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.

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For intervals x and y, the sum of x + y is bounded by the sum of the lower 
bounds and the sum of the upper bounds, which is the sum of the widths of x 
and y.

width(x) = (upper(x) - lower(x)) / 2
width(y) = (upper(y) - lower(y)) / 2

width(x+y) = (upper(x) + upper(y) - (lower(x) + lower(y))) / 2
           = width(x) + width(y)

Since y and -y have the same width, this applies equally to subtraction.

In the case of multiplication, width(x * y), assuming x and y are positive,
is equal to upper(x) * upper(y) - lower(x) * lower(y).  This width depends 
on the magnitude of x and y, not only their widths.

If we multiply two pairs of intervals all having width 1, we can get 
different results:

width((1,2) * (3,4)) = width(3,8) = 5/2
width((1,2) * (4,5)) = width(4,10) = 3

Likewise for division:

width((2,3) / (1,2)) = width(1,3) = 1
width((3,4) / (2,3)) = width(1,2) = 1/2