A solution to Exercise 2.9:
Some preliminary definitions:
(upper-bound (make-interval a b)) = a [1]
(lower-bound (make-interval a b)) = b [2]
(width i) = (/ (- (upper-bound i) (lower-bound i)) 2) [3]
(add i1 i2) = (make-interval (+ (upper-bound i1) (upper-bound i2))
(+ (lower-bound i1) (lower-bound i2))) [4]
Rearranging [3]:
(* 2 (width i)) = (- (upper-bound i) (lower-bound i))
(+ (* 2 (width i)) (lower-bound i)) = (upper-bound i) [5]
From [3]:
(width (add i1 i2)) = (/ (- (upper-bound (add i1 i2))
(lower-bound (add i1 i2)))
2)
Simplifying with [1], [2] and [4]:
(width (add i1 i2)) = (/ (- (+ (upper-bound i1)
(upper-bound i2))
(+ (lower-bound i1)
(lower-bound i2)))
2)
Substituting in [5]:
= (/ (- (+ (+ (* 2 (width i1)) (lower-bound i1))
(+ (* 2 (width i2)) (lower-bound i2)))
(+ (lower-bound i1)
(lower-bound i2)))
2)
= (/ (- (+ (* 2 (width i1)) (* 2 (width i2))
(lower-bound i1)
(lower-bound i2))
(+ (lower-bound i1)
(lower-bound i2)))
2)
All the (lower-bound x)s cancel out, leaving:
= (/ (+ (* 2 (width i1))
(* 2 (width i2)))
2)
So do the 2s:
(width (add i1 i2)) = (+ (width i1) (width i2))
Clearly the width of the sum is a function only of the width of the operands.
Ken Dyck 