Function with period (!)

[Tolaso J Kos]

Give an example of a function \( f:\mathbb{R} \rightarrow \mathbb{R} \) such that any rational number is its period but any irrational is not. Also, prove that there exists no function \( g:\mathbb{R} \rightarrow \mathbb{R} \) such that any irrational is its period and any rational is not.

[Papapetros Vaggelis]

We define \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}\) by

$$f(x)=\begin{cases}
1\,\,,x\in\mathbb{Q}\\
0\,\,,x\in\mathbb{R}-\mathbb{Q}
\end{cases}$$

which is better known as $\rm Dirichlet$ function. Here is a graph:

If \(\displaystyle{T\in\mathbb{Q}}\), then : \(\displaystyle{f(x+T)=f(x)\,\,,\forall\,x\in\mathbb{R}}\) since :

$$x\in\mathbb{Q}\implies x+T\in\mathbb{Q}\implies f(x+T)=1=f(x)$$ and
$$x\in\mathbb{R}-\mathbb{Q}\implies x+T\in\mathbb{R}-\mathbb{Q}\implies f(x+T)=0=f(x)$$

Now, if \(\displaystyle{K\in\mathbb{R}-\mathbb{Q}}\), then :\(\displaystyle{f(1+K)=0\neq 1=f(1)}\) .

For the second part, suppose that there exists a function \(\displaystyle{g:\mathbb{R}\longrightarrow \mathbb{R}}\) such that any irrational is its period and any rational is not. Since \(\displaystyle{\pi}\) and \(\displaystyle{1-\pi}\) are irrational numbers, we get:

$$\forall\,x\in\mathbb{R}: g(x+1)=g((x+1-\pi)+\pi)=g(x+1-\pi)=g(x)$$

which means that the rational number \(\displaystyle{q=1}\) is a period of \(\displaystyle{g}\), a contradiction.