Let $f:\mathbb R \to \mathbb R$ be a function such that for any irrational number $r$ and any real number $x$ we have $f(x)=f(x+r)$. Show that $f$ is a constant function.

It's easy to see any constant function satisfies the original property. But I don't see how to show this is the only solution.


2 Answers 2


If $y$ is irrational, the set $x = 0$, $r = y$ to get $f(y) = f(0)$.

If $y$ is rational, then set $x = y$, $r = \pi-y$ to get $f(y) = f(\pi)$

But by the first statement, $f(\pi) = f(0)$. Hence, $f$ is constant.


Wrong question.

The general solution of $f(x)=f(x+r)$ is $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic functions with period $r$ .

Adding the condition $f:\mathbb{R}\to\mathbb{R}$ the general solution will be modified to $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic $\mathbb{R}\to\mathbb{R}$ functions with period $r$ .

For example a particular solution $f(x)=\sin\dfrac{2\pi x}{r}$ , it is a $\mathbb{R}\to\mathbb{R}$ function, a periodic function with period $r$ and satisfies $f(x)=f(x+r)$ .

  • 4
    $\begingroup$ $r$ is an arbitrary irrational number; this function is periodic with every irrational period. $\endgroup$
    – Ian
    Aug 3, 2014 at 4:21

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