Membership of Equivalence Class of m mod pi is Replicative Function

From ProofWiki
Jump to navigation Jump to search

Theorem

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

$\forall x \in \R: \map f x = \sqbrk {\exists r \in \Q, \exists m \in \Z: x = r \pi + m}$

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


Then $f$ is a replicative function.


Proof

Let $\map f x = 1$.

Then:

$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$

and:

\(\displaystyle x\) \(=\) \(\displaystyle r \pi + m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle n x\) \(=\) \(\displaystyle \paren {r n} \pi + m n\)

But $r n \in Q$ and $m n \in \Z$.

So:

$\map f x = 1 \implies \map f {n x} = 1$


Let $n \in \Z_{>0}$.

Let $k \in \Z: 0 \le k < n$.

Then:

\(\displaystyle x + \frac k n\) \(=\) \(\displaystyle r \pi + m + \frac k n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x + \frac k n\) \(=\) \(\displaystyle \paren {r + \frac {k \pi} n} \pi + m\)

Thus, when $0 \le k < n$ it follows that the coefficient of $\pi$ cannot be rational.

Thus:

$\map f x = 1 \implies \map f {x + \dfrac k n} = 0$

and so:

$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = \map f {n x}$

$\Box$


Now let $\map f {n x} = 1$.

Then:

$r_n \pi + m_n = n x$

for some rational $r_n$ and integer $m_n$.

Thus exactly one of $m_n + k$ where $0 \le n < k$ is a multiple of $n$.

Therefore there exists exactly one $k \in \Z$ such that $0 \le n < k$ such that:

$\map f {x + \dfrac k n} = 1$

Therefore:

$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = 1 = \map f {n x}$


Hence the result by definition of replicative function.

$\blacksquare$


Sources