# 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 real number plane.

## Proof

$f(q) = q f(1)$ for all $q\in\Q$.

Without loss of generality, let

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

Since $f$ is not linear, let $\alpha\in\R\setminus\Q$ be such that

$f(\alpha) = \alpha+\delta$ for some $\delta \neq 0$.

Consider an arbitrary nonempty circle in the plane.

Let its centre be

$(x,y)$ where $x\neq y$ and $x,y\in\Q$

and its radius be $r>0$.

We will show how to find a point of the graph of $f$ inside this circle.

As $x\neq y$ and $r$ can be arbitrarily small, this will prove the theorem.

Since $\delta\neq0$, let

$\beta = \frac{y - x}{\delta}$

Since $x\neq y$,

$\beta\neq0$.

As Rationals are Everywhere Dense in Topological Space of Reals, there exists a rational number $b\neq 0$ such that:

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

As Rationals are Everywhere Dense in Topological Space of Reals, there also exists a rational number $a$ such that:

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

Now put:

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

Then:

$|X-x| = |b (\alpha - a)| < \frac{r}{2}$

so $X$ is in the circle.

Then:

 $\displaystyle Y$ $=$ $\displaystyle f(x + b (\alpha - a))$ Definition of $Y$ and $X$ $\displaystyle$ $=$ $\displaystyle f(x) + f(b \alpha) - f( b a)$ Cauchy functional equation $\displaystyle$ $=$ $\displaystyle x + b f(\alpha) - b f(a)$ Additive Function is Linear for Rational Factors $\displaystyle$ $=$ $\displaystyle y - \delta \beta + b f(\alpha) - b f(a)$ Definition of $y$ $\displaystyle$ $=$ $\displaystyle y - \delta \beta + b (\alpha + \delta) - b a$ Additive Function is Linear for Rational Factors $\displaystyle$ $=$ $\displaystyle y + b (\alpha - a) - \delta (\beta - b)$

Therefore

$|Y-y| = |b (\alpha - a) - \delta (\beta - b)| \le |b (\alpha - a)| + |\delta (\beta - b)| \le r$

so $Y$ is in the circle as well.

Hence the point $(X, Y)$ is inside the circle.

$\blacksquare$