Problem
ALG-B1-M08-P010 Zero of an injective additive function
#10
★★☆☆☆ Level 2 of 5
Let \(f:\mathbb R\to\mathbb R\) be additive, meaning \(f(x+y)=f(x)+f(y)\), and injective. Prove that if \(f(a)=0\), then \(a=0\).
First find \(f(0)\).
By additivity, \(f(0)=0\). If \(f(a)=0\), then \(f(a)=f(0)\). By injectivity, \(a=0\).
The problem reinforces how injectivity turns equality of values into equality of arguments.