A function is additive if f(x+y) = f(x) + f(y). Intuitively, you might think that an additive function on R is necessarily linear, specifically of the form f(x) = kx. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of R as a vector space over Q, and then you define your function f to be different in at least two distinct elements of the basis. But my question is, does the same conclusion hold in the absence of the axiom of choice? Or does the negation of choice force the function to be linear? Any help would be greatly appreciated. Thank You in Advance.