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Let $p$ be 2 or 7 in the following. For certain reasons I need to find a "simple" rational and irrational numbers whose $p-$adic expansion is a Laurent series with order $-5$ $$a_{-5}p^{-5}+a_{-4}p^{-4}+a_{-3}p^{-3}+a_{-2}p^{-2}+a_{-1}p^{-1}+a_{0}p^{0}+a_{1} p^{1}+a_{2}p^{2}+\ldots$$

with $a_{n}\neq 0$ for all $n\geq -5$, with $a_k=0$ for all $k\leq -6$. Some $a_n$'s with $n\geq 0$ may be zeroes, but not on a tail. I also need a justification that the four numbers (ir)rationals/$p$ is 2 or 7 are as desired.

Michael Hardy
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1 Answers1

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A simple fact which should give you plenty of examples is that a $p$-adic number is $\in \mathbb Q$ if and only if its $p$-adic expansion

$$\sum_{i \gg -\infty} a_i p^i, \qquad\qquad a_i \in \lbrace 0,...,p-1\rbrace$$

is eventually periodic (or finite, i.e. has a tail of $0$'s); exactly as it is for $b$-adic expansions of real numbers, and basically with the same proof. Compare e.g. Relating the base-p periodic expansion of a rational to its p-adic representation, p-adic expansion of a rational number.

Now I'm sure you can come up with many eventually periodic and many eventually non-periodic sequences $a_{-5}, a_{-4}, a_{-3}, ...$.

Torsten Schoeneberg
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  • Yes, but I cannot determine from the periodic sequence $a_{-5}, a_{-4}, a_{-3}, ...$ the number itself ... –  Oct 24 '19 at 17:09
  • To determine what rational fraction a periodic sequence represents, the answer by Lubin to the second question I linked to should give a complete answer. For example if $p=2, 1=a_{-5}=a_{-3}=a_{-1}=a_1=..., 0=a_{-4}=a_{-2}=a_0=...$, we get $p^{-5} \cdot( p^0+p^2+p^4+...) = 2^{-5}\cdot \dfrac{1}{1-2^2}= \dfrac{1}{32\cdot (-3)}=-\dfrac{1}{96}$. – Torsten Schoeneberg Oct 24 '19 at 17:43