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 <matth...@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 <foruman...@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? >> >> ____________________ >> Racket Users list: >> http://lists.racket-lang.org/users > > > ____________________ > Racket Users list: > http://lists.racket-lang.org/users ____________________ Racket Users list: http://lists.racket-lang.org/users
