[Attempting to resend]

Hi - I think your logic is:

I can define these two:

fFrac :: Fractional a => a -> a

fFrac n = n * (n+1) / 2

fInt :: Integral a => a -> a

fInt n = n * (n+1) `div` 2

so that, e.g.

fFrac (5.0 :: Float) == 15.0 :: Float

fInt (5 :: Integer) == 15 :: Integer

And all number types are either Integral or Fractional, so surely I should be able to define a single function of type:

f :: Num a => a -> a

This would seem reasonable, but I think there’s a problem with the last assumption. It is indeed possible for other types to be instances of Num, but not of Integral or Fractional.

For example, I could define:

instance Num Bool where

  fromInteger 0 = False

  fromInteger _ = True

  (+) = (&&)

  (*) = (||)

  abs = id

  signum _ = True

  negate = not

Now this would probably be pretty dumb (and probably doesn’t comply with expectations), but is possible. (And also pretty dumb to define it without also define an instance Integral Bool where ..., but still possible).

So I don’t think

f :: Num a => a -> a

could be possible, since Num by itself (& the dumb Bool instance) has no way to do the division. (At least that I can think of, but would be very interested to hear if there is).

Regards, David.

From: Beginners <beginners-bounces@haskell.org> on behalf of M Douglas McIlroy <m.douglas.mcilroy@dartmouth.edu>
Sent: Friday, December 11, 2020 1:03:03 PM
To: beginners@haskell.org <beginners@haskell.org>
Subject: [Haskell-beginners] unwanted Fractional constraint

For rational functions that take on integer values at integer
arguments, for example n*(n+1)/2, is there a way to doctor the
corresponding Haskell definition

    f n = n*(n+1)/2

so that the type signature becomes

    f :: Num a => a -> a

rather than

    f :: Fractional a => a -> a

Doug McIlroy
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