Modulo Addition is Closed

Integers
Let $$m \in \Z$$ be an integer.

Then addition modulo $m$ on the set of integers modulo $m$ is closed:


 * $$\forall \left[\!\left[{x}\right]\!\right]_m, \left[\!\left[{y}\right]\!\right]_m \in \Z_m: \left[\!\left[{x}\right]\!\right]_m +_m \left[\!\left[{y}\right]\!\right]_m \in \Z_m$$.

Real Numbers
Let $$z \in \R$$ be a real number.

Then addition modulo $z$ on the set of all residue classes modulo $z$ is closed:


 * $$\forall \left[\!\left[{x}\right]\!\right]_z, \left[\!\left[{y}\right]\!\right]_z \in \R_z: \left[\!\left[{x}\right]\!\right]_z +_z \left[\!\left[{y}\right]\!\right]_z \in \R_z$$.

Proof for Integers
From the definition of addition modulo $m$, we have:
 * $$\left[\!\left[{x}\right]\!\right]_m +_m \left[\!\left[{y}\right]\!\right]_m = \left[\!\left[{x + y}\right]\!\right]_m$$

By the Division Theorem:
 * $$x + y = q m + r$$ where $$0 \le r < m$$

Therefore $$\left[\!\left[{x + y}\right]\!\right]_m = \left[\!\left[{r}\right]\!\right]_m, 0 \le r < m$$.

Therefore $$\left[\!\left[{x + y}\right]\!\right]_m \in \Z_m$$, from the definition of integers modulo $m$.

Proof for Real Numbers
From the definition of addition modulo $z$, we have:
 * $$\left[\!\left[{x}\right]\!\right]_z +_z \left[\!\left[{y}\right]\!\right]_z = \left[\!\left[{x + y}\right]\!\right]_z$$

As $$x, y \in R$$, we have that $$x + y \in \R$$ as Real Addition is Closed.

Hence by definition, $$\left[\!\left[{x + y}\right]\!\right]_z \in \R_z$$.