Problem
ALG-B3-M04-P001 Cancelling the Outside Function
#1
★★☆☆☆ Level 2 of 5
Let \(f:\mathbb R\to\mathbb R\) be injective and suppose that \(f(f(x)+y)=f(f(y)+x)\) for all \(x,y\). Prove that there is a constant \(c\) such that \(f(x)=x+c\) for all \(x\).
First cancel the outside \(f\) legally.
Since \(f\) is injective, the equality gives \(f(x)+y=f(y)+x\). Hence \(f(x)-x=f(y)-y\) for all \(x,y\). Fixing \(y=0\), we get \(f(x)-x=f(0)\), so \(f(x)=x+f(0)\). Take \(c=f(0)\).
Basic injectivity task: the student should see that the outside function, not an argument, is being cancelled.