Hi - I am trying to do Exercise 9.9 in HSOE; and I've come up with an answer that works but I'm not sure if it answers the question properly. The problem is: The Question: ------------- Suppose we define a function "fix" as: fix f = f (fix f) Suppose further we have a recursive function: remainder :: Integer -> Integer -> Integer remainder a b = if a < b then a else remainder (a - b) b Rewrite this function using "fix" so that it's not recursive. My Answer: ---------- wierdFunc x y z = if y - z > z then ((\a b c -> a b c) x (y-z) z) else ((\a b c -> a b c) (\d e -> d) (y-z) z) myRemainder = fix wierdFunc My Question: ------------ Does the ((\a b c -> a b c) x (y-z) z) mean that I've not removed the recursion? I was assuming that I was returning a function that is to be evaluated and not actually doing any recursion. That's why I thought I answered the question. However, I have a headache now and would like another opinion. thanks, -andrew