Floor of x+m over n/Proof 2

Theorem

Let $m, n \in \Z$ such that $n > 0$.

Let $x \in \R$.

Then:

$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$

where $\floor x$ denotes the floor of $x$.

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \dfrac {x + m} n$

It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$.

Let $\dfrac {x + m} n \in \Z$.

Then:

$\ds \exists s \in \Z: \, $

$\ds \dfrac {x + m} n$

$=$

$\ds s$

$\ds \leadsto \ \ $

$\ds x + m$

$=$

$\ds n s$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds n s - m$

$\ds $

$\in$

$\ds \Z$

Thus:

$\forall x \in \R: \map f x \in \Z \implies x \in \Z$

So the conditions are fulfilled for McEliece's Theorem (Integer Functions) to be applied:

$\floor {\map f x} = \floor {\map f {\floor x} } \iff \paren {\map f x \in \Z \implies x \in \Z}$

Hence the result.

$\blacksquare$

1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ (Answers to Exercises)