Graph of Nonlinear Additive Function is Dense in the Plane

Theorem
Let $f: \R \to \R$ be an additive function which is not linear.

Then the graph of $f$ is dense in the plane.

Proof
From Additive Function is Linear for Rational Factors:
 * $f(q) = q f(1)$ for all $q\in\Q$.

Without loss of generality, let


 * $f(q) = q$ for all $q\in\Q$

and


 * $f(r) = r+\delta$ for some $r\in\R\setminus\Q$ and $\delta \neg 0$.

Suppose without loss of generality that $f(q) = q \ \forall q \in \mathbb{Q}$, and $f(\alpha) \neq \alpha$ for some $\alpha \in \mathbb{R}$.

Then put $f(\alpha) = \alpha + \delta, \delta \neq 0$.

We now show how to find a point in an arbitrary circle, centre $(x,y)$, radius $r$ where $x,y,r \in \mathbb{Q}, r > 0, x \neq y$.

Put $\beta = \frac{y - x}{\delta}$ and choose a rational number $b\neq 0$ close to $\beta$ with:


 * $\left| \beta - b  \right| < \frac{r}{2 \left|\delta\right|}$

Then choose a rational number $a$ close to $\alpha$ with:


 * $\left| \alpha - a  \right| < \frac{r}{2\left|b\right|} $

Now put:


 * $X = x + b (\alpha - a) \ $
 * $ Y = f(X) \ $

Then using the functional equation, we get:


 * $ Y = f(x + b (\alpha - a)) \ $
 * $ = x + b f(\alpha) - b f(a) \ $
 * $ = y - \delta \beta + b f(\alpha) - b f(a) \ $
 * $ = y - \delta \beta + b (\alpha + \delta) - b a \ $
 * $ = y + b (\alpha - a) - \delta (\beta - b) \ $

Because of our choices above, the point $(X, Y)$ is inside the circle.