Double of Antiperiodic Element is Periodic

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f: \R \to \R$ be a real function.

Let $L \in \R_{>0}$ be an anti-periodic element of $f$.


Then $2 L$ is a periodic element of $f$.

In other words, every anti-periodic function is also periodic.


Proof

By Non-Zero Real Numbers Closed under Multiplication we have that $2 L \in \R_{>0}$.

Then:

\(\ds \map f {x + 2 L}\) \(=\) \(\ds \map f {x + \paren {L + L} }\)
\(\ds \) \(=\) \(\ds \map f {\paren {x + L} + L}\) Real Addition is Associative
\(\ds \) \(=\) \(\ds -\map f {x + L}\)
\(\ds \) \(=\) \(\ds \map f x\) Negative of Negative Real Number

$\blacksquare$