# Answer ## Complete family of solutions All solutions are exactly the additive involutions: - f additive: f(x+y)=f)+f(y) for all x. - f is an involution: f(f(x))=x for all x. Equivalently (and more concretely): - Choose any decomposition of ℝ as a direct sum of two ℚ-linear subspaces ℝ = U ⊕ V. - Define f(u+v) = u − v for u ∈ U, v ∈ V. Every such f satisfies f(x+f(y)) = f(x)+y, and every solution arises this way. ## Common “regular” cases - f(x) = x (take U=ℝ, V={}). - f(x) = −x (take U={}, V=ℝ). If you additionally impose continuity, measurability, or monotonicity, these two are the only solutions.