Modulo Multiplication is Commutative

Theorem

Multiplication modulo $m$ is commutative:

$\forall \left[\!\left[{x}\right]\!\right]_m, \left[\!\left[{y}\right]\!\right]_m \in \Z_m: \left[\!\left[{x}\right]\!\right]_m \times_m \left[\!\left[{y}\right]\!\right]_m = \left[\!\left[{y}\right]\!\right]_m \times_m \left[\!\left[{x}\right]\!\right]_m$

Proof

Follows directly from the definition of multiplication modulo $m$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_m \times_m \left[\!\left[{y}\right]\!\right]_m\)	\(=\)	\(\displaystyle \left[\!\left[{x y}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{y x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{y}\right]\!\right]_m \times_m \left[\!\left[{x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

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