Find all increasing functions $f: \mathbb{R} \rightarrow \mathbb{R}$, such that $f(1)=1$ and, for all $x, y \in \mathbb{R}, f(x+y)=f(x)+f(y)$.
By setting $x=y=0$ we find $f(0)=0$. By setting $y=-x$ we have $f(0)=f(x)+f(-x)$, so $f(-x)=-f(x)$. Let $x \in \mathbb{R}$. We have $f(2 x)-f(x)+f(x)$. By induction we have for all natural $n$,

$$
f((n) x)=n f(x)
$$

For all $m \in \mathbb{N}$, we have $f\left(1 \cdot \frac{1}{m}=m f\left(\frac{1}{m}\right)\right)$, so $f\left(\frac{1}{m}\right)=\frac{1}{m}$. By substituting into (III.8) we get $f(q)=q$ for all $q \in \mathbb{Q}_{+}$. Since $f$ is odd, $f(q)=q$ for all rational $q$.

Let $r \in \mathbb{R}$ and suppose $f(r) \neq r$. Assume $f(r)>r$, the case $f(r)<r$ being analogous. Since between any two real numbers there is a rational number, we take $q_{0}$ such that $f(r)<q_{0}<r$. Since $f$ is increasing, we have $f(f(r))<f\left(q_{0}\right) \leqslant f(r)$. Contradiction. Thus, $f(x)=x$ for all $x$.