Boundedness of a discontinuous function satisfying $f(x+y)=f(x)+f(y)$ Suppose $f:\mathbb R\rightarrow \mathbb R$ s.t. $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb R$ and $f$ is not continuous on $\mathbb R$. Prove that
(a).$f$ is not bounded below (or above) on any subinterval $(a,b)$ of $\mathbb R$.
(b). $f$ is not monotone.
On plugging $x=y=0$; $f(0)=f(0)+f(0)$ which gives $f(0)=0$.
Also, plug $y=-x$ to get $f(-x)=-f(x)$ i.e. $f$ is an odd function and thus have graph in opposite quadrants. Now, I am stuck how to show that graph will be unbounded? Please help!

Here's a route:
$\qquad(1)$: Show that $f$ is $\mathbb{Q}$-linear. In particular, show that $f(q\cdot x)=q\cdot f(x)$ for all $q\in\mathbb{Q}$.
$\qquad(2)$: Using $(1)$, show that if $f$ is bounded on any interval $[a,b]$ (why can you assume it is closed?), then $f$ is continuous at $x=0$.
$\qquad(3)$: Establish a contradiciton by showing that if $f$ is continuous at $0$, it is continuous everywhere.
$\qquad(4)$: Show that $f$ cannot be monotone on any given interval because it is not bounded on that interval, by $(3)$.

Suppose without loss of generality that $0<a<b$ (why can you assume both are positive?) and that $|f(x)|\leq C$ for all $x\in \mathbb[a,b]$ and some $C>0$.
We will show that
$$\forall \epsilon>0,\,\exists\delta>0,\,\forall x \text{ with $0<x<\delta$},\, |f(x)|\leq\epsilon$$
Observe that because $f$ is odd and $f(0)=0$, this suffices.
Notice that $f(x)=\frac1q f(qx)$ for any $q\in\mathbb{Q}\setminus\{0\}$, because of point $(1)$.
It follows that for all $q\in\mathbb{Q}_+$ and all $x\in[a,b]$, $|f(qx)|\leq qC$. In other words, for all $q\in\mathbb{Q}_+$ and all $x\in[qa,qb]$, $|f(x)|\leq qC$.
Taking $q=\frac{\epsilon}C$, we find that $|f(x)|\leq \epsilon$ for all $x\in\left[\frac{a\epsilon}C,\frac{b\epsilon}C\right]$.
Now, if $0<x<\frac{a\epsilon}C$, there is some $q>1$ with $qx\in\left[\frac{a\epsilon}C,\frac{b\epsilon}C\right]$.
Then:
$$|f(x)|\leq q|f(x)|=|qf(x)|=|f(qx)|\leq \epsilon$$
It follows that whenever $0<x\leq\frac{b\epsilon}C$, $|f(x)|\leq\epsilon$, so we may take $\delta=\frac{b\epsilon}C$ which completes the proof.
