Discontinuous homomorphism from $\Bbb R$ to $\Bbb R$ Does there exist a function $f:\Bbb R \to \Bbb R$ such that $f$ is discontinuous at a point and satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$ in $\Bbb R$?
 A: Yes. The reason is that we have a lot of $\mathbb{Q}$-linear independence over $\mathbb{R}$, i.e. $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}$.
First note we must have $f(0) = 0$.  Then we can define $f(1)$ to be anything. Once we have $f(1)$ determined, it's easy to see we must have $f(k) = kf(1)$ for $k \in \mathbb{Z}$ and then $f(q) = qf(1)$ for $q \in \mathbb{Q}$. However, there are no other restrictions this puts on $f$! Note if $f$ were continuous everywhere, then we would know what $f$ is completely, since we know what it is on rationals.
Therefore, we can define $f(\sqrt{2})$ to be whatever we want, and then once again, we can deduce what $f(x)$ is for $x \in \mathbb{Q}[\sqrt{2}]$. And then we can define $f(\alpha)$ to be whatever we want, for some $\alpha \not \in \mathbb{Q}[\sqrt{2}]$. And we can keep doing this on and on, infinitely many times since $\mathbb{R}$ has infinite dimension as a vector space over $\mathbb{Q}$. 
It should be clear that the resulting $f$ need not be continuous. To see this directly, note that $\sqrt{2}$ is a limit of rational numbers so once $f$ is defined on $\mathbb{Q}$, if it were continuous, it would be defined at $\sqrt{2}$ already. But we chose $f(\sqrt{2})$ to be whatever we wanted.
The moral of the story is that continuity is more of a metric property of the reals whereas the equation $f(x+y) = f(x)+f(y)$ only uses the algebraic properties of the real line and allows freedom metrically. 
