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

Then it is not necessarily the case that either:
 * $\forall x \in \R: f \left({x}\right) = \left \lfloor{x}\right \rfloor$

or:
 * $\forall x \in \R: f \left({x}\right) = \left \lceil{x}\right \rceil$

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:
 * $f \left({x}\right) = \left \lfloor{h \left({x}\right)}\right \rfloor$

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 $(2)$: By induction, we have
 * (5) $h(nx) = nh(x)$

and
 * (6) $h(x/n) = h(x)/n$

for all $n \in \Z_{> 0}$.

In particular,
 * (7) $h(n) = nh(1) = n$ for all $n \in \Z_{\ge 0}$

as
 * $1 = h(1) = h(1+0) = h(1) + h(0) = 1 + 0$

so
 * $h(0) = 0.$

Let $x\in\R$.

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

Then

In the last line, we have also used
 * $0\le\beta<1 \implies 0\le n\beta < n \implies 0\le \left\lfloor{n \beta }\right \rfloor\le n-1.$

Consider $\R$ as a vector space over $\Q$.

Let $B$ be a basis of $\R$ which includes 1 and $\sqrt{2}$ (for example).

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
 * $f(x) = \sum_{i=1}^n f(b_i) x_i.$

Then
 * $f(x)+f(y) = f(x+y)$

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

Define
 * $f(1) = 1$
 * $f(\sqrt{2}) = 4$.

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

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