Exercise 1.1

10 => 10 (+ 5 3 4) => 12 (- 9 1) => 8 (/ 6 2) => 3 (+ ( 2 4) (- 4 6)) => 6 (define a 3) => a (define b (+ a 1)) => b (+ a b ( a b)) => 19 (= a b) => false (if (and (> b a) ( 4 (cond ((= a 4) 6) ((= b 4) (+ 6 7 a)) => 16(+) (else 25)) (+ 2 (if (> b a) b a)) => 6 (* (cond ((> a b) a)
(( 16

Exercise 1.2:

(/ (+ 5 t (- 2 (- 3 (+ 6 (/ 1 5))))) 3 (- 6 2) (- 2 7))

Exercise 1.3:

efine (sum-of-squares a b) (+ (square a) (square b))) (define (sum-of-squares-of-larger-two a b c) (cond ((and ( a c)) (sum-of-squares a b)) ((and (= b c) (> b a)) (sum-of-squares b c)) ((and (= c a) (> c b)) (sum-of-squares c a)) ((and (= a b) ( b 0) + -) a b))

(a-plus-abs-b a b) expands to ((if (> b 0) + -) a b) for -ve b => (- a b) for +ve b => (+ a b)

Exercise 1.5:

??

Exercise 1.6

Infinite loop due to Applicative-order.

Exercise 1.7

For large numbers, The first iteration produces a very large number (will be slow?) For small numbers invariably good-enough? will return a false +ve due to the tolerance variable.