Associates in $\mathbb{Z}$: integers each of which divides the other

Integers $a$ and $b$ are associates, written $a \sim b$, when each divides the other (Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$):

$a \sim b \quad :\Longleftrightarrow \quad a \mid b \ \text{ and } \ b \mid a .$

As a binary relation in the sense of Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ this is the subset

$R \;=\; \{\, (a,b) \in \mathbb{Z} \times \mathbb{Z} \;:\; a \mid b \ \text{ and } \ b \mid a \,\}$

of $\mathbb{Z} \times \mathbb{Z}$, and $a \sim b$ abbreviates $(a,b) \in R$.

Nothing is claimed here beyond the definition. That $R$ is an equivalence relation — reflexive, symmetric and transitive — is a statement about $R$ that has to be proved, and it is proved next, in For integers $a$ and $b$ the following are equivalent: $a \mid b$ and $b \mid a$; $b = ua$ for a unit $u$; $|a| = |b|$. Being associates is an equivalence relation whose class of $a$ is $\{a, -a\}$, together with the identification of the class of $a$ as $\{a, -a\}$. Until then the symbol $\sim$ is notation for membership of $R$ and carries no further content; in particular the language of equivalence classes is not used above.

• Why the notion is worth naming. Divisibility does not distinguish an integer from its negative (Divisibility is reflexive and transitive on $\mathbb{Z}$, and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$; also $d \mid a$ implies $d \mid ac$, $-d \mid a$ and $d \mid -a$), so it is a preorder rather than an order: $2 \mid -2$ and $-2 \mid 2$ while $2 \ne -2$. Associates are exactly the pairs that divisibility cannot tell apart, and naming them is what lets the greatest common divisor be pinned down by a sign convention rather than left ambiguous up to that failure.

• The classes have at most two elements, which is special to $\mathbb{Z}$ and comes from its group of units being $\{1,-1\}$ ($(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$). The class of $0$ is $\{0\}$ alone.