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

Am Dienstag 29 September 2009 09:02:19 schrieb 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.

Quite. *MMlouds> givenSum 11 [True,False,False,False,True,False,False,False,False,False,False] The code tests whether a*a+2*(num+1) is a square, equivalently, whether there's a b such that 2*(num+1) == b^2 - a^2 Now, the difference of two squares is either odd or a multiple of 4. 2*(num+1) is never odd, so it must be a multiple of 4, i.e. num must be odd for any number of the form a^2 + 2*(num+1) to be a square. isIntegral x = snd (properFraction x) == 0 works for rational x and for small enough doubles/floats, but beware: *MMlouds> [n | n <- [1 .. 100], not $ isIntegral (sqrt $ 4^n+222)] [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29]

Thomas

On Mon, Sep 28, 2009 at 11:54 PM, Magicloud Magiclouds

wrote:

...

*Main> givenSum 10 [False,False,False,False,False,False,False,False,False,False]