Periodic Element is Multiple of Period

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f: \R \to \R$ be a real periodic function with period $P$.

Let $L$ be a periodic element of $f$.


Then $P \divides L$.


Proof

Aiming for a contradiction, suppose that $P \nmid L$.

Then by the Division Theorem we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$.

And so:

\(\displaystyle \map f {x + L}\) \(=\) \(\displaystyle \map f {x + \paren {q P + r} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\paren {x + r} + q P}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {x + r}\) General Periodicity Property
\(\displaystyle \) \(=\) \(\displaystyle \map f x\) Definition of Periodic Element


But then $r$ is a periodic element of $f$ that is less than $P$.

Therefore $P$ cannot be the period of $f$.

The result follows from Proof by Contradiction.

$\blacksquare$