Antiperiodic Element is Multiple of Antiperiod

Theorem
Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.

Let $L$ be an anti-periodic element of $f$.

Then $A \divides L$.

Proof
that $A \nmid L$.

By the Division Theorem we have:
 * $\exists! q \in \Z, r \in \R: L = qA + r, 0 \lt r \lt A$

By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd.

Case 1
Suppose that $q$ is even.

Then:

And so $r$ is an anti-periodic element of $f$ that's less than $A$.

But then $A$ cannot be the anti-period of $f$.

Therefore by contradiction $q$ cannot be even.

Case 2
Now suppose that $q$ is odd.

Then:

And so $- \map f x = - \map f {x + r} \implies \map f x = \map f {x + r}$.

It is seen that $r$ is a periodic element of $f$ such that $0 \lt r \lt A$.

But consider $0 \lt A - r \lt A$:

This contradicts the fact that $A$ is the anti-period of $f$.

Hence the result.