1991 IMO Problems/Problem 6
Problem
An infinite sequence of real numbers is said to be bounded if there is a constant such that for every .
Given any real number , construct a bounded infinite sequence such that for every pair of distinct nonnegative integers .
Solution 1
Since , the series is convergent; let be the sum of this convergent series. Let be the interval (or any bounded subset of measure ).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose such that holds for all distinct . The only new cases are when one number (WLOG ) is equal to , so we must guarantee that for all .
Let be the interval , of length . The points that are valid choices for are precisely the points of , so we must show that this set is nonempty. The total length is at most the sum of the lengths . This is .
Therefore the total measure of is , so has positive measure and thus is nonempty. Choosing any and continuing by induction constructs the desired sequence.
Solution 2
The argument above would not work for , since only converges for . But Osmun Nal argues in this video that satisfies the stronger inequality for all distinct ; in other words, this sequence simultaneously solves the problem for all simultaneously.
Solution 3
Let , then we have
Thus
For any non-negative integer , represent it as base- positional notation:
, where
We directly construct
We allow leading zeros for representing due to the leading zeros do not effect the value of
Then
Thus , which is bounded
For any two distinct non-negative integers , represent them as base- positional notation:
where
Let be the minimum number such that , that is,
Then we have
Without loss of generality, assume that , then we have
Thus we have
~Joseph Tsai, mgtsai@gmail.com
See Also
1991 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
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