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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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This document was uploaded on 03/25/2011.
 Spring '10
 Math

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