Module representing rational numbers

Question

Taking inspiration from SICP Exercise 2.1, I decided to implement a module in Haskell for rational numbers.

Some of my concerns include:

• Is this idiomatic Haskell?
• Besides Num, Eq, Ord, Show, and Read, are there other typeclasses I should be implementing? I've got "free-floating" functions toInteger' and rationalDiv. Is there some typeclass that would accommodate them?
• Is this a good way to handle division by zero?
• How is the code layout and organization?

module Rational'
( toInteger'
, rationalDiv
) where

data Rational' = Rational' { numer :: Integer
                           , denom :: Integer
                           }

instance Num Rational' where
  (+)           = rationalAdd
  (-)           = rationalSub
  (*)           = rationalMul
  abs           = rationalAbs
  signum        = rationalSignum
  fromInteger   = rationalFromInteger

instance Eq Rational' where
  (==)          = rationalEq

instance Ord Rational' where
  compare       = rationalCompare

instance Show Rational' where
  show          = rationalShow

instance Read Rational' where
  readsPrec     = rationalRead



divByZero n = error ("Division by zero: " ++ (show n) ++ "/0")

rationalAdd :: Rational' -> Rational' -> Rational'
a `rationalAdd` b = reduce $ Rational' (an * bd + bn * ad) (ad * bd)
  where (Rational' an ad) = a
        (Rational' bn bd) = b

rationalSub :: Rational' -> Rational' -> Rational'
a `rationalSub` b = a + (Rational' (-bn) bd)
  where bn = numer b
        bd = denom b

rationalMul :: Rational' -> Rational' -> Rational'
a `rationalMul` b = reduce $ Rational' (an * bn) (ad * bd)
  where (Rational' an ad) = a
        (Rational' bn bd) = b

rationalDiv :: Rational' -> Rational' -> Rational'
a `rationalDiv` b = a * (Rational' bd bn)
  where bn = numer b
        bd = denom b

rationalAbs :: Rational' -> Rational'
rationalAbs (Rational' n d) = reduce $ Rational' (abs n) (abs d)

{- You might expect the result of signum to be an Int (namely -1 | 0 | 1), but
 - the Num typeclass requires it to be the same type as the input. -}
rationalSignum :: Rational' -> Rational'
rationalSignum (Rational' n d)
  | n == 0              = 0
  | (n > 0) == (d > 0)  = 1
  | otherwise           = Rational' (-1) 1

rationalFromInteger :: Integer -> Rational'
rationalFromInteger n = Rational' n 1

{- Truncates towards 0 -}
toInteger' :: Rational' -> Integer
toInteger' (Rational' n d) = n `div` d

rationalEq :: Rational' -> Rational' -> Bool
a `rationalEq` b = (a `rationalCompare` b) == EQ

rationalCompare :: Rational' -> Rational' -> Ordering
a `rationalCompare` b
  | (ad == 0)                   = divByZero a
  | (bd == 0)                   = divByZero b
  | (signum ad) == (signum bd)  = (an * bd) `compare` (bn * ad)
  | otherwise                   = (bn * ad) `compare` (an * bd)
  where (Rational' an ad) = a
        (Rational' bn bd) = b

rationalShow :: Rational' -> [Char]
rationalShow a = (show $ numer a) ++ "/" ++ (show $ denom a)

rationalRead :: Int -> [Char] -> [(Rational', [Char])]
rationalRead _ r = [(Rational' (read ns) (read ds), "")]
  where (ns, (_:ds)) = break (=='/') r

reduce :: Rational' -> Rational'
reduce a = Rational' (n `div` common) (d' `div` common)
  where (Rational' n d) = a
        d'
          | (d == 0)    = divByZero n
          | otherwise   = d
        common = (gcd n d') * (signum d')

Answer 1

Not too bad.

1. You do not export Rational' from the module so no outside code can make one directly
2. numer and denom will conflict with a similar function name in any other code which imports your module. Perhaps rational'Numer and rational'Denom?
3. See http://www.haskell.org/tutorial/numbers.html for a good intro to numbers in Haskell. You missed the Fractional class which defines div, recip, and fromRational.
4. Why are rationalSub and rationalDiv not defined like add and mult? Be consistent.
5. In rationalSignum your comment says the reader would expect a 0 or 1 Int but the return needs to be of the Rational' type and then you actually return 0 or 1. Be consistent and return Rational' 0 1 and Rational' 1 1
6. rationalFromInteger could be rationalFromInteger = (flip Rational') 1

7. I would define reduce like this

   reduce (Rational' n 0) = divByZero n
   reduce (Rational' n d) = Rational' (n `div` common) (d `div` common)
     where common = (gcd n d) * (signum d)