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Let $U=\{\frac{1}{n}: n\in\mathbb{N}\} \cup \{-\frac{1}{n}: n\in\mathbb{N}\}$ be the set of positive and negative unit fractions.

Are there positive integers $m<n \in \mathbb{N}$, such that for every $u,v\in U$ there are $u',v'\in U$ such that $|\frac{m}{n} - (u'+v')| < |\frac{m}{n} - (u+v)|$?

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    $\begingroup$ As it's now, for $\ u=v=1\ $ the difference is always the smallest (for any real, not just for positive rational numbers). You're possibly thinking about the absolute value of each side of the inequality. But the sums of elements of $U$ form a discrete set. Thus the answer is still NO (again for arbitrary reals). $\endgroup$ Aug 28, 2014 at 9:08
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    $\begingroup$ This is not what one means by "approximation theory". So I added "diophantine-approximation" ... maybe that's what this is? $\endgroup$ Aug 28, 2014 at 12:34
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    $\begingroup$ Something very strange is written. Looks like a misprint. $\endgroup$ Aug 28, 2014 at 15:57
  • $\begingroup$ That's correct - I put the absolute value signs in the wrong place and edited the post now accordingly. $\endgroup$ Sep 2, 2014 at 12:53

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I suppose the intended inequality is $$ \left| \frac{m}{n} - |u+v| \, \right| < \left| \frac{m}{n} - |u'+v'| \, \right| $$ as Wlodzimierz Holsztynski suggests. But there are no such $m,n$. Indeed for any real $r>0$, rational or not, there exists a unique element of $$ U + U := \{ u_1 + u_2 \mid u_1,u_2 \in U \} $$ closest to $r$.

Proof: If $r\in U$ then $r \in U+U$, because $r = \frac1n = \frac1{2n} + \frac1{2n}$. If not, choose $n$ such that $|r| \geq 1/n$. Then if $u_1,u_2 \in U$ with $|r - (u_1+u_2)| < \left|r - \frac1n\right|$ then at least one of $u_1,u_2$ exceeds $\frac1{2n}$; without loss of generality assume it is $u_1$. Then there are finitely many candidates for $u_1$, and for each of them there is an optimal choice of $u_2$ (because $r \neq u_1$ by hypothesis). The optimal approximation is then the best one of these as $u_1$ varies over the finite list of fractions larger than $1/2n$.

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  • $\begingroup$ Sorry - I really got my $|\cdot|$ signs wrong. What I intended is the inequality $|\frac{m}{n} - (u'+v')| < |\frac{m}{n} - (u+v)|$. The goal is to approximate $\frac{m}{n}$ as good as it gets by a sum or difference of unit fractions, so I want to minimize $|\frac{m}{n} - (u+v)|$ where $u,v\in U$. $\endgroup$ Sep 2, 2014 at 12:51

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