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 $d\mid -a$

Let $a,b,c,d,x,y\in \mathbb{Z}$ and let $\mid$ be divisibility (Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$). Then

1. reflexivity: $a\mid a$;
2. transitivity: $d\mid a$ and $a\mid b$ imply $d\mid b$;
3. linearity: $d\mid a$ and $d\mid b$ imply $d\mid ax+by$; in particular $d\mid a+b$ and $d\mid a-b$;
4. $d\mid a$ implies $d\mid ac$;
5. $d\mid a$ implies $-d\mid a$ and $d\mid -a$.

Consequently the four statements $d\mid a$, $-d\mid a$, $d\mid -a$ and $-d\mid -a$ are equivalent.

• Linearity is the workhorse of the whole page. Every later argument that a common divisor of $a$ and $b$ divides some combination of them — Bézout's identity, the Euclidean step, the coprimality criterion — is clause 3 applied once.

• Divisibility ignores signs, and that is why $\gcd$ can be normalised. Clause 5 says the divisors of $a$ and of $-a$ are the same set, so $d\mid a$ holds exactly when $d\mid \mathrm{\mid }a\mathrm{\mid }$, whichever of the two values $\mathrm{\mid }a\mathrm{\mid }$ takes (The absolute value $\mathrm{\mid }a\mathrm{\mid }$ of an integer). That is the observation behind $\gcd (a,b)=\gcd (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$ in $\gcd$ is symmetric and unchanged by signs: $\gcd (a,b)=\gcd (b,a)=\gcd (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$; moreover $\gcd (a,0)=\mathrm{\mid }a\mathrm{\mid }$, $\gcd (a,1)=1$, $\gcd (a,a)=\mathrm{\mid }a\mathrm{\mid }$, and $\gcd (a,b)\ge 1$ unless $a=b=0$.