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

Then it is not necessarily the case that either:
 * $\forall x \in \R: f \paren x = \floor x$

or:
 * $\forall x \in \R: f \paren x = \ceiling x$

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

Consider the integer-valued function $f: \R \to \Z$ defined as:
 * $f \paren x = \floor {h \paren x}$

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

Proof for $(1)$:

We have that:
 * $h \paren {x + 1} = h \paren x + h \paren 1 = h \paren x + 1$

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

It follows that:

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
 * $(5): \quad h \paren x = x$

for all $x \in \Q$.

Let $x \in \R$.

Define $\alpha$ and $\beta$ by
 * $(6): \quad \alpha := \floor {h \paren x} = f \paren x$
 * $(7): \quad \beta := h \paren x - \alpha$

Then:

and:

Thus $h$ satisfies $(2)$.

We have that:
 * Rational Numbers form Subfield of Real Numbers
 * Field is Vector Space over Subfield

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

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

Let $f$ be defined as:
 * $f \paren x = \displaystyle \sum_{i \mathop = 1}^n f \paren {b_i} x_i$

From Expression of Vector as Linear Combination from Basis is Unique, we have:
 * $f \paren x + f \paren y = f \paren {x + y}$

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

Let $f$ be further defined as:
 * $f \paren 1 = 1$

and, for example:
 * $f \paren {\sqrt 2} = 4$

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

But:
 * $f \paren {\sqrt 2} \notin \set {1, 2}$