Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$

Let $d,a\in \mathbb{Z}$ (The integers as equivalence classes of pairs of naturals). We say $d$ divides $a$, and write $d\mid a$, when

$a=dq\text{for some }q\in \mathbb{Z},$

the product being that of Arithmetic on the integers. We write $d\nmid a$ when this fails. In this situation $d$ is called a divisor, or a factor, of $a$, and $a$ is called a multiple of $d$.

This is the relation the library already has, not a second one. The published Division with remainder in $\mathbb{Z}$: for $a\in \mathbb{Z}$ and $b>0$ there are unique $q,r\in \mathbb{Z}$ with $a=qb+r$ and $0\le r<b$ introduces it in its own Statement, in these words: "We say $b$ divides $a$, written $b\mid a$, when $a=qb$ for some $q\in \mathbb{Z}$." Since multiplication on $\mathbb{Z}$ is commutative (The integers form a commutative ring), $a=qd$ and $a=dq$ are the same condition, so the definition above is that relation verbatim and the two usages agree everywhere. The theorem defined it for use on its own page and left the systematic theory to a later page; this is that page, and this item records the agreement rather than introducing a rival notion.

The remainder test. For $b>0$ the same Statement records that $b\mid a$ holds exactly when the remainder $r$ in $a=qb+r$, $0\le r<b$, is $0$.

Boundary values. Each is one line from the ring axioms, and each is used below, so all three are recorded here rather than assumed:

• $d\mid 0$ for every integer $d$, including $d=0$, since $0=d\cdot 0$;
• $0\mid a$ only for $a=0$, since $a=0\cdot q$ forces $a=0$;
• $1\mid a$ and $a\mid a$ for every $a$, since $a=1\cdot a$ and $a=a\cdot 1$.

• The remainder test for a negative divisor is Division with remainder for any nonzero divisor: for $a\in \mathbb{Z}$ and $b\mathrm{\ne }0$ there are unique $q,r\in \mathbb{Z}$ with $a=qb+r$ and $0\le r<\mathrm{\mid }b\mathrm{\mid }$, proved next: for every $b\mathrm{\ne }0$, $b\mid a$ holds exactly when the remainder in $a=qb+r$, $0\le r<\mathrm{\mid }b\mathrm{\mid }$, is $0$.

• $d\mid a$ says a quotient exists; it says nothing on its own about the size or the sign of $d$. Signs are irrelevant to it — $2\mid -4$, $-2\mid 4$ and $-2\mid -4$ all hold — and that is 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 $d\mid -a$. The one place where divisibility does constrain size is If $d\mid a$ and $a\mathrm{\ne }0$ then $d\mathrm{\ne }0$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; hence the set of divisors of a nonzero integer is bounded above by $\mathrm{\mid }a\mathrm{\mid }$, and it needs the hypothesis $a\mathrm{\ne }0$: without it $d\mid 0$ holds for arbitrarily large $\mathrm{\mid }d\mathrm{\mid }$.

• Notation. The bar in $d\mid a$ is a relation symbol read from left to right: the divisor is on the left. The reversed reading is a common slip, and the two directions are genuinely different, since $2\mid 4$ holds while $4\nmid 2$ (by If $d\mid a$ and $a\mathrm{\ne }0$ then $d\mathrm{\ne }0$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; hence the set of divisors of a nonzero integer is bounded above by $\mathrm{\mid }a\mathrm{\mid }$, $4\mid 2$ with $2\mathrm{\ne }0$ would force $4=\mathrm{\mid }4\mathrm{\mid }\le \mathrm{\mid }2\mathrm{\mid }=2$).