Is the relation $R$ on $\Bbb N$ given by $(a,b)\in R\iff a\mid b$ an equivalence relation?

Question

$R \subset \Bbb N \times \Bbb N$

Is this an equivalence relation?

$$R=\{(a,b)\in \Bbb N\times \Bbb N\,:\,a\mid b\}$$

I would argue that it is reflexive because $a\mid a$, but it is not symmetric because $a\mid b$ and $b\mid a$ are different.

I am not sure why it was claimed to be an equivalence relation in a previous math exam.

Answer 1

It is not an equivalence relation because it isn't symmetric (despite the symmetric nature of the divisibility notation). This can be seen through a variety of counterexample, such as: $$2 \mid 4 \quad \text{but} \quad 4 \not\mid 2.$$

In fact, it is never symmetric (apart from when $a=b$) because $a \mid b$ requires $a \leq b$ whilst $b \mid a$ requires $b \leq a$.

Answer 2

As pointed out by Zain Patel, divisibility is not an equivalence relation because it is not symmetric.

However, since divisibility is reflexive, antisymmetric, and transitive, then it is a partial order.