Additive Function is Odd Function

Theorem

Let $f: \R \to \R$ be an additive function.

Then $f$ is an odd function.

Proof

From Additive Function of Zero is Zero:

$\map f 0 = 0$

Thus, for all $x \in \R$, we have:

\(\ds 0\)

\(=\)

\(\ds \map f 0\)

\(\ds \)

\(=\)

\(\ds \map f {x + \paren {-x} }\)

\(\ds \)

\(=\)

\(\ds \map f x + \map f {-x}\)

It follows that the function $f$ is odd:

$\forall x \in \R: \map f {-x} = -\map f x$.

$\blacksquare$