Digits in an algebraic irrational number












6












$begingroup$


I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).



I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).



However, my question is in the opposite direction:



Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?



Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).



Any help will be welcomed.






nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution







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    6












    $begingroup$


    I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).



    I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).



    However, my question is in the opposite direction:



    Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?



    Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).



    Any help will be welcomed.






    nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution







    share|cite|improve this question









    $endgroup$















      6












      6








      6


      1



      $begingroup$


      I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).



      I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).



      However, my question is in the opposite direction:



      Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?



      Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).



      Any help will be welcomed.






      nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution







      share|cite|improve this question









      $endgroup$




      I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).



      I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).



      However, my question is in the opposite direction:



      Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?



      Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).



      Any help will be welcomed.






      nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution




      nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution






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      asked 12 hours ago









      JeanJean

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          $begingroup$

          What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
          Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
          $$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
          so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.



          Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).






          share|cite|improve this answer









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            $begingroup$

            What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
            Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
            $$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
            so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.



            Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).






            share|cite|improve this answer









            $endgroup$


















              14












              $begingroup$

              What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
              Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
              $$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
              so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.



              Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).






              share|cite|improve this answer









              $endgroup$
















                14












                14








                14





                $begingroup$

                What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
                Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
                $$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
                so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.



                Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).






                share|cite|improve this answer









                $endgroup$



                What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
                Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
                $$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
                so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.



                Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 11 hours ago









                Robert IsraelRobert Israel

                42.6k51122




                42.6k51122






























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