Modulo Multiplication has Identity

Theorem

Multiplication modulo $m$ has an identity:

$\forall \left[\!\left[{x}\right]\!\right]_m \in \Z_m: \left[\!\left[{x}\right]\!\right]_m \times_m \left[\!\left[{1}\right]\!\right]_m = \left[\!\left[{x}\right]\!\right]_m = \left[\!\left[{1}\right]\!\right]_m \times_m \left[\!\left[{x}\right]\!\right]_m$

Proof

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_m \times_m \left[\!\left[{1}\right]\!\right]_m\)	\(=\)	\(\displaystyle \left[\!\left[{x \times 1}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{1 \times x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{1}\right]\!\right]_m \times_m \left[\!\left[{x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus $\left[\!\left[{1}\right]\!\right]_m$ is the identity for multiplication modulo $m$.

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 19.1$