  SICP exercise 2.09
;
  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.

  Let 's have two intervals :
;
; i₁ = (l₁, u₁)
  i₂ = ( l₂ , u₂ )
;
; With respective widths:
;
  w₁ = (   
; w₂ = (u₂ - l₂)/2
;
; When adding i₁ to i₂, we get:
;
  i₃ = i₂ + i₁ = ( l₁ + l₂ , u₁ + u₂ )
  w₃ = ( u₁ + u₂)/2 - ( l₁ + l₂)/2 = (     ( u₂ - l₂)/2 = ( w₁ + w₂)/2
;
; When subtracting, we get:
;
  i₃ = i₂ - i₁ = ( l₂ - u₁ , u₂ - l₁ )
  w₃ = ( u₂ - l₁)/2 - ( l₂ - u₁)/2 = ( u₁ + u₂)/2 - ( l₁ + l₂)/2 = ( w₁ + w₂)/2
;
  Now , let 's take a look at multiplication . Let 's take two pairs :
;
  ( 1 , 2 ) x ( 3 , 4 ) = ( 3 , 8)
  ( 3 , 4 ) x ( 5 , 6 ) = ( 15 , 24 )
;
; Both pairs have intervals with the same width, but the the resulting
; intervals have different widths.
;
; Let's take a look at division:
;
  ( 1 , 2 ) ÷ ( 3 , 4 ) = ( 1/4 , 2/3 )
  ( 3 , 4 ) ÷ ( 5 , 6 ) = ( 1/2 , 4/5 )
;
; Again, the intervals have different widths.


 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
 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
 intervals being added (or subtracted). Give examples to show that this is not
 true for multiplication or division.

 i₁ = (l₁, u₁)

 With respective widths:

 w₂ = (u₂ - l₂)/2

 When adding i₁ to i₂, we get:


 When subtracting, we get:




 Both pairs have intervals with the same width, but the the resulting
 intervals have different widths.

 Let's take a look at division:


 Again, the intervals have different widths.
  SICP exercise 2.09
  The width of an interval is half of the difference between its upper and
  result of combining two intervals is a function only of the widths of the
  sum ( or difference ) of two intervals is a function only of the widths of the

  Let 's have two intervals :
  i₂ = ( l₂ , u₂ )
  w₁ = (   
  i₃ = i₂ + i₁ = ( l₁ + l₂ , u₁ + u₂ )
  w₃ = ( u₁ + u₂)/2 - ( l₁ + l₂)/2 = (     ( u₂ - l₂)/2 = ( w₁ + w₂)/2
  i₃ = i₂ - i₁ = ( l₂ - u₁ , u₂ - l₁ )
  w₃ = ( u₂ - l₁)/2 - ( l₂ - u₁)/2 = ( u₁ + u₂)/2 - ( l₁ + l₂)/2 = ( w₁ + w₂)/2
  Now , let 's take a look at multiplication . Let 's take two pairs :
  ( 1 , 2 ) x ( 3 , 4 ) = ( 3 , 8)
  ( 3 , 4 ) x ( 5 , 6 ) = ( 15 , 24 )
  ( 1 , 2 ) ÷ ( 3 , 4 ) = ( 1/4 , 2/3 )
  ( 3 , 4 ) ÷ ( 5 , 6 ) = ( 1/2 , 4/5 )
