# Additive Function is Linear for Rational Factors

## Theorem

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

Then

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

## Proof

Trivially, we have:

$\forall x \in \R: f \left({1 x}\right) = 1 f \left({x}\right)$

Next, suppose that:

$f \left({n x}\right) = n f \left({x}\right)$

By additivity of $f$, we have:

$f \left({\left({n + 1}\right) x}\right) = f \left({n x + x}\right) = f \left({n x}\right) + f \left({x}\right) = n f \left({x}\right) + f \left({x}\right) = \left({n+1}\right) f \left({x}\right)$

Now induction gives us:

$\forall n \in \N, x \in \R: f \left({n x}\right) = n f \left({x}\right)$

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

$\forall p \in \Z, x \in \R: f \left({p x}\right) = p f \left({x}\right)$

Let $p \neq 0$.

By substituting $y = p x$, the above gives

$\forall p \in \Z \setminus \set{0}, y \in \R: f \left({y}\right) = p f \left(\dfrac{y}{p}\right)$

In other words,

$\forall p \in \Z \setminus \set{0}, y \in \R: f \left(\dfrac{y}{p}\right) = \dfrac{1}{p} f \left({y}\right)$.

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

$f \left({\dfrac p q} x\right) = f \left({p \dfrac x q}\right) = p f \left({\dfrac x q}\right) = p \dfrac{f(x)}{q} = \dfrac p q f(x)$

Therefore we conclude:

$\forall r \in \Q, x\in \R: f \left({r x}\right) = r f(x).$

$\blacksquare$