[Haskell-cafe] How to decide if a number is an integer?

[Thomas DuBuisson]

Unless I missed something, the function in question is:

sqrt (a * a + 2 + 2 * num) - fromIntegral a
where num = 10

1 -> sqrt (1 * 1 + 2 + 2 * 10) - 1 -> sqrt (1 + 2 + 20) - 1 -> sqrt
(23) - 1 -> 3.79xxxxx

the fractional will only ever come from the sqrt function.  Do any of
the following actually look like square values to you?

26
31
38
47
58
71
86
103
122

IMO, the code works and your expectations are a bit off.

Thomas

On Mon, Sep 28, 2009 at 11:54 PM, Magicloud Magiclouds
<magicloud.magiclouds at gmail.com> wrote:
> The original code is
> givenSum num = map (\a ->
>                      let l = (sqrt $ fromIntegral (a * a + 2 + 2 *
> num)) - (fromIntegral a) in
>                      case properFraction l of
>                        (_, 0) ->
>                          True
>                        _ ->
>                          False
>                   ) $ take num [1..]
> :t l is (Floating a) => a
> Well, in ghci
> *Main> givenSum 10
> [False,False,False,False,False,False,False,False,False,False]
>
> On Tue, Sep 29, 2009 at 2:45 PM, Thomas DuBuisson
> <thomas.dubuisson at gmail.com> wrote:
>> On Mon, Sep 28, 2009 at 11:35 PM, Magicloud Magiclouds
>> <magicloud.magiclouds at gmail.com> wrote:
>>> It never matches to (_, 0.0)....
>>> I mean
>>> case properFraction l of
>>>  (_, 0) -> l
>>>  _ -> 0 -- always goes here.
>>
>> Odd, it works fine for me.
>>
>> f x =
>>        case properFraction x of
>>                (_,0) -> True
>>                _     -> False
>>
>>
>> *Main> f 5
>> True
>> *Main> f 5.5
>> False
>> *Main> f 4.0
>> True
>> *Main> f 4.00000001
>> False
>>
>>
>> Thomas
>>
>
>
>
> --
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> 山高哪阻野云飞
>