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Fedor Petrov
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  1. No, it can not unless it is integer.
  2. This is an open problem known as Pisot-Vijayaraghavan problem. For algebraic numbers the answer is positive, that implies the answer to 1). See https://en.m.wikipedia.org/wiki/Pisot-Vijayaraghavan_number

Short self-contained answer to 1): if $\alpha^n=A_n+\delta_n$ for positive integer $A_n$ and $\delta_n\to 0$, and $\alpha=p/q$, we get $qA_{n+1}-pA_n=-q\delta_{n+1}+p\delta_n\to 0$, but $qA_{n+1}-pA_n$ is integer, hence for large enough $n$ we have $A_{n+1}=\frac pq A_n$, this is impossible since such a sequence takes non-integer values for coprime $p,q>1$.

  1. No, it can not unless it is integer.
  2. This is an open problem known as Pisot-Vijayaraghavan problem. For algebraic numbers the answer is positive, that implies the answer to 1). See https://en.m.wikipedia.org/wiki/Pisot-Vijayaraghavan_number
  1. No, it can not unless it is integer.
  2. This is an open problem known as Pisot-Vijayaraghavan problem. For algebraic numbers the answer is positive, that implies the answer to 1). See https://en.m.wikipedia.org/wiki/Pisot-Vijayaraghavan_number

Short self-contained answer to 1): if $\alpha^n=A_n+\delta_n$ for positive integer $A_n$ and $\delta_n\to 0$, and $\alpha=p/q$, we get $qA_{n+1}-pA_n=-q\delta_{n+1}+p\delta_n\to 0$, but $qA_{n+1}-pA_n$ is integer, hence for large enough $n$ we have $A_{n+1}=\frac pq A_n$, this is impossible since such a sequence takes non-integer values for coprime $p,q>1$.

Source Link
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

  1. No, it can not unless it is integer.
  2. This is an open problem known as Pisot-Vijayaraghavan problem. For algebraic numbers the answer is positive, that implies the answer to 1). See https://en.m.wikipedia.org/wiki/Pisot-Vijayaraghavan_number