Modulo Multiplication is Well-Defined/Proof 1

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


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

is a well-defined operation.

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 $x \equiv x' \pmod m$ and $y \equiv y' \pmod m$.

By definition, we have:


 * $x \equiv x' \pmod m \implies \exists k_1 \in \Z: x = x' + k_1 m$
 * $y \equiv y' \pmod m \implies \exists k_2 \in \Z: y = y' + k_2 m$

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 $x y \equiv \left({x' y'}\right) \pmod m$.

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

Warning
This result does not hold when $x, y, m \notin \Z$.

We get to this stage in the above proof:
 * $x y = \left({x' + k_1 m}\right) \left({y' + k_2 m}\right) = x' y' + \left({x' k_2 + y' k_1}\right) m + k_1 k_2 m^2$

and we note that:
 * $\left({x' k_2 + y' k_1}\right) m + k_1 k_2 m^2$

is not necessarily an integer.

In fact, $\left({x' k_2 + y' k_1}\right) m + k_1 k_2 m^2$ can only be guaranteed to be an integer if each of $x', y', m \in \Z$.

Hence $x' y'$ is not necessarily congruent to $x y$.