Modulo Multiplication is Well-Defined

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


By definition, we have:

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


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


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

$\blacksquare$


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 $x y$.


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