The Question:
Regard $\mathbb{R}$ as a vector space over $\mathbb{Q}$, where vector addition and scalar multiplication is just addition and multiplication as they would usually be done in the real numbers.
Assuming Zorn's lemma, and hence that $\mathbb{R}$ has a basis, show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is such that $f(x+y)=f(x)+f(y)$, then $f$ is a linear transformation.
From this, show that there is a function $g: \mathbb{R} \rightarrow\mathbb{R}$ where $g(x+y)=g(x)+g(y)$, and where $g$ is not of the form $g(x)=kx$, for any real number $k$.
What I have done so far:
So we need to show that for any rational $q$, that $f(qx)=qf(x)$. Since $\mathbb{R}$ has a basis, $f(qx)=f(q(p_1x_1+...+p_rx_r))=f(qp_1x_1)+...+f(qp_rx_r)$, where the $x_i's$ are basis vectors.
So I tried to show that $f(qy)=qf(y)$ for any basis vector $y$ by defining a set $P$ of linearly independent subsets of $\mathbb{R}$ such that every vector $x$ in the subsets has the property $f(qx)=qf(x)$ for all $q$, and then using Zorn's lemma to get the maximal subset which will be a basis with the basis vectors having the desired property.
But I'm thinking $P$ may be empty because $\{0\}$ is not linearly independent.
You don't need the fact that $\Bbb R$ has a basis over $\Bbb Q$ to deal with the first part. All you need is $$f\left(\frac pqx\right)=f\left(\underbrace{\frac1qx\ +\ldots +\frac1qx}_{p\text{ times}}\right)=p\cdot f\left(\frac 1qx\right)$$ where the last equality follows from the hypothesis and by induction, and then by the same principle $$f\left(\frac qqx\right)=q\cdot f\left(\frac 1qx\right)\Longrightarrow f \left(\frac 1qx\right)=\frac1q\cdot f\left(x\right)$$
which altogether imply $$f\left(\frac pqx\right)=\frac pq\cdot f(x)$$