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Approximating rational values in ]0$]0,1[1[$ by a sum or difference of unit fractions

I had the |..| bars for absolute value wrong
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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|$$|\frac{m}{n} - (u'+v')| < |\frac{m}{n} - (u+v)|$?

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 $\frac{m}{n} - |u'+v'| < \frac{m}{n} - |u+v|$?

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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Gerald Edgar
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