Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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3.33. The largest length for a subinterval as described in the exercise is n - 1 (consider [1 ,n ] itself) and the smallest is 0 (obtained by the intervals of the form [ i,i ] , 1 i n ). Fixing i N , 0 i n - 1, we wish to count the number of subintervals of length i . We can do this by the following process. We start the count at 1 by considering the interval [1 , 1 + i ]. At each stage, we will have a count of j and consider the interval [ j,j + i ]. There are two possibilities: 1. j + i < n , in which case we can increment the count by one (to j + 1) and consider the interval [ j + 1 , ( j + 1) + i ]. 2. j + i = n , in which case we have counted all the subintervals of length i and determined that there are j of them. (Intuitively we are just taking the interval [1 , 1 + i ] and repeatedly moving it to the right one unit until the right end point of the interval is n .) From this standpoint it is easy to see that the number of subintervals of length i is n - i . Thus the number
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HW7solns - 3.33 The largest length for a subinterval as described in the exercise is n 1(consider[1 n itself and the smallest is 0(obtained by the

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