Membership of Set of Integers is Replicative Function

Theorem
Let $f: \R \to \R$ be the real function defined as:


 * $\forall x \in \R: \map f x = \sqbrk {x \in \Z}$

where $\sqbrk {\cdots}$ is Iverson's convention.

Then $f$ is a replicative function.

Proof
First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$.

Thus there can be no more than one $k$ such that $0 \le k < n$ such that:
 * $x + \dfrac k n \in \Z$

Hence:
 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$

First let:
 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$

Then:

Let:
 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 0$

$\sqbrk {n x \in \Z} = 1$.

Then:

But at least one $n x + k$ such that $0 \le k < n$ is a multiple of $n$.

Hence:
 * $\exists k \in \Z: 0 \le k < n: \sqbrk {x + \dfrac k n \in \Z} = 1$

So:
 * $\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$

which contradicts the supposition that $\sqbrk {n x \in \Z} = 1$.

Hence the result by definition of replicative function.