I may be missing something here, but this is what I intended. An expression of the form 1 a1 + ------ 1 a2 + ------ 1 a3 + -- a4 + ... Where the ai's are positive integers is called a continued fraction. Function cf should take [1,2,6,5] to produce 1 1 + ----- 1 2 + ----- 1 6 + -- 5 Michael --- On Sat, 3/28/09, Brandon S. Allbery KF8NH <allbery@ece.cmu.edu> wrote: From: Brandon S. Allbery KF8NH <allbery@ece.cmu.edu> Subject: Re: [Haskell-cafe] type Rational and the % operator To: "michael rice" <nowgate@yahoo.com> Cc: "Brandon S. Allbery KF8NH" <allbery@ece.cmu.edu>, "Duane Johnson" <duane.johnson@gmail.com>, haskell-cafe@haskell.org Date: Saturday, March 28, 2009, 10:39 PM On 2009 Mar 28, at 22:36, michael rice wrote:import Data.Ratio cf :: [Integer] -> Rational cf (x:xs) = (toRational x) + (1 % (cf xs)) cf (x:[]) = toRational x cf [] = toRational 0 Data.Ratio> :load cf.hs ERROR "cf.hs":3 - Type error in application *** Expression : toRational x + 1 % cf xs *** Term : toRational x *** Type : Ratio Integer *** Does not match : Ratio (Ratio Integer) Your function cf produces a Rational (Ratio Int); you're using it in the denominator of another Ratio, which makes that Ratio's type Ratio (Ratio Int). This is almost certainly not what you intended, but I couldn't say what you actually want. -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allbery@kf8nh.comsystem administrator [openafs,heimdal,too many hats] allbery@ece.cmu.eduelectrical and computer engineering, carnegie mellon university KF8NH