# Modulo Operation as Integer Difference by Quotient

## Theorem

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

Let $y > 0$.

Let $0 \le z < y$.

Let:

$\dfrac {x - z} y = k$

for some integer $k$.

Then:

$z = x \bmod y$

where $\bmod$ denotes the modulo operation.

## Proof

We have:

 $\ds \dfrac {x - z} y$ $=$ $\ds k$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds z - k y$

We also have:

$0 \le z < y$

Hence:

$0 \le \dfrac z y < 1$

and so by definition of floor function:

$(2): \quad \floor {\dfrac z y} = 0$

Thus:

 $\ds x \bmod y$ $=$ $\ds x - y \floor {\dfrac x y}$ Definition of Modulo Operation: note that $y \ne 0$ by hypothesis $\ds$ $=$ $\ds \paren {z - k y} - y \floor {\dfrac {z - k y} y}$ from $(1)$ $\ds$ $=$ $\ds z - k y - y \floor {\dfrac z y - k}$ $\ds$ $=$ $\ds z - k y - y \paren {\floor {\dfrac z y} - k}$ Floor of Number plus Integer $\ds$ $=$ $\ds z - k y + k y - y \floor {\dfrac z y}$ $\ds$ $=$ $\ds z - y \times 0$ from $(2)$

The result follows.

$\blacksquare$