[racket] HTDP 2nd edition exercise 4.2.3 - floating point issues

Hi BoGus, 

fwiw, this exercise is NOT in HTDP 2nd edition but HTDP 1st edition, 1-4th printing. 
Just in case you really want to work through HtDP 2e. 

-- Matthias






On Oct 22, 2013, at 8:53 AM, Matthias Felleisen <matthias at ccs.neu.edu> wrote:

> 
> ;; equation3 : number  ->  boolean
> ;; to determine whether n is a solution for 2n^2  =  102
> (define (equation3 n)
>  (=~ (* 2 n n) 102 .001))
> 
> 
> -- Matthias
> 
> 
> 
> On Oct 22, 2013, at 8:40 AM, Bo Gus <forumangus at gmail.com> wrote:
> 
>> equation 2 is 2n^2 = 102 so I implement like this:
>> 
>> ;; equation3 : number  ->  boolean
>> ;; to determine whether n is a solution for 2n^2  =  102
>> (define (equation3 n)
>>  (= (* 2 n n) 102))
>> 
>> And my answer is the same as per the online answer.  so great.
>> 
>> But how can I check a valid answer.
>> 
>> Eg if I do:
>> 
>> (equation3 (sqrt 51))
>>> false
>> 
>> same using - square root 51.  
>> 
>> How can I fix this?  Is the only way to do a range check?  Eg have some sort of tolerance - eg between 0.01 above and below answer?  Any other ideas?
>> 
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