# Periodic Element is Multiple of Period

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## 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.

$\blacksquare$