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

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 $(1)$: In fact $h$ satisfies (1), as
 * $h(x+1) = h(x) + h(1) = h(x) + 1$

by (4) and (3). It follows that
 * $f(x+1) = \left \lfloor{h \left({x+1}\right)}\right \rfloor

= \left \lfloor{h \left({x}\right)+1}\right \rfloor = \left \lfloor{h \left({x}\right)}\right \rfloor +1 = f(x)+1$.

Proof for $(2)$: Since $h$ satisfies (4), it is an additive function.

Let $x\in\R$.

Define $\alpha$ and $\beta$ by
 * (5) $\alpha:=\left \lfloor{h \left({x}\right)}\right \rfloor = f(x)$
 * (6) $\beta:=h \left({x}\right)-\alpha$.

Then
 * (7) $0\le\beta<1 \implies 0\le n\beta < n \implies 0\le \left\lfloor{n \beta }\right \rfloor\le n-1$

and

Since Rational Numbers form Subfield of Real Numbers, and Field is Vector Space over Subfield, we can consider $\R$ as a vector space over $\Q$.

As Square Root of 2 is Irrational, the set $\set {1, \sqrt{2}}$ is a linearly independent set in the vector space $\R$.

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

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

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

As Expression of Vector as Linear Combination from Basis is Unique, we have
 * $f(x)+f(y) = f(x+y)$

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

Define:
 * $f(1) = 1$
 * $f(\sqrt{2}) = 4$ (for example)

Then the function $f$ still satisfies (1) and (2).

But $f(\sqrt{2}) \notin \set {1, 2}$.