# 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

Aiming for a contradiction, suppose that $A \nmid L$.

By the Division Theorem we have:

$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$

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

### Case 1

Suppose $q$ is even.

Then:

 $\ds \map f {x + L}$ $=$ $\ds \map f {x + \paren {q A + r} }$ $\ds$ $=$ $\ds \map f {\paren {x + r} + q A}$ $\ds$ $=$ $\ds \map f {x + r}$ General Antiperiodicity Property $\ds$ $=$ $\ds -\map f x$

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

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

Therefore by contradiction $q$ cannot be even.

### Case 2

Suppose $q$ is odd.

Then:

 $\ds - \map f x$ $=$ $\ds \map f {x + L}$ $\ds$ $=$ $\ds \map f {x + \paren {q A + r} }$ $\ds$ $=$ $\ds \map f {\paren {x + r} + q A}$ $\ds$ $=$ $\ds - \map f {x + r}$ General Antiperiodicity Property

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 $0 < r < A$.

But consider $0 < A - r < A$:

 $\ds \map f {x + \paren {A - r} }$ $=$ $\ds \map f {\paren {x + A} - r}$ $\ds$ $=$ $\ds \map f {x + A}$ $\ds$ $=$ $\ds -\map f x$

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

Hence the result.

$\blacksquare$