Types in a function in Haskell

Question

I am trying to write a function called check that when applied to n returns a Bool indicating whether the ∑r=0 to n (n choose r) = 2^n

I have made the factorial and choose functions which work but I am having difficulty understanding what the type would be for the check function. I have written what I have done for it and I think logically it might work but I am not sure. The error I get when running says 'couldn't match type a -> Bool to actual type Bool' but I am not entirely sure how to make it an actual Bool type.

factorial :: (Integral a) => a -> a
factorial 0 = 1 
factorial n = n * factorial (n - 1)

choose :: (Integral a) => a -> a -> a
n `choose` r
    | r < 0     = 0
    | r > n     = 0
    | otherwise = factorial n `div` (factorial r * factorial (n-r))

check :: (Integral a) => a -> a -> Bool 
check n = (if (factorial (choose n)) == 2^n then True else False)

Answer 1

The correct type declaration for check is a -> Bool

The types are separated by -> and grouped using parentheses.

The last type is the returned type, here Bool.

The previous types are the input types, here a.

Your error message is couldn't match type a -> Bool to actual type Bool. What's happening here is that Haskell matches the a to variable n, but is then left with a -> Bool, when actually the return type is Bool.