Characteristics of Floor and Ceiling Function/Real Domain

From ProofWiki
Jump to navigation Jump to search


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:

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


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


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

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

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:

\(\ds \map f {x + 1}\) \(=\) \(\ds \floor {\map h {x + 1} }\)
\(\ds \) \(=\) \(\ds \floor {\map h x + 1}\)
\(\ds \) \(=\) \(\ds \floor {\map h x} + 1\)
\(\ds \) \(=\) \(\ds \map f x + 1\)

Thus $h$ satisfies $(1)$.


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

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


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


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

Thus $h$ satisfies $(2)$.


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 $\R$.

From Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set:

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:

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

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

Let $f$ be defined as:

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

From Expression of Vector as Linear Combination from Basis is Unique, we have:

$\map f x + \map f y = \map f {x + y}$

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 $(2)$.


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


Historical Note

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