Additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.

A function $T:V \rightarrow W$ is additive if $T(x+y) = T(x) + T(y)$ for every $x, y \in V$. Prove that there exists an additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.

My attempt: Let $T$ be the function $T: \mathbb{R}$ (over the field $\mathbb{Q}$) $\rightarrow \mathbb{R}$ (over the field $\mathbb{R}$). The set $\{1, \sqrt{2}\} \subseteq \mathbb{R}$ is linearly independent for the vector space $\mathbb{R}$ over $\mathbb{Q}$. Then, there must exist a linearly independent set $W \subseteq \mathbb{R}$ (over $\mathbb{Q}$) such that $\{1, \sqrt{2}\} \subseteq W \subseteq \text{span}(W) = \mathbb{R}$ (over $\mathbb{Q}$).

I have been told that the function defined as $T(1) = 1$ and $T(w) = 0$ for all $w \in \text{span}(W) \setminus \{1\}$ is additive but not linear, but I cannot see why this is? I can see why it is not linear, clearly $T(\sqrt{2} \cdot 1) = 0$ but $\sqrt{2} T(1) = \sqrt{2}$. But, why is $T$ additive? For example, $T(1+1) = T(2) =0$ but $T(1) + T(1) = 1+1 = 2$? Is there a mistake somewhere?

You can indeed construct an additive, non-linear function $T:\Bbb R\to \Bbb R$ (as vetor spaces over $\Bbb R$, since additive maps of $\Bbb Q$-vector spaces necessarily become linear automatically) by starting with what you have: $T(p) = p$ for $p\in \Bbb Q$, and $T(r) = 0$ for all $r$ with $r/\sqrt2 \in \Bbb Q$. However, constructing it is not easy, and requires the so-called axiom of choice in order to be finalised.

So, we start out with what we already have: some values that $T$ sends to themself, and some values that $T$ sends to $0$. By additivity, this forces $T$ to be defined for a slightly greater class of numbers, namely $T(p + q\sqrt2) = p$, for all rational $p, q$.

But we haven't defined $T$ for all real numbers yet. Note that whatever else we define $T$ to be, it already cannot be linear. So we just build on it as best we can and ensure that it stays additive. We do this by picking a number for which $T$ isn't already defined, like $\pi$ or $\sqrt 3$ or $\ln 2$ or $e$ or anything else. I'll pick $\pi$ for now. We can make $T(\pi)$ to be whatever we want it to be. I'll pick $3$ because I like $3$. By additivity, $T$ is now forcibly defined to be $T(p+q\sqrt2+r\pi) = p+3r$ for any rational $r$.

However, we are not done definining $T$ yet; there are still many more real numbers to cover. And this is where the axiom of choice comes in: there is no finite way, or even countably infinite way to finalise this definition of $T$. We just have to keep going and going, picking one new real number each step, until we've covered the entire real number line, which requires an uncountably infinite number of steps. The axiom of choice is exactly what allow us to say that even though we cannot finalise our $T$, some finalised $T$ does indeed exist somewhere out there.

You need some key elements here.

First of all, note that if we consider $\mathbb R$ as a vector space over $\mathbb Q$, then it is infinite dimensional. This means that a "Hamel" basis $E$ for $\mathbb R$ over $\mathbb Q$ is at least countably infinite, and exists by the axiom of choice. Note : Without the axiom of choice, this is a rather interesting question, one that I don't know much about.

Choose two elements $e_1,e_2$ from this Hamel basis. Define the "linear transformation" $f$ from $\mathbb R \to \mathbb R$ as vector spaces over $\mathbb Q$, by defining $f(e_1) = 1$ and $f(e) = 0$ for all $e \neq e_1 \in E$. Note that because we are extending $f$ from it's basis construction, it obviously is additive, but then it isn't of the said form, since $f(e_2) = 0 \neq \frac{e_2}{e_1}f(e_1)$.

Hence $f$ is additive, but not linear.

(Read discussion below for field-related query)

Such a function cannot be defined explicitly. It only can be shown (assuming the axiom of choice), that such a function exists. First note that each such function $T:\mathbb R\to\mathbb R$ is a solution to the so called Cauchy functional equation. It can be shown that $T$ is automatically linear, if $T$ is Lebesgue-measurable or continuous or bounded. In particular, non-linear examples have to be really pathological.

For a more detailed discussion and a proof of the existence of a nonlinear example see

https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation