### Additive order

A ring whose operations are "multiplication" and "addition" can have additive orders and multiplicative orders for each of its elements. The additive order of a given element r in a ring R is the number of times n which r must be 'added' to itself (under the ring's "additive" operation) to get the zero of R. In basic mathematical notation: `n ∈ `**N** such that n*r = 0_{R}. (Note: even though integers and ring elements are being mixed/multiplied together, we define this kind of 'multiplication' as 'multiple additions', so something like `3*r` is understood to mean `r+r+r` under the ring's addition operation.) If no finite n exists such that `n*r = 0`_{R}, we let `n = ∞`.

### In the context of the integers

The set of integers form a ring (in fact, an integral domain) over the multiplication and addition operations. The additive order of any given integer within the context of the integers is ∞, so we usually talk about the additive order of integers modulo a specifc integer.

Given integers a and n (n positive), the **additive order** of `a mod n` is the smallest positive integer m such that `m*a=0(mod n)`. In fact, the additive order m of `a mod n` is always `m=n/gcd(a,n)`.

Proof (`m=n/gcd(a,n)` is the additive order of `a mod n`):

Let a and n be integers, n positive. Let `m=n/gcd(a,n)`. Then `m*a=a*n/gcd(a,n)=n*(a/gcd(a,n))=0(mod n)` (since gcd(a,n) divides a).

Let a and n be integers, n positive. Notice that `n*a=0(mod n)`, since n|n*a. But `n|n*(a/gcd(a,n))` as well, which is the same as saying `n|(n/gcd(a,n))*a`. Then d=gcd(a,n) is the greatest integer such that d|a and d|n, so n/gcd(a,n) is the smallest number with the property `n|n*a/gcd(a,n)`. Thus `m=n/gcd(a,n)` is the additive order of `a mod n`.