Hi, I have a new 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.


Again, please provide a full solution so i may understand how to do it. 


Thanks for the help! Much appreciated.  

FencingKat  Nov 15, 2018

\(\text{The problem is to show that }f(x) = f(x+q),~q \in \mathbb{Q}\\ r \in \mathbb{R}-\mathbb{Q} \Rightarrow q-r \in \mathbb{R}-\mathbb{Q}\\ \text{i.e. }q-r \text{ is irrational}\\ f(x) = f(x+q-r)\\ x+q-r \in \mathbb{R} \\ f(x) = f(x+q-r) = f(x+q-r + r) = f(x+q)\)


\(\text{so }f(x) = f(x+q)=f(x+r),~q \in \mathbb{Q},~r \in \mathbb{R}-\mathbb{Q} \\ \mathbb{Q} \cup \left( \mathbb{R}-\mathbb{Q} \right) = \mathbb{R}, \text{ thus}\\ f(x) = f(x + y),~\forall y \in \mathbb{R},~\text{ i.e. }f \text{ is constant}\)

Rom  Nov 15, 2018

Thank you so much!

FencingKat  Nov 15, 2018

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