stuck on understanding how to prove transitivity.

Here is my proof so far:

$\forall x, ||x-x|| = 0 \leq 3$, so $\sim$ is reflexive.

$||x-y|| \leq 3$ $\implies$ $||y-x|| \leq 3$, so $\sim$ is symmetric.

Let $x,y,z \in \mathbb{N}$, $x \sim y$ gives $||x - y|| \leq 3$, $y \sim z$ gives $||y-z|| \leq 3$.

I am not sure that I've actually proved the properties so much as stated that they are true and am unsure on how to prove transitivity.

Any help would be appreciated.

Answer 1

One counterexample is enough to refute transitivity.

So note that $1 \sim 4$ holds, and $4 \sim 7$ holds, but $1 \sim 7$ does not hold. So $\sim$ is not transitive and so $\sim$ is not an equivalence relation.

Your problem was not of "prove" type but of "prove or refute" type, presumably.