Characterization of group homomorphisms from $R$ to $R$ I am trying to characterize all the group homomorphisms from $R$ to $R$. 
I have characterized all the "continuous" group homomorphisms from $R$ to $R$. They are of the form $f(x) = f(1) x$. Now I claim that all the group homomorphisms from $R$ to $R$ are necessarily continuous. Is this claim correct? I provide my justifications below. 
Let $f$ be any group homomorphism from $R$ to $R$. Since $f(x+y) = f(x) + f(y)$ would hold for all $x,y$ in $Q$, f(restricted to $Q$) is a group homomorphism from $Q$ to $R$.
However, all the group homomorphisms from $Q$ to $R$ are continuous as they are of the form $f(x) = f(1)x$. Not only that, they are uniformly continuous as $|f(x)-f(y)| = |f(1)(x-y)|< \epsilon$ if I choose $\delta$ as $\epsilon/|f(1)|$.
Since $f$ is uniformly continuous on a dense subspace of $R$, it can be extended to a unique continuous function in its closure. (Rud Exercise in Ch4. Continuity). Thus, every group homomorphism from $R$ to $R$ is continuous and hence are of the from $f(x) = f(1)x$.
 A: You're correct that every homomorphism $f:\mathbb Q\rightarrow \mathbb R$ is of the form $f(x)=ax$. Moreover, you're correct that every such $f$ can be uniquely extended to a continuous $f:\mathbb R\rightarrow \mathbb R$. Your error is that you assume that every homomorphism $g:\mathbb R\rightarrow\mathbb R$ is the (continuous) extension of one $\mathbb Q\rightarrow \mathbb R$ - and this is circular. It is entirely equivalent to having assumed that all homomorphisms are continuous. What you'd need to do would be to show that any $f$ has a unique extension as a homomorphism rather than as a continuous function.
However, this runs into difficulty. Note that functions like $f(a+b\sqrt{2})=a+b$ for rational $a,b$ defined on $\mathbb Q[\sqrt{2}]\rightarrow\mathbb R$ form a group homomorphism from a subfield of $\mathbb R$ to $\mathbb R$ itself which is not continuous (even though its restriction to $\mathbb Q$ is). The question of whether such functions can be extended (as homomorphisms) to all of $\mathbb R$ requires some more machinery.
In particular, note that $\mathbb R$ is a vector space over $\mathbb Q$. It follows** that it has some basis $S$ such that every $x\in \mathbb R$ is writable (uniquely) as a finite sum of elements of $S$ with rational coefficients. Then, letting $s\in S$ we can see the map $f(as+a_1s_1+a_2s_2+\ldots+a_ns_n)=a$ to be a discontinuous group homomorphism (where the $s_1$ through $s_n$ are elements of $S$ other than $s$ and the $a_i$ are rational coefficients).
(*If you've not seen it before, it suffices to know that $\mathbb Q[\sqrt{2}]$ is the field of elements of the form $a+b\sqrt{2}$)
(**We invoke the axiom of choice here. Without this, it is consistent that all group homomorphisms $\mathbb R\rightarrow\mathbb R$ are continuous)
