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Inspiration for this question: Can every number be represented as a sum of different reciprocal numbers?

My question: Can every positive rational number $\frac{m}{n}\in\mathbb{Q}$ be written as the finite sum of distinct (i.e. all different) reciprocal primes:

$$\frac{m}{n} = \sum_k \frac{1}{k},\quad k \text{ is a prime number } ?$$

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    $\begingroup$ Well, how would you write $\frac 14$? Note that, if $p$ is one of your primes, then $p$ divides the denominator of $\sum \frac 1k$. $\endgroup$
    – lulu
    Commented Apr 24, 2022 at 22:40
  • $\begingroup$ I gave the reason. If $p$ were one of your primes we'd need to have $p\,|\,4$, hence $p=2$. $\endgroup$
    – lulu
    Commented Apr 24, 2022 at 22:42
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    $\begingroup$ To elaborate: if $\{p_i\}$ is a collection of distinct primes then the denominator of $\sum \frac 1{p_i}$ (in least terms) is $\prod p_i$. Indeed, if $p$ is one of the $p_i$ then every term in the numerator is divisible by $p$, except for one. $\endgroup$
    – lulu
    Commented Apr 24, 2022 at 22:43
  • $\begingroup$ Ok - so $\sum \frac{1}{k}$ cannot be equal to a rational number whose denominator is even. But can the denominator of the sum be odd? $\endgroup$
    – Adam Rubinson
    Commented Apr 24, 2022 at 22:51
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    $\begingroup$ When you return to it: as far as your original question goes, note that you don't actually need to show that the denominator of $\sum \frac 1{p_i}$ is $\prod p_i$ (though it is not hard to show that). It is entirely clear that the denominator is at least a divisor of that, hence any representable fraction in least terms has a square free denominator. (to be sure, that condition is necessary for representablility, but it is not sufficient.) $\endgroup$
    – lulu
    Commented Apr 24, 2022 at 22:59

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