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Let set $C$: $C \subset \mathbb{Z}^+$, and give $c \in \mathbb{Z}^+, c > 1$. if $\sum_{k \in C}\frac{1}{c^k}$ is an algebraic number, for other $p \in \mathbb{Z}^+, p > 1$, is the number $\sum_{k \in C}\frac{1}{p^k}$ an algebraic number too?

If $C$ is a finite set, it is obvious. And infinite set?

xunitc
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  • The case of an infinite sum is drastically different from a finite sum, and it is usually extremely difficult to determine whether any particular infinite sum is algebraic. Do you know of some reason that the algebraicness of the first sum should be related to algebraicness of the second sum? – Eric Wofsey Oct 03 '18 at 06:06
  • Notice in particular that in the case $c=2$, the first sum can be any real number in $[0,1]$ by choosing $C$ appropriately. Probably for most algebraic numbers in $[0,1]$, the corresponding sums for most other $p$ will not be algebraic, but this is probably very hard to prove. – Eric Wofsey Oct 03 '18 at 06:14
  • @EricWofsey Thank you for your attention. I just think about that, but have no idea. – xunitc Oct 03 '18 at 06:19

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If $x=\sum_{k\in C} 1/c^k$ for some integer $c\ge 2$, then the base-$c$ expansion of $x$ has digits $1$ for $k \in C$ and $0$ otherwise.

If $x$ is rational, the base-$c$ expansion of a rational number is eventually periodic, and this implies that the sum is rational for all $p$.

It is conjectured that every irrational algebraic number is normal in every base. This would imply that your sum can't be an irrational algebraic for $c > 2$. But we are very far from being able to prove that conjecture.

Robert Israel
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