Double of Antiperiodic Element is Periodic
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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 |
$\blacksquare$