1 (define (add-interval x y)
2   (make-interval (+ (lower-bound x) (lower-bound y))
3 		 (+ (upper-bound x) (upper-bound y))))
4 (define (mul-interval x y)
5   (let ((p1 (* (lower-bound x) (lower-bound y)))
6 	(p2 (* (lower-bound x) (upper-bound y)))
7 	(p3 (* (upper-bound x) (lower-bound y)))
8 	(p4 (* (upper-bound x) (upper-bound y))))
9     (make-interval (min p1 p2 p3 p4)
10 		   (max p1 p2 p3 p4))))
11 
12 (define (div-interval x y)
13   (mul-interval x
14 		(make-interval (/ 1.0 (upper-bound y))
15 			       (/ 1.0 (lower-bound y)))))
16 
17 (define (make-interval lower upper)
18   (cons lower upper))
19 (define (upper-bound interval)
20   (cdr interval))
21 (define (lower-bound interval)
22   (car interval))
23 
24 
25 ;; Exercise 2.8.  Using reasoning analogous to Alyssa's, describe how the difference of two intervals may be computed. Define a corresponding subtraction procedure, called sub-interval. 
26 
27 (define (sub-interval x y)
28   (make-interval (- (lower-bound x) (upper-bound y))
29 		 (- (upper-bound x) (lower-bound y))))
30 
31 ;; Exercise 2.9.  The width of an interval is half of the difference between its upper and lower bounds. The width is a measure of the uncertainty of the number specified by the interval. For some arithmetic operations the width of the result of combining two intervals is a function only of the widths of the argument intervals, whereas for others the width of the combination is not a function of the widths of the argument intervals. Show that the width of the sum (or difference) of two intervals is a function only of the widths of the intervals being added (or subtracted). Give examples to show that this is not true for multiplication or division.