Problem
ALG-B3-M01-P013 What can be proved without regularity
#13
★★★★★ Level 5 of 5
Let \(f:\mathbb R\to\mathbb R\) be additive: \(f(x+y)=f(x)+f(y)\), and let \(f(1)=0\). Prove that \(f(q)=0\) for all \(q\in\mathbb Q\). Explain why this does not yet imply \(f\equiv0\) on \(\mathbb R\).
Hint. First prove \(f(n)=0\), then use \(n f(m/n)=f(m)\).
For integers \(n\), \(f(n)=n f(1)=0\). If \(q=m/n\), then \(n f(q)=f(m)=0\), so \(f(q)=0\). But on \(\mathbb R\), without regularity, additive functions exist that are not determined by their values on \(\mathbb Q\); therefore one cannot conclude \(f\equiv0\) without extra assumptions.
Goal: separate guessing the answer from a complete proof and domain check.