In the following, $\mathbb Z$ will denote the set of all integers, i.e. $\mathbb Z = \{\dots, -3, -2, -1, 0, 1, 2, 3, 4, \dots\}$, and $\mathbb N$ will denote the set of all natural numbers (aka positive integers), $\mathbb N = \{1, 2, 3, 4, 5, \dots\}$.

Let $b$ be an integer and $a$ be a natural number. Then there exists a unique pair of integers $(q, r)$, such that $b = qa + r$ and $0 \leq r < a$.

For example, consider the number 6. It divides, e.g., 6, 12, -18, 24, 0 and is divided by 6, -6, 3, -2, 1.

Since $0=0\cdot1$, we can write $0 \mid 0$, i.e. zero divides itself. However, we can not write $0 \mid b$ where $b\neq 0$ since there is no such integer $c$ that could provide the equality $b=0\cdot c$.

This is standard criterion for checking divisibility.

In the following let us restrict ourselves to natural numbers.

For instance, consider a graph formed by all the divisors of $12$.

Such a graph is called a lattice of divisibility.

Apriori it may not be clear that a lattice of divisibility exists for any given number. One might imagine a divisor $a$ such that for any other divisor $b$ there exists $c$ sitting inbetween $a$ and $b$, i.e., satisfying $a\mid c$ and $c\mid b$.

So, why there is no such divisor? If such a divisor $a$ existed, then taking $c$ in place of $b$ we would find $d$ sitting between $a$ and $c$, then $e$ between $a$ and $d$ and so on. We could go on infinitely. But in practice we cannot go on infinitely.

For instance, if we start with a pair $1\mid 12$ we can quickly discover $1\mid 6$ and $6 \mid 12$ and then discover $1\mid 3$ and $3\mid 6$. Then there is nothing to be put between $1$ and $3$, so there must be an arc from $1$ to $3$.

To understand why this process is finite it is enough to observe that $a$ can divide $b$ only if $a\leq b$ and for any number $b$ there are only finitely many numbers less than $b$ and greater than $1$, so there are only finitely many divisors of $b$.

Thus, indeed any given number has a lattice of divisibility.

This follows from the transitivity of divisibility. Namely, if $a$ divides $b$ and $b$ divides $c$ then $a$ also divides $c$.

Transitivity guarantees that there are no directed cycles in the graph. Indeed, let us observe a cycle with an arc from $a$ to $b$ ($a \neq b$):

Since it is a cycle, there is path from $b$ to $a$, i.e. $b \mid a$. And there is arc from $a$ to $b$ meaning $a \mid b$. Recall that from $a \mid b$ follows that $a \leq b$. We have also $b \mid a$ meaning $a \geq b$. Thus $a=b$ which is contradiction.

The divisibility relation is a partial order. See Ordered sets for more details on this type of structures.