Equivalence of Definitions of Integer Congruence

Theorem
Let $m \in \Z_{> 0}$.

Proof
Let $x_1, x_2, z \in \Z$.

Let $x_1 \equiv x_2 \pmod z$ as defined by the equal remainder after division:
 * $\mathcal R_z = \set {\tuple {x, y} \in \Z \times \Z: \exists k \in \Z: x = y + k z}$

Let $\tuple {x_1, x_2} \in \mathcal R_z$.

Then by definition:
 * $\exists k \in \Z: x_1 = x_2 + k z$

So, by definition of the modulo operation:

So:
 * $x_1 \equiv x_2 \pmod z$

in the sense of definition by modulo operation.

Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation.

That is:
 * $x_1 \equiv x_2 \pmod z \iff x_1 \mod z = x_2 \mod z$

Let $z = 0$.

Then by definition, $x_1 \mod 0 = x_1$ and $x_2 \mod 0 = x_2$.

So as $x_1 \mod 0 = x_2 \mod 0$ we have that $x_1 = x_2$.

So:
 * $x_1 - x_2 = 0 = 0.z$

and so $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.

Now suppose $z \ne 0$.

Then from definition of the modulo operation:
 * $x_1 \mod z = x_1 - z \floor {\dfrac {x_1} z}$
 * $x_2 \mod z = x_2 - z \floor {\dfrac {x_2} z}$

Thus:
 * $x_1 - z \floor {\dfrac {x_1} z} = x_2 - z \floor {\dfrac {x_2} z}$

and so:
 * $x_1 - x_2 = z \paren {\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z} }$

From the definition of the floor function, we see that both $\floor {\dfrac {x_1} z}$ and $\floor {\dfrac {x_2} z}$ are integers.

Therefore, so is $\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z}$ an integer.

So $\exists k \in \Z: x_1 - x_2 = k z$.

Thus $x_1 - x_2 = k z$ and:
 * $x_1 \equiv x_2 \pmod z$

in the sense of definition by integer multiple.

Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.

That is, $\exists k \in \Z: x_1 - x_2 = k z$.

Then $x_1 = x_2 + k z$ and so $\tuple {x_1, x_2} \in \mathcal R_z$ where:
 * $\mathcal R_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$

and so
 * $x_1 \equiv x_2 \pmod z$

in the sense of definition by equivalence relation.

So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$.