Exercise 2.13 of SICP

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.

$(a-\epsilon_1,a+\epsilon_1) \cdot (b-\epsilon_2,b+\epsilon_2)$
$(a+\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2$
$(a+\epsilon_1)\cdot(b-\epsilon_2)=ab-a\epsilon_2+b\epsilon_1-\epsilon_1\epsilon_2$
$(a-\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2-b\epsilon_1-\epsilon_1\epsilon_2$
$(a-\epsilon_1)\cdot(b-\epsilon_2)=ab-a\epsilon_2-b\epsilon_1+\epsilon_1\epsilon_2$

Since a>0 and b>0:
$(a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+ a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2$
$(a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab -a\epsilon_2-b\epsilon_1+\epsilon_1\epsilon_2$
Assuming $\epsilon_1$ and $\epsilon_2$ are small, the quantity $\epsilon_1\epsilon_2$ can be discarded.
$(a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+a\epsilon_2+b\epsilon_1$
$(a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab-a\epsilon_2-b\epsilon_1$

Since we are interested in the tolerance terms:
$(a+\epsilon_1)\cdot(b+\epsilon_2) \approx a\epsilon_2+b\epsilon_1$
$(a-\epsilon_1)\cdot(b-\epsilon_2) \approx -a\epsilon_2-b\epsilon_1$

The new product interval in terms of small tolerance terms looks like this:
$[ab-(a\epsilon_2+b\epsilon_1),ab+(a\epsilon_2+b\epsilon_1)]$