## zmudz one year ago 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.

1. Zarkon

what have you tried

2. Zarkon

the proof is quite short

3. zmudz

I am having trouble even getting started. I don't know what I'm supposed to do. Any help is greatly appreciated.

4. Zarkon

choose x=0 then f(0)=f(0+r)=f(r) for all r irrational so f is constant on the irrational numbers now you need to extend this to the rational numbers consider x=a-r where a is a rational number