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Problem: Prove that all positive rational numbers can be expressed as the finite sum of different numbers $\displaystyle \frac {1} {n}$ ($n$ is a natural number).

Example: $\displaystyle \frac {19}{16}=1+ \frac {1}{8} + \frac {1}{16}.$

*We cant sum numbers as $\displaystyle \frac {3}{16}$ (denominator > 1) but we can sum $\displaystyle \frac {1}{8}+ \frac {1}{16}.$

Any solutions? Suggestions?

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So, Egyptian fractions? – J. M. May 15 '11 at 14:32
You might want to say positive rational numbers since negative rationals obviously cannot be formed with numerator 1 and natural-number denominators. (The Putnam problem cited by Chandru1 specifies that the numbers be positive.) – Fixee May 15 '11 at 15:17
@Fixee, natural numbers are positive. – quanta May 15 '11 at 15:18
@quanta: Exactly. Which is why it's unlikely you can form a negative rational number via a sum of fractions with 1 over a natural number. – Fixee May 15 '11 at 15:56

This is a putnam problem. For a complete solution please look here.

  • Take a look at this article as well: J.C.Owings, American Mathematical Monthly Vol. 75 (1968), Pages $777-778$.
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Thank you very much! – tomerg May 15 '11 at 20:23
For those that cannot follow the link, it essentially says that a greedy algorithm works: simply keep subtracting the largest reciprocal not already used which is less than the residual, and eventually this will terminate because, once you stop using the initial terms of the harmonic series, the numerator of the residual will strictly reduce with each step until it reaches zero. – Henry May 15 '11 at 21:41
@tomerg: Please accept an answer so that this question is not in the unanswered list. – user9413 May 17 '11 at 6:38
Jstor link for that article: – Martin Sleziak Nov 27 '11 at 12:17

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