Is that true : Every positive rational number $q$ can be written as $q = \sum_{i=0}^{k}1/n_i$ , where $n_i,k$ are positive intergers and $n_i\not=n_j$ if $i\not=j$.
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Express the fraction $\dfrac{a}{b}$ as a sum of $a$ copies of $\dfrac{1}{b}$. (For technical reasons, if $b=1$, use $2a$ copies of $\dfrac{1}{2}$.) Major flaw: the denominators are not all different. Using repeatedly the identity $$\frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)},\tag{$1$}$$ we can express any $\dfrac{1}{n}$ as a sum of distinct unit fractions with all denominators as large as we wish. So leave the first $\dfrac{1}{b}$ alone. Express the second one as $\dfrac{1}{b+1}+\dfrac{1}{b(b+1}$. For the third $\dfrac{1}{b}$, use Identity $(1)$ repeatedly to express $\dfrac{1}{b}$ as a sum of distinct unit fractions with denominators all greater than $b(b+1)$. Continue. The algorithm we have described is quite inefficient. There are much better algorithms available, going back to Fibonacci's greedy algorithm. |
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From Wikipedia. Read here: http://en.wikipedia.org/wiki/Egyptian_fraction |
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