# Modulo Addition is Well-Defined/Proof 1

## Theorem

Let $m \in \Z$ be an integer.

Let $\Z_m$ be the set of integers modulo $m$.

The modulo addition operation on $\Z_m$, defined by the rule:

$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$

That is:

If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a + x \equiv b + y \pmod m$.

## Proof

We need to show that if:

$\eqclass {x'} m = \eqclass x m$
$\eqclass {y'} m = \eqclass y m$

then:

$\eqclass {x' + y'} m = \eqclass {x + y} m$

Since:

$\eqclass {x'} m = \eqclass x m$

and:

$\eqclass {y'} m = \eqclass y m$

it follows from the definition of set of integers 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:

$x + y = x' + k_1 m + y' + k_2 m = x' + y' + \paren {k_1 + k_2} m$

As $k_1 + k_2$ is an integer, it follows that, by definition:

$x + y \equiv \paren {x' + y'} \pmod m$

Therefore, by the definition of integers modulo $m$:

$\eqclass {x' + y'} m = \eqclass {x + y} m$

$\blacksquare$