Additive Function is Linear for Rational Factors

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

Then
 * $\forall x \in \Q: f\left({x}\right) = x f(1)$

Proof
Denote $a = f \left({1}\right)$.

Trivially, we have:


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

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:


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

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


 * $a = f \left({1}\right) = f \left({\dfrac q q}\right) = f \left({q \dfrac 1 q}\right) = q f \left({\dfrac 1 q}\right)$

It follows that:


 * $\forall q \in \Z, q \ne 0: f \left({\dfrac 1 q}\right) = \dfrac a q$

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


 * $f \left({\dfrac p q}\right) = f \left({p \dfrac 1 q}\right) = p f \left({\dfrac 1 q}\right) = p \dfrac a q = a \dfrac p q$

Therefore we conclude:


 * $ \forall r \in \Q: f \left({r}\right) = a r = r f(1).$