> [!NOTE] Lemma > Let $f:\mathbb{Q}\to \mathbb{Q}$ be [[Additive Function|additive]], i.e. for all $x,y\in \mathbb{Q},$ $f(x+y)=f(x)+f(y),$then $f$ is $\mathbb{Q}$-linear, i.e. $f(x)=xf(1).$ **Proof** Let $n$ be a positive integer. We have inductively, $f(1)=\underbrace{f\left( \frac{1}{n} \right)+\dots+f\left( \frac{1}{n} \right)}_{n \text{ times}}$so $f\left( \frac{1}{n} \right)=\frac{1}{n}f(1).$ For any positive integer $m,$ $f\left( \frac{m}{n} \right)=\underbrace{f\left( \frac{1}{n} \right)+\dots+f\left( \frac{1}{n} \right)}_{m\text{ times}}=\frac{m}{n}f(1).$Substituting $x=y=0$ yields $f(0)=0$ while substituting $x=\frac{m}{n}$ and $y=-\frac{m}{n}$ yields $f\left( -\frac{m}{n} \right)=-f\left( \frac{m}{n} \right).$ So indeed for all rationals $r,$ $f(r)=rf(1).$ # Applications **Consequences**: ...