Problem
ALG-B3-M04-P002 An Arbitrary Image
#2
★★☆☆☆ Level 2 of 5
Find all surjective functions \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+f(y)\) for all \(x,y\).
Replace \(f(y)\) by an arbitrary number \(t\).
By surjectivity, for every \(t\in\mathbb R\) there is \(y\) such that \(t=f(y)\). Hence \(f(x+t)=f(x)+t\) for all \(x,t\). Taking \(x=0\), we get \(f(t)=f(0)+t\). Thus \(f(x)=x+c\), where \(c=f(0)\). Checking: \(f(x+f(y))=x+f(y)+c=f(x)+f(y)\). The answer is all \(f(x)=x+c\), \(c\in\mathbb R\).
Practises the main technique for using surjectivity.