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.
Solution
The classic example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that any rational number is its period but any irrational is not is:
$$f(x)= \left\{\begin{matrix}
1 &, & x \in \mathbb{Q}\\
0& , & x \in \mathbb{R} \setminus \mathbb{Q}
\end{matrix}\right.$$
better known as Dirichlet function.
On the contrary there does not exist a function that any irrational number is its period while any rational is not. Suppose on the contrary that such function exists. Then since $\pi$ and $1-\pi$ are periods of $f$ we have that:
$$f(x+1)=f( (x+1-\pi)+\pi) = f(x + (1-\pi))=f(x)$$
a contradiction.
Solution
The classic example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that any rational number is its period but any irrational is not is:
$$f(x)= \left\{\begin{matrix}
1 &, & x \in \mathbb{Q}\\
0& , & x \in \mathbb{R} \setminus \mathbb{Q}
\end{matrix}\right.$$
better known as Dirichlet function.
$$f(x+1)=f( (x+1-\pi)+\pi) = f(x + (1-\pi))=f(x)$$
a contradiction.
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