Periodic Element is Multiple of Period

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
that $P \nmid L$.

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

And so:

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

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

The result follows from Proof by Contradiction.