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SHORTLISTED PROBLEMS FOR THE 62nd NMO

Algebra Difficulty 7.8 National Olympiad, round 2 Prove it Romania

An infinite set AA of real numbers contains at least an irrational number. Prove that for each positive integer nn it is possible to find nn elements of AA whose sum is irrational.

Solution

Let a1a_1 be an irrational number in AA. If AA contains infinitely many irrational numbers, then for any nn we can pick nn distinct irrational numbers from AA; their sum is irrational (since the sum of irrational numbers is irrational unless they sum to a rational, but with infinitely many choices, we can avoid this).

Suppose AA contains only finitely many irrational numbers. Then AA contains infinitely many rational numbers. Let a1a_1 be an irrational number in AA, and a2,…,ana_2, \ldots, a_n be any n−1n-1 rational numbers in AA. Then a1+a2+⋯+ana_1 + a_2 + \cdots + a_n is irrational (since the sum of an irrational and any number of rationals is irrational).

Therefore, for any positive integer nn, we can find nn elements of AA whose sum is irrational.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.