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Suppose $(q_{n})_{n\in\mathbb{Z}_{\gt 0}}$ is a decreasing sequence of positive rational numbers such that $Q:=\displaystyle{\sum_{n>0}q_{n}}$ is finite. Let's denote by $n_{i}$ and $d_{i}$ the numerator and the denominator of $q_{i}$ such that $(n_{i},d_{i})=1$. If $i\ne j\Rightarrow (n_{i}d_{i},n_{j}d_{j})=1$, is $Q$ necessarily irrational?

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  • $\begingroup$ What if $q_i = \frac{1}{2^i}$. $\endgroup$ – Mustafa Said Oct 28 '14 at 18:44
  • $\begingroup$ @Mustafa: then $d_i = 2^i$ so $(n_i d_i, n_j d_j) = (2^i, 2^j) \neq 1$ in general. $\endgroup$ – Qiaochu Yuan Oct 28 '14 at 18:45
  • $\begingroup$ I misread the question. Thx @Qiaochu Yuan. $\endgroup$ – Mustafa Said Oct 28 '14 at 18:45
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No. In fact, every positive real number $Q$ can be written this way. Here is the algorithm: pick $q_1$ to be any positive rational which lies in the interval $\left( \frac{Q}{2}, Q \right)$. Once $q_1, q_2, \dots q_n$ have been picked, pick $q_{n+1}$ to be any positive rational whose numerator and denominator are coprime to the numerators and denominators of $q_1, q_2, \dots q_n$ and which lies in the interval $\left( \frac{Q_{n+1}}{2}, Q_{n+1} \right)$, where

$$Q_{n+1} = Q - q_1 - q_2 - \dots - q_n.$$

The key point is that the set of rational numbers whose numerators and denominators do not contain a fixed finite set of primes is still dense in $\mathbb{R}$. To prove this it suffices to observe that the logarithms of the primes are linearly independent over $\mathbb{Q}$, so any two such logarithms generate a dense subgroup of $\mathbb{R}$: in fact this shows that the set of rational numbers whose numerators and denominators are divisible by two fixed (arbitrarily large) primes is still dense in $\mathbb{R}$.

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No. You can use a greedy algorithm to construct a counterexample. For example, let $n_1/d_1 = 2/3$. Given $n_1/d_1,\dots,n_k/d_k$ whose sum is $S_k$, choose a rational number $n_{k+1}/d_{k+1}$ in the interval $[\frac12(1-S_k),\frac23(1-S_k)]$ whose numerator and denominator are relatively prime to $n_1\dots n_kd_1\dots d_k$. Then $\sum_{k=1}^\infty n_k/d_k$ equals $1$ by construction.

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  • $\begingroup$ I see Qiaochu Yuan beat me to this! $\endgroup$ – Greg Martin Oct 28 '14 at 18:51

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