Problem
ALG-B1-M08-P007 Additivity on rationals
#7
★★☆☆☆ Level 2 of 5
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(1)=3\). Prove that \(f(q)=3q\) for all \(q\in\mathbb Q\).
For \(q=\frac{m}{n}\), use \(nq=m\).
For an integer \(m\), \(f(m)=3m\). Let \(q=\frac{m}{n}\), \(n>0\). Then \(n f(q)=f(nq)=f(m)=3m\), so \(f(q)=\frac{3m}{n}=3q\).
Clearly separates \(\mathbb Q\) from \(\mathbb R\), where additional conditions are needed.