Constant Function Question

Question

Problem:

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.

I have seen the other posts where the answers say that $f(a)=f(0+a)=f(0)$ given $a$ is irrational, but I don't understand how this is derived. Please help on this problem!

We are told that it is true for all $x$ and for all irrational $r$, so we picked particular ones to apply it to. — Ross Millikan, May 28 '17 at 17:11

score 5 · Accepted Answer · answered May 28 '17 at 16:57

First of all, if $a$ is irrational, then $$ f(a) = f(0 + a) = f(0) $$ (let $x=0$ and $r=a$).

If $a$ is rational, then $$ f(a) = f((a + \pi) - \pi) = f(a + \pi) = f(0) $$ (here $x = a + \pi$ and $r = \pi$ and using that $a + \pi$ is irrational).

Hence $f(a) = f(0)$ for all $a \in \mathbb R$.

Constant Function Question

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