Problem
ALG-B1-M08-P021 Integer values on an interval
Let \(f:\mathbb R\to\mathbb R\) be additive and take integer values on the whole interval \([0,1]\). Prove that \(f(x)=0\) for all \(x\).
For \(t\in[0,1]\), consider \(f\left(\frac{t}{n}\right)\).
Let \(t\in[0,1]\). Then \(\frac{t}{n}\in[0,1]\), so \(f\left(\frac{t}{n}\right)\) is an integer. But \(n f\left(\frac{t}{n}\right)=f(t)\). If \(f(t)\neq0\), then for \(n>|f(t)|\), the integer \(f\left(\frac{t}{n}\right)=\frac{f(t)}{n}\) cannot be an integer. Hence \(f(t)=0\) on \([0,1]\).
For any real \(x\), choose a positive integer \(N>|x|\). Then \(\frac{x}{N}\in[-1,1]\). If \(x<0\), use \(f(-u)=-f(u)\), so \(f\left(\frac{x}{N}\right)=0\). Therefore \(f(x)=N f\left(\frac{x}{N}\right)=0\).
A nonstandard regularity condition instead of monotonicity.