Number minus Modulo is Integer Multiple

Theorem

Let $x, y \in \R$ be real numbers.

Let $x \bmod y$ denote the modulo operation:

$x \bmod y := \begin{cases}

x - y \floor {\dfrac x y} & : y \ne 0 \\ x & : y = 0 \end{cases}$ where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.

Let $y < 0$.

Then:

$x - \paren {x \bmod y}$

When $y = 0$ we have:

$x \bmod y := x$

Thus:

$x - \paren {x \bmod y} = 0$

$x - \paren {x \bmod y}$

Let $y \ne 0$.

Then:

$\ds x \bmod y$

$=$

$\ds x - y \floor {\dfrac x y}$

$\ds \leadsto \ \ $

$\ds x - \paren {x \bmod y}$

$=$

$\ds y \floor {\dfrac x y}$

From Floor Function is Integer, $\floor {\dfrac x y}$ is an integer.

Thus:

$\exists n \in \Z: x - \paren {x \bmod y} = n y$

where in the case $n = \floor {\dfrac x y}$

Hence the result by definition of integer multiple.

$\blacksquare$