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Heights and transcendence of $p$--adic continued fractions

Ignazio Longhi, Nadir Murru, Francesco Maria Saettone

Abstract

Special kinds of continued fractions have been proved to converge to transcendental real numbers by means of the celebrated Subspace Theorem. In this paper we study the analogous $p$--adic problem. More specifically, we deal with Browkin $p$--adic continued fractions. First we give some new remarks about the Browkin algorithm in terms of a $p$--adic Euclidean algorithm. Then, we focus on the heights of some $p$--adic numbers having a periodic $p$--adic continued fraction expansion and we obtain some upper bounds. Finally, we exploit these results, together with $p$--adic Roth-like results, in order to prove the transcendence of two families of $p$--adic continued fractions.

Heights and transcendence of $p$--adic continued fractions

Abstract

Special kinds of continued fractions have been proved to converge to transcendental real numbers by means of the celebrated Subspace Theorem. In this paper we study the analogous --adic problem. More specifically, we deal with Browkin --adic continued fractions. First we give some new remarks about the Browkin algorithm in terms of a --adic Euclidean algorithm. Then, we focus on the heights of some --adic numbers having a periodic --adic continued fraction expansion and we obtain some upper bounds. Finally, we exploit these results, together with --adic Roth-like results, in order to prove the transcendence of two families of --adic continued fractions.
Paper Structure (8 sections, 16 theorems, 84 equations)

This paper contains 8 sections, 16 theorems, 84 equations.

Table of Contents

  1. Introduction
  2. Preliminaries
  3. Some remarks on Browkin's approach to $p$-adic continued fractions
  4. On the axiomatization of $s$
  5. Auxiliary results
  6. Properties of partial quotients, convergents and height bounds
  7. $p$-adic Subspace Theorem
  8. Main results

Key Result

Theorem A

Let be a non--periodic Browkin $p$--adic continued fraction. Assume that one of the following holds: Then $\alpha$ is transcendental. Moreover, if the hypothesis of (a) is weakened to $b_{h+k_i} = b_{h}$, for $n_i \leq h \leq n_i + (\lambda_i-1)k_i -1$, for every $i$ (with small changes in the technical conditions, see Theorem ooto), then $\alpha$ is either transcendental or quadratic irrationa

Theorems & Definitions (37)

  • Theorem A
  • Proposition 1
  • proof
  • Remark 1
  • Proposition 2
  • proof
  • Remark 2
  • Lemma 1
  • proof
  • Lemma 2
  • ...and 27 more