Additive Function is Linear for Rational Factors

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

Then:
 * $\forall r \in \Q, x \in \R: \map f {x r} = r \map f x$

Proof
Trivially, we have:


 * $\forall x \in \R: \map f {1 \cdot x} = 1 \map f x$

Next, suppose that:


 * $\map f {n x} = n \map f x$

By additivity of $f$, we have:

Hence by the Principle of Mathematical Induction:


 * $\forall n \in \N, x \in \R: \map f {n x} = n \map f x$

As Additive Function is Odd Function and Odd Function of Zero is Zero, we conclude:


 * $\forall p \in \Z, x \in \R: \map f {p x} = p \map f x$

Let $p \ne 0$.

By substituting $y = p x$, the above gives:
 * $\forall p \in \Z \setminus \set 0, y \in \R: \map f y = p \map f {\dfrac y p}$

In other words:
 * $\forall p \in \Z \setminus \set 0, y \in \R: \map f {\dfrac y p} = \dfrac 1 p \map f y$

Given $p, q \in \Z, q \ne 0$, we have:

Therefore we conclude:


 * $\forall r \in \Q, x \in \R: \map f {r x} = r \map f x$