Congruence modulo an integer: $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ when $n\mid (a-b)$, including the moduli $0$ and $1$

Let $a,b,n\in \mathbb{Z}$. We say that $a$ is congruent to $b$ modulo $n$, and write

$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n),$

when $n\mid (a-b)$ in the sense of Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$. Thus there is an integer $q$ with $a-b=nq$. The modulus is any integer; it is not required to be positive.

At modulus $0$, the condition is $0\mid (a-b)$, which holds exactly when $a-b=0$. Hence

$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}0)⟺a=b\mathrm{.}$

At modulus $1$, every two integers are congruent because $1$ divides every integer. Changing the sign of the modulus changes nothing: $n\mid (a-b)$ exactly when $-n\mid (a-b)$.

• Congruence is a relation on integers. The quotient sets and operations below use nonnegative or positive moduli only where their statements require it.
• The notation $a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>n$ is not used for a chosen remainder here. The relation $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ is defined even for $n=0$, where a remainder operation would not be available.