functional equations - Proof that solutions to Cauchy F.E. over $mathbb{Q}$ are linear

I would like to prove that the solutions to Cauchy's functional equation over $\mathbb{Q}$ are linear, that is, all solutions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ have the property $f(x)=cx$ for some $c\in\mathbb{Q}$.

The proof is to be done in a very specific order: First, we must prove that it works for all $x\in\mathbb{N}$, then for all $x\in\mathbb{Z}$. Third we must show it works for all $\frac{1}{x}$ where $x\in\mathbb{Z}$. Finally, we must show that it works for all $\frac{p}{q}$ where $p,q\in\mathbb{Z}$, $q\neq 0$.

I am having trouble with the third part, where I have to show it works for $\frac{1}{x}$.
Here is what I have done up to this point:

Let $f$ be a solution to the cauchy functional equation.
We know that $f(0)=c\cdot 0 = 0$, as $f(x)=f(0+x)=f(0)+f(x)$ for some number $x$.
Suppose $x\in\mathbb{N}$ and $x\neq 0$.
Then,
$$\begin{align*} f(x) & =\underbrace{f(1)+f(1)+\dots+f(1)}_{\text{$x$ times}} \\ & = x\cdot f(1) \\ & =x\cdot c \end{align*}$$
If $x\in\mathbb{Z}$ and $x\neq 0$, then we have two cases: $x>0$ or $x<0$.
For $x>0$, the proof with $x\in\mathbb{N}$ suffices.

If $x<0$, then we have
$$\begin{align*} f(x) & =\underbrace{f(-1)+f(-1)+\dots+f(-1)}_{\text{$x$ times}} \\ & = x\cdot f(-1) \\ & =x\cdot c \end{align*}$$

I don't know how to proceed with this method to show that $f(\frac{1}{x})=\frac{c}{x}$.

Answer

The right way to proceed lies in the realization that your argument for $x \in \mathbb{N}$ can be extended slightly so say that if $x \in \mathbb{N}$, then $f(x k) = x f(k)$. Why? For the same reason you pointed out. If $x \in \mathbb{N}$, then

$$f(xk) = f(k + k + \dots +k) = f(k) + \dots + f(k)=x f(k)$$ This result is also what provides the last part of the question.

With this in hand, you're in fine shape! This is because you know $f(x* \frac{1}{x}) = f(1) = c = x f(\frac{1}{x})$.