TITLE: If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$
QUESTION [6 upvotes]: For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?
I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly 
$$
f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right)
$$
but I can't seem to finish it, any help?

REPLY [12 votes]: Try computing $bf(x/b){}{}{}$.

REPLY [4 votes]: Let's show that $f(x/b) = \frac{f(x)}{b}$. Denote $t= f(x/b)$. Now sum $t$ with itself exactly $b$ times: 
$$b\cdot t=t+t+\dots+t=f(x/b)+f(x/b)+\dots + f(x/b)$$
$$=f(x/b+\dots+x/b)=f((bx)/b)=f(x)$$
Divide both sides by $b$ and you have $t=\frac{f(x)}{b}$.

REPLY [4 votes]: For $c \in \mathbb{Z}^*$:
$$
f(cx) = cf(x)
$$
Let's do a variable substitution: $\hat{x} = cx$:
\begin{align*}
f(\hat{x}) &= cf\left(\frac{\hat{x}}{c}\right) \\
\Rightarrow f\left(\frac{\hat{x}}{c}\right) &= \frac{1}{c} f(\hat{x})
\end{align*}
Combine this with your initial findings to get the proof for $c \in \mathbb{Q}$.