Modulo Multiplication is Well-Defined

Theorem
The multiplication operation on $$\Z_m$$, the set of integers modulo $m$, defined by the rule:


 * $$\left[\!\left[{a}\right]\!\right]_m \times_m \left[\!\left[{b}\right]\!\right]_m = \left[\!\left[{a b}\right]\!\right]_m$$

is a well-defined operation.

This operation is called multiplication modulo $$m$$.

Proof
We need to show that if:


 * $$\left[\!\left[{x'}\right]\!\right]_m = \left[\!\left[{x}\right]\!\right]_m$$ and
 * $$\left[\!\left[{y'}\right]\!\right]_m = \left[\!\left[{y}\right]\!\right]_m$$

then $$\left[\!\left[{x' y'}\right]\!\right]_m = \left[\!\left[{x y}\right]\!\right]_m$$.

Since $$\left[\!\left[{x'}\right]\!\right]_m = \left[\!\left[{x}\right]\!\right]_m$$ and $$\left[\!\left[{y'}\right]\!\right]_m = \left[\!\left[{y}\right]\!\right]_m$$, it follows from the definition of residue class modulo $m$ that $$x \equiv x' \left({\bmod\, m}\right)$$ and $$y \equiv y' \left({\bmod\, m}\right)$$.

By definition, we have:


 * $$x \equiv x' \left({\bmod\, m}\right) \Longrightarrow \exists k_1 \in \Z: x = x' + k_1 m$$
 * $$y \equiv y' \left({\bmod\, m}\right) \Longrightarrow \exists k_2 \in \Z: y = y' + k_2 m$$

which gives us $$x y = \left({x' + k_1 m}\right) \left({y' + k_2 m}\right) = x' y' + \left({x' k_2 + y' k_1}\right) m + k_1 k_2 m^2$$.

Thus by definition $$x y \equiv \left({x' y'}\right) \left({\bmod\, m}\right)$$.

Therefore, by the definition of residue class modulo $m$, $$\left[\!\left[{x' y'}\right]\!\right]_m = \left[\!\left[{x y}\right]\!\right]_m$$.

Comment
Although the operation of multiplication modulo $$m$$ is denoted by the symbol $$\times_m$$, if there is no danger of confusion, the conventional multiplication symbols $$\times, \cdot$$ etc. are often used instead.

In fact, the notation for multiplication of two integers modulo $m$ is not usually $$\left[\!\left[{a}\right]\!\right]_m \times_m \left[\!\left[{b}\right]\!\right]_m$$.

What is more normally seen is $$a b \left({\bmod\, m}\right)$$.

Using this notation, what this result says is:

$$ $$ $$

and it can be proved in the same way.