Difference between revisions of "1987 IMO Problems/Problem 4"
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− | Let <math>N</math> be the set of non-negative integers. Put <math>A = N - f(N)</math> (the set of all <math>n</math> such that we cannot find m with f(m) = n). Put <math>B = f(A)</math>. | + | Let <math>N</math> be the set of non-negative integers. Put <math>A = N - f(N)</math> (the set of all <math>n</math> such that we cannot find <math>m</math> with <math>f(m) = n</math>). Put <math>B = f(A)</math>. |
Note that <math>f</math> is injective because if <math>f(n) = f(m)</math>, then <math>f(f(n)) = f(f(m))</math> so <math>m = n</math>. We claim that <math>B = f(N) - f(f(N))</math>. Obviously <math>B</math> is a subset of <math>f(N)</math> and if <math>k</math> belongs to <math>B</math>, then it does not belong to <math>f(f(N))</math> since <math>f</math> is injective. Similarly, a member of <math>f(f(N))</math> cannot belong to <math>B</math>. | Note that <math>f</math> is injective because if <math>f(n) = f(m)</math>, then <math>f(f(n)) = f(f(m))</math> so <math>m = n</math>. We claim that <math>B = f(N) - f(f(N))</math>. Obviously <math>B</math> is a subset of <math>f(N)</math> and if <math>k</math> belongs to <math>B</math>, then it does not belong to <math>f(f(N))</math> since <math>f</math> is injective. Similarly, a member of <math>f(f(N))</math> cannot belong to <math>B</math>. |
Revision as of 01:13, 15 March 2008
Problem
Prove that there is no function from the set of non-negative integers into itself such that for every .
Solution
We prove that if for all , where is a fixed positive integer, then must be even. If , then we may take .
Suppose with . Then by an easy induction on we find , . We show this leads to a contradiction. Suppose , so for some . Then . But , so . Contradiction. So we must have , so for some . But now . But , so . Contradiction.
So if , then and have different residues . Suppose they have and respectively. Then the same induction shows that all sufficiently large have , and that all sufficiently large have . Hence if has a different residue , then cannot have residue or . For if had residue , then the same argument would show that all sufficiently large numbers with residue had . Thus the residues form pairs, so that if a number is congruent to a particular residue, then of the number is congruent to the pair of the residue. But this is impossible for odd.
Other Solution
Solution by Sawa Pavlov:
Let be the set of non-negative integers. Put (the set of all such that we cannot find with ). Put .
Note that is injective because if , then so . We claim that . Obviously is a subset of and if belongs to , then it does not belong to since is injective. Similarly, a member of cannot belong to .
Clearly and are disjoint. They have union which is . But since is injective they have the same number of elements, which is impossible since has an odd number of elements.
1987 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
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