Theorem: Characteristics of Floor and Ceiling Function/Real Domain

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 $x, y \in \R$:
{{begin-eqn}}
{{eqn | n = 3
      | l = \map h 1
      | r = 1
      | c = 
}}
{{eqn | n = 4
      | l = \map h x + \map h y
      | r = \map h {x + y}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map f {x + 1}
      | r = \floor {\map h {x + 1} }
      | c = 
}}
{{eqn | r = \floor {\map h x + 1}
      | c = 
}}
{{eqn | r = \floor {\map h x} + 1
      | c = 
}}
{{eqn | r = \map f x + 1
      | c = 
}}
{{end-eqn}}
Thus $h$ satisfies $(1)$.
{{qed|lemma}}
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:
{{begin-eqn}}
{{eqn | n = 8
      | l = 0
      | o = \le
      | r = \beta < 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \le
      | r = n \beta < n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \le
      | r = \floor {n \beta} \le n - 1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map f {\dfrac {\map f n x} n}
      | r = \floor {\map h {\dfrac {\floor {\map h {n x} } } n} }
}}
{{eqn | r = \floor {\dfrac {\floor {n \map h x} } n}
      | c = $(5)$, Additive Function is Linear for Rational Factors
}}
{{eqn | r = \floor {\dfrac {\floor {n \paren {\alpha + \beta} } } n}
      | c = $(6)$, $(7)$
}}
{{eqn | r = \floor {\dfrac {n \alpha + \floor {n \beta} } n}
      | c = as $\alpha \in \Z$
}}
{{eqn | r = \alpha + \floor {\dfrac 1 n \floor {n \beta} }
      | c = 
}}
{{eqn | r = \floor {\map h x}
      | c = $(8)$
}}
{{eqn | r = \map f x
      | c = Definition of $\map f x$
}}
{{end-eqn}}
Thus $h$ satisfies $(2)$.
{{qed|lemma}}
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 $(2)$.
But:
:$\map f {\sqrt 2} \notin \set {1, 2}$
{{qed}}