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 $q$):

$a\sim b:⟺a\mid b\text{ and }b\mid a\mathrm{.}$

As a binary relation in the sense of Equivalence relation, equivalence class, and the quotient set $A\mathrm{/}\sim$ 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 $u$; $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$. 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 $2\mathrm{\ne }-2$. 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.