# Exercise 2.9

We takes two intervals represented by a double similarly to the representation in our program:

\(a=(x_1,y_1)(a)=\)

\(b=(x_2,y_2)(b)=\)

Now we can do the sum of the two intervals:

\(a+b=(x_1+x_2,y_1+y_2)\)

\((a+b)==+=(a)+(b)\)

We can see that the width function is a linear map so the same thing works for the difference of two intervals.

It’s not true for the multiplication as for example:

\(c = (0,1)(c)=\)

\(d = (-1,1)(d)=1\)

\((c)(d)=\)

But we have \(cd=(-1,1)\) using the `mul-interval`

function and \((cd)=1\)

Using the same \(c\) and \(d\) and the `div-interval`

function we show as example:

\((c)/(b)=1 = (c/d)\)

Therefore the width of the division of two intervals is not equal to the division of the width of the two intervals.