The exercise is short. First, he defines a "fix" function: fix f = f (fix f) Which has the type (a -> a) -> a Then the function "remainder": remainder :: Integer -> Integer -> Integer remainder a b = if a < b then a else remainder (a - b) b The function fix, has far as i understand the matter, has no base case. So if someone wants to evaluate it, it will indefinitely apply the function f to itself, leading the hugs interpreter to its end. That's ok. But next he asks the reader to rewrite the function remainder using fix and i cannot find out how. I tried something like: remainder2 a b = fix (\x -> x - b) but there is no base case too. As soon as the lambda expression wants to evaluate its argument, the interpreter crashes. Can someone help me to solve the problem ? TIA Ludovic Kuty