Representing positive integers as floor of integer powers of real number.












2














Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?






exponentiation







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  • 1




    You mean a single $a$ for all $b,c$?
    – gammatester
    Nov 27 '18 at 10:09










  • I've edited the question, sorry for my mistake. Yes, one $a$.
    – user4201961
    Nov 27 '18 at 10:11


















2














Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?






exponentiation







share|cite|improve this question




















  • 1




    You mean a single $a$ for all $b,c$?
    – gammatester
    Nov 27 '18 at 10:09










  • I've edited the question, sorry for my mistake. Yes, one $a$.
    – user4201961
    Nov 27 '18 at 10:11
















2












2








2


2





Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?






exponentiation







share|cite|improve this question















Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?






exponentiation




exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 10:11

























asked Nov 27 '18 at 10:06









user4201961

629411




629411








  • 1




    You mean a single $a$ for all $b,c$?
    – gammatester
    Nov 27 '18 at 10:09










  • I've edited the question, sorry for my mistake. Yes, one $a$.
    – user4201961
    Nov 27 '18 at 10:11
















  • 1




    You mean a single $a$ for all $b,c$?
    – gammatester
    Nov 27 '18 at 10:09










  • I've edited the question, sorry for my mistake. Yes, one $a$.
    – user4201961
    Nov 27 '18 at 10:11










1




1




You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09




You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09












I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11






I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11












1 Answer
1






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oldest

votes


















2














No.



The simplest way to see why the answer is no is the fact that, for $a>1$, you have



$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$



(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)



The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$






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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    2














    No.



    The simplest way to see why the answer is no is the fact that, for $a>1$, you have



    $$lim_{btoinfty} (a^b-a^{b-1}) = infty$$



    (which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)



    The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$






    share|cite|improve this answer


























      2














      No.



      The simplest way to see why the answer is no is the fact that, for $a>1$, you have



      $$lim_{btoinfty} (a^b-a^{b-1}) = infty$$



      (which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)



      The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$






      share|cite|improve this answer
























        2












        2








        2






        No.



        The simplest way to see why the answer is no is the fact that, for $a>1$, you have



        $$lim_{btoinfty} (a^b-a^{b-1}) = infty$$



        (which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)



        The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$






        share|cite|improve this answer












        No.



        The simplest way to see why the answer is no is the fact that, for $a>1$, you have



        $$lim_{btoinfty} (a^b-a^{b-1}) = infty$$



        (which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)



        The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 10:21









        5xum

        89.5k393161




        89.5k393161






























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