Equivalence of Definitions of Congruence

Theorem
The definitions of congruence (in the context of Number Theory):


 * $(1) \qquad x \equiv y \left({\bmod\, z}\right) \iff \left({x, y}\right) \in \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$


 * $(2) \qquad x \equiv y \left({\bmod\, z}\right) \iff x \,\bmod\, z = y \,\bmod\, z$


 * $(3) \qquad x \equiv y \left({\bmod\, z}\right) \iff \exists k \in \Z: x - y = k z$

... are equivalent.

Integer definition
Also, if $x, y, z$ are all integers, the definition:
 * $x \equiv y \left({\bmod\, z}\right) \iff z \backslash \left({x - y}\right)$

is likewise equivalent to the other given definitions.

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

Let $x_1 \equiv x_2 \left({\bmod\, z}\right)$ in the sense of Definition by Equivalence Relation.

That is, let $\mathcal R_z$ be the relation on the set of all $x, y \in \R$:
 * $\mathcal R_z = \left\{{\left({x, y}\right) \in \R \times \R: \exists k \in \Z: x = y + k z}\right\}$

Let $\left({x_1, x_2}\right) \in \mathcal R_z$.

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

So, by definition of the modulo operation, we have:

So:
 * $x_1 \equiv x_2 \left({\bmod\, z}\right)$

in the sense of Definition by Modulo Operation.

Now let $x_1 \equiv x_2 \left({\bmod\, z}\right)$ in the sense of Definition by Modulo Operation.

That is, :$x_1 \equiv x_2 (\bmod\, z) \iff x_1 \,\bmod\, z = x_2 \,\bmod\, z$.

Let $z = 0$.

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

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

So $x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \left({\bmod\, z}\right)$ in the sense of Definition by Integral Multiple.

Now suppose $z \ne 0$.

Then from definition of the modulo operation:
 * $x_1 \,\bmod\, z = x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor$
 * $x_2 \,\bmod\, z = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$

Thus:
 * $x_1 - z \left \lfloor {\dfrac {x_1} z}\right \rfloor = x_2 - z \left \lfloor {\dfrac {x_2} z}\right \rfloor$

and so:
 * $x_1 - x_2 = z \left({\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor}\right)$

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

Therefore, so is $\left \lfloor {\dfrac {x_1} z}\right \rfloor - \left \lfloor {\dfrac {x_2} z}\right \rfloor$ an integer.

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

Thus $x_1 - x_2 = k z$ and:
 * $x_1 \equiv x_2 \left({\bmod\, z}\right)$

in the sense of Definition by Integral Multiple.

Now let $x_1 \equiv x_2 \left({\bmod\, z}\right)$ in the sense of Definition by Integral Multiple.

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

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

and so
 * $x_1 \equiv x_2 \left({\bmod\, z}\right)$

in the sense of Definition by Equivalence Relation.

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

Proof for Integer Definition
By definition of divisor, we have that:
 * $z \backslash x - y \iff x - y = k z$

which is equivalent to the third definition.