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: :$x := \displaystyle \sum_{i \mathop = 1}^n b_i x_i$ where $b_i \in B$, $x_i \in \Q$ and $n$ depends on $x$. Let $f$ be defined as: :$\map f x = \displaystyle \sum_{i \mathop = 1}^n \map f {b_i} x_i$ 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}}