Modulo Addition is Commutative

Theorem

Modulo addition is commutative:

$\forall x, y, z \in \R: x + y \left({\bmod\,z}\right) = y + x \left({\bmod\,z}\right)$

Proof

From the definition of modulo addition, this is also written:

$\forall z \in \R: \forall \left[\!\left[{x}\right]\!\right]_z, \left[\!\left[{y}\right]\!\right]_z \in \R_z: \left[\!\left[{x}\right]\!\right]_z +_z \left[\!\left[{y}\right]\!\right]_z = \left[\!\left[{y}\right]\!\right]_z +_z \left[\!\left[{x}\right]\!\right]_z$

Hence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_z +_z \left[\!\left[{y}\right]\!\right]_z\)	\(=\)	\(\displaystyle \left[\!\left[{x + y}\right]\!\right]_z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of addition modulo $m$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{y + x}\right]\!\right]_z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Addition of Numbers is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{y}\right]\!\right]_z +_z \left[\!\left[{x}\right]\!\right]_z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of addition modulo $m$

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $2.7$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 19.1$