Problem
ALG-B1-M08-P011 Injectivity from the equation
#11
★★★☆☆ Level 3 of 5
A function \(f:\mathbb R\to\mathbb R\) satisfies \(f(x+f(y))=x+y\) for all \(x,y\). Prove that \(f\) is injective.
Assume \(f(a)=f(b)\) and substitute \(y=a\), \(y=b\) with the same \(x\).
Let \(f(a)=f(b)\). For any \(x\), we have \(f(x+f(a))=x+a\) and \(f(x+f(b))=x+b\). The left sides are equal because \(f(a)=f(b)\). Hence \(x+a=x+b\), so \(a=b\). The function is injective.
Here injectivity is not given; it is extracted from the structure of the equation.