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Exercise 2.9

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



Now we can do the sum of the two intervals:



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:

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

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


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.