The absolute value $\mathrm{\mid }a\mathrm{\mid }$ of an integer

Let $a\in \mathbb{Z}$ (The integers as equivalence classes of pairs of naturals). The absolute value of $a$ is

$\mathrm{\mid }a\mathrm{\mid }:=\{\begin{array}{ll}a& \text{if }a\ge 0,\\ -a& \text{if }a<0,\end{array}$

where $\le$ is the order of Order on the integers and $-a$ is the additive inverse supplied by The integers form a commutative ring for the operations of Arithmetic on the integers.

Why the two clauses define a function. The order on $\mathbb{Z}$ is total and antisymmetric (The integers form a totally ordered ring), so for each $a$ exactly one of $a\ge 0$ and $a<0$ holds: totality gives at least one of $0\le a$ and $a\le 0$, and if both hold then $a=0$ by antisymmetry, in which case $a\ge 0$ and $a<0$ fails, since $a<0$ means $a\le 0$ together with $a\mathrm{\ne }0$. The two clauses therefore never both apply and never both fail, and $-a$ is a single element of $\mathbb{Z}$ because additive inverses are unique in a commutative ring: if $a+u=0$ and $a+v=0$ then $u=u+0=u+(a+v)=(u+a)+v=0+v=v$.

At the boundary, $\mathrm{\mid }0\mathrm{\mid }=0$ by the first clause, and the second clause also gives $-a$ whenever $a\le 0$: for $a<0$ that is the definition, and for $a=0$ both readings give $0$.

• This is not a second notion of absolute value. The library's other absolute value, Absolute value in an ordered field, is stated for an ordered field. The structure the construction of $\mathbb{Z}$ supplies is that of a totally ordered commutative ring (The integers form a commutative ring, The integers form a totally ordered ring), and multiplicative inverses are no part of it, so Absolute value in an ordered field does not apply here and the definition above is a new object rather than a redefinition of an existing one. (That $\mathbb{Z}$ is genuinely not a field, rather than merely not presented as one, is $(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$ below: its only invertible elements are $1$ and $-1$.) A general form for an ordered ring would cover both cases at once, but the definition of an ordered ring comes much later in the reading order than this page, so it is not available to reach for.

• The two agree along the embedding of $\mathbb{Z}$ in $\mathbb{Q}$. Write $j:\mathbb{Z}\to \mathbb{Q}$ for the injective, order-preserving and arithmetic-preserving map of The integers embed in the rationals. Then $j(\mathrm{\mid }a\mathrm{\mid })$ is the absolute value of $j(a)$ in the sense of Absolute value in an ordered field, the check being the same case split: $j$ preserves the order and reflects it (if $j(a)\le j(b)$ but not $a\le b$, then $b\le a$ by totality, so $j(b)\le j(a)$, whence $j(a)=j(b)$ and $a=b$ by injectivity), so $a\ge 0$ exactly when $j(a)\ge 0$, and $j(-a)=-j(a)$ because $j$ preserves addition. Nothing on this page rests on this observation; it is recorded so that a reader meeting $\mathrm{\mid }\cdot \mathrm{\mid }$ twice knows the two notations are consistent, and every result below is proved for $\mathbb{Z}$ from the integer definition alone.

• Why an absolute value is load-bearing here and not a convenience. The published division algorithm 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 $0\le r<b$ is stated for a positive divisor, so every use of it below must first replace a divisor by a positive integer, and $\mathrm{\mid }b\mathrm{\mid }$ is what that replacement produces.