Exercise 2.13.  Show that under the assumption of small percentage tolerances 
there is a simple formula for the approximate percentage tolerance of the 
product of two intervals in terms of the tolerances of the factors. You may 
simplify the problem by assuming that all numbers are positive.

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Let x and y be the center points of two intervals, and tx and ty be their 
tolerances.

Then the width (defined as in Exer. 2.9 as half the difference between 
the upper and lower bounds) of the result is:

((x + xtx) * (y + yty) - (x - xtx) * (y - yty)) / 2

= (xy + x yty + xtx y + xtx yty - xy + x yty + xtx y - xtx yty) / 2

= (2 x yty + 2 xtx y) / 2

= x yty + xtx y

And the center point of the result is:

( (x + xtx)(y + yty) + (x - xtx)(y - yty) )/2

= ( xy + x yty + xtx y + xtx yty + xy - x yty - xtx y + xtx yty ) / 2

= ( 2xy + 2 xtx yty ) / 2

= xy + xtx yty

The tolerance of the result is the width over the center point:

x yty + xtx y
—————————————
xy + xtx yty

Which we can simplify to:

 tx + ty
—————————
1 + tx ty

Assuming small tolerances, tx*ty will be negligible, so we can 
approximate the error tolerance as tx + ty.