Is it possible to represent an improper fraction as a finite sum of unique unit fractions (Egyptian fractions)?

  • $\begingroup$ See this question. $\endgroup$ – Dietrich Burde Mar 27 '17 at 18:53
  • $\begingroup$ You know the result for rational numbers less than $1$ and you ask if it holds above $1$? $\endgroup$ – lulu Mar 27 '17 at 18:57
  • $\begingroup$ @DietrichBurde, thank you for the link, but the answer provided works for 0 < x < 1, whereas I am asking for improper fractions. $\endgroup$ – Alderson Mar 27 '17 at 18:57
  • $\begingroup$ @lulu, yes, that's exactly what I am asking, thanks $\endgroup$ – Alderson Mar 27 '17 at 19:01

Yes. Start with $\alpha \in \mathbb Q$, $\alpha >1$. Then let $n$ be the greatest integer such that $$H_n=\sum_{i=1}^n\frac 1i<\alpha $$

Of course $n$ exists because the infinite Harmonic series diverges.

It follows that $\alpha - H_n<\frac 1n$ so none of the fractions in the standard Egyptian decomposition of $\alpha - H_n$ can appear in $H_n$.

  • $\begingroup$ Thank you, it did not occur to me that uniqueness can be proven by Harmonic series too! $\endgroup$ – Alderson Mar 27 '17 at 19:08
  • $\begingroup$ This doesn't show uniqueness! Uniqueness isn't even true... $\frac 32 = 1 +\frac 12=1+\frac 13 +\frac 16$ $\endgroup$ – lulu Mar 27 '17 at 19:11
  • $\begingroup$ Or did you mean that each of the fractions that appears is guaranteed to occur only once? That is true (and is demonstrated by what I wrote). $\endgroup$ – lulu Mar 27 '17 at 19:13
  • $\begingroup$ Yes, I meant that in this representation each fraction occurs only once $\endgroup$ – Alderson Mar 27 '17 at 19:21

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