How to do modulo in scheme

Question

How would I do the following in sicp/scheme/dr. racket?

(define (even? n) (= (% n 2) 0))

Currently it seems like that's not a primitive symbol: %: unbound identifier in: %.

This may be the stupidest way in the world to do it, but without a % or bitwise-&1 I am doing (without logs or anything else):

(define (even? n)
  (if (< (abs n) 2)
       (= n 0)
       (even? (- n 2))))

Answer 1

I think it's a good practice to get comfortable writing your own procedures when it feels like they are "missing". You could implement your own mod as -

(define (mod a b)
  (if (< a b)
      a
      (mod (- a b) b)))

(mod 0 3) ; 0
(mod 1 3) ; 1
(mod 2 3) ; 2
(mod 3 3) ; 0
(mod 4 3) ; 1
(mod 5 3) ; 2
(mod 6 3) ; 0
(mod 7 3) ; 1
(mod 8 3) ; 2

But maybe we make it more robust by supporting negative numbers and preventing caller from divi

(define (mod a b)
  (if (= b 0)
      (error 'mod "division by zero")
      (rem (+ b (rem a b)) b)))

(define (rem a b)
  (cond ((= b 0)
         (error 'rem "division by zero"))
        ((< b 0)
         (rem a (neg b)))
        ((< a 0)
         (neg (rem (neg a) b)))
        ((< a b)
         a)
        (else
         (rem (- a b) b))))