Proof of Cauchy's functional equation for rational arguments

Question

We have thesis that for every $c\in\mathbb{Q}$ every additive function has form of $f(x)=cx$. In the proof we're showing that $f(nx)=nf(x)$. Then we're supposed to replace $nx$ by $\frac{1}{n}x$. Why can we do this?

Answer 1

It follows from $x = \underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}$

$$f(x) = f(\underbrace{\frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \frac{1}{n}x + \cdots}_{n\text{ times}}) = \underbrace{f(\frac{1}{n}x) + f(\frac{1}{n}x) + f(\frac{1}{n}x) + \cdots}_{n\text{ times}} = nf\left(\frac{1}{n}x\right)$$