# Characteristics of Floor and Ceiling Function/Real Domain

## Theorem

Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:

$(1): \quad \map f {x + 1} = \map f x + 1$
$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$

Then it is not necessarily the case that either:

$\forall x \in \R: \map f x = \floor x$

or:

$\forall x \in \R: \map f x = \ceiling x$

## Proof

Let $h: \R \to \R$ be a real function such that for all $x, y \in \R$:

 $(3):\quad$ $\displaystyle \map h 1$ $=$ $\displaystyle 1$ $(4):\quad$ $\displaystyle \map h x + \map h y$ $=$ $\displaystyle \map h {x + y}$

Consider the integer-valued function $f: \R \to \Z$ defined as:

$\map f x = \floor {\map h x}$

We claim that $f$ satisfies $(1)$ and $(2)$.

Proof for $(1)$:

We have that:

$\map h {x + 1} = \map h x + \map h 1 = \map h x + 1$

by $(4)$ and $(3)$.

It follows that:

 $\displaystyle \map f {x + 1}$ $=$ $\displaystyle \floor {\map h {x + 1} }$ $\displaystyle$ $=$ $\displaystyle \floor {\map h x + 1}$ $\displaystyle$ $=$ $\displaystyle \floor {\map h x} + 1$ $\displaystyle$ $=$ $\displaystyle \map f x + 1$

Thus $h$ satisfies $(1)$.

$\Box$

Proof for $(2)$:

Since $h$ satisfies $(4)$, it is an additive function.

By $(3)$ and since Additive Function is Linear for Rational Factors, this implies

$(5): \quad \map h x = x$

for all $x \in \Q$.

Let $x \in \R$.

Define $\alpha$ and $\beta$ by

$(6): \quad \alpha := \floor {\map h x} = \map f x$
$(7): \quad \beta := \map h x - \alpha$

Then:

 $(8):\quad$ $\displaystyle 0$ $\le$ $\displaystyle \beta < 1$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $\le$ $\displaystyle n \beta < n$ $\displaystyle \leadsto \ \$ $\displaystyle 0$ $\le$ $\displaystyle \floor {n \beta} \le n - 1$

and:

 $\displaystyle \map f {\dfrac {\map f n x} n}$ $=$ $\displaystyle \floor {\map h {\dfrac {\floor {\map h {n x} } } n} }$ $\displaystyle$ $=$ $\displaystyle \floor {\dfrac {\floor {n \map h x} } n}$ $(5)$, Additive Function is Linear for Rational Factors $\displaystyle$ $=$ $\displaystyle \floor {\dfrac {\floor {n \paren {\alpha + \beta} } } n}$ $(6)$, $(7)$ $\displaystyle$ $=$ $\displaystyle \floor {\dfrac {n \alpha + \floor {n \beta} } n}$ as $\alpha \in \Z$ $\displaystyle$ $=$ $\displaystyle \alpha + \floor {\dfrac 1 n \floor {n \beta} }$ $\displaystyle$ $=$ $\displaystyle \floor {\map h x}$ $(8)$ $\displaystyle$ $=$ $\displaystyle \map f x$ Definition of $\map f x$

Thus $h$ satisfies $(2)$.

$\Box$

We have that:

Rational Numbers form Subfield of Real Numbers
Vector Space on Field Extension is Vector Space

Thus we can consider $\R$ as a vector space over $\Q$.

We also have that Square Root of 2 is Irrational

Hence the set $\set {1, \sqrt 2}$ is a linearly independent set in the vector space $\R$.

there exists a basis $B$ of $\R$ which includes $1$ and $\sqrt 2$.

Then each $x \in \R$ can be written as a finite sum:

$x := \displaystyle \sum_{i \mathop = 1}^n b_i x_i$

where $b_i \in B$, $x_i \in \Q$ and $n$ depends on $x$.

Let $f$ be defined as:

$\map f x = \displaystyle \sum_{i \mathop = 1}^n \map f {b_i} x_i$
$\map f x + \map f y = \map f {x + y}$

no matter how $\map f b$ is defined for $b \in B$.

Let $f$ be further defined as:

$\map f 1 = 1$

and, for example:

$\map f {\sqrt 2} = 4$

Then $f$ satisfies $(1)$ and $(2)$.

But:

$\map f {\sqrt 2} \notin \set {1, 2}$

$\blacksquare$

## Historical Note

The limitation of Characteristics of Floor and Ceiling Function to rational numbers was demonstrated by Georg Karl Wilhelm Hamel in $1905$.