Prove that $forall n, , exists N,x :lfloor{x^{N}}rfloor =n , land ,lfloor{x^{N+1}}rfloor =n+1$












1














The question is related to the interesting problem raised by following OP.



Notably I'm trying to prove the following fact




For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that



$$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$




My idea for the proof is to consider




  • $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$


and then we need that





  • $x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$


therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that



$$sqrt[N]{n}=1+frac{1+alpha}{n}$$



but I'm totally stuck here and I can't find any method to prove that.



Thanks in advance for any idea or suggestion about that!






real-analysis







share|cite|improve this question



























    1














    The question is related to the interesting problem raised by following OP.



    Notably I'm trying to prove the following fact




    For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that



    $$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$




    My idea for the proof is to consider




    • $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$


    and then we need that





    • $x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$


    therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that



    $$sqrt[N]{n}=1+frac{1+alpha}{n}$$



    but I'm totally stuck here and I can't find any method to prove that.



    Thanks in advance for any idea or suggestion about that!






    real-analysis







    share|cite|improve this question

























      1












      1








      1


      0





      The question is related to the interesting problem raised by following OP.



      Notably I'm trying to prove the following fact




      For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that



      $$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$




      My idea for the proof is to consider




      • $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$


      and then we need that





      • $x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$


      therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that



      $$sqrt[N]{n}=1+frac{1+alpha}{n}$$



      but I'm totally stuck here and I can't find any method to prove that.



      Thanks in advance for any idea or suggestion about that!






      real-analysis







      share|cite|improve this question













      The question is related to the interesting problem raised by following OP.



      Notably I'm trying to prove the following fact




      For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that



      $$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$




      My idea for the proof is to consider




      • $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$


      and then we need that





      • $x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$


      therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that



      $$sqrt[N]{n}=1+frac{1+alpha}{n}$$



      but I'm totally stuck here and I can't find any method to prove that.



      Thanks in advance for any idea or suggestion about that!






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 27 '18 at 21:57









      gimusigimusi

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          Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.



          We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired



          $$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$






          share|cite|improve this answer





















          • Thanks! I need to study that a little bit in order to digest all step! Bye
            – gimusi
            Nov 27 '18 at 23:00










          • Fell free to ask if anything is unclear.
            – Ingix
            Nov 27 '18 at 23:01










          • Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
            – gimusi
            Nov 27 '18 at 23:07










          • Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
            – gimusi
            Nov 27 '18 at 23:12










          • Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
            – Ingix
            Nov 27 '18 at 23:27











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          Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.



          We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired



          $$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$






          share|cite|improve this answer





















          • Thanks! I need to study that a little bit in order to digest all step! Bye
            – gimusi
            Nov 27 '18 at 23:00










          • Fell free to ask if anything is unclear.
            – Ingix
            Nov 27 '18 at 23:01










          • Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
            – gimusi
            Nov 27 '18 at 23:07










          • Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
            – gimusi
            Nov 27 '18 at 23:12










          • Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
            – Ingix
            Nov 27 '18 at 23:27
















          1














          Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.



          We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired



          $$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$






          share|cite|improve this answer





















          • Thanks! I need to study that a little bit in order to digest all step! Bye
            – gimusi
            Nov 27 '18 at 23:00










          • Fell free to ask if anything is unclear.
            – Ingix
            Nov 27 '18 at 23:01










          • Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
            – gimusi
            Nov 27 '18 at 23:07










          • Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
            – gimusi
            Nov 27 '18 at 23:12










          • Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
            – Ingix
            Nov 27 '18 at 23:27














          1












          1








          1






          Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.



          We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired



          $$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$






          share|cite|improve this answer












          Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.



          We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired



          $$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 22:49









          IngixIngix

          3,354145




          3,354145












          • Thanks! I need to study that a little bit in order to digest all step! Bye
            – gimusi
            Nov 27 '18 at 23:00










          • Fell free to ask if anything is unclear.
            – Ingix
            Nov 27 '18 at 23:01










          • Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
            – gimusi
            Nov 27 '18 at 23:07










          • Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
            – gimusi
            Nov 27 '18 at 23:12










          • Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
            – Ingix
            Nov 27 '18 at 23:27


















          • Thanks! I need to study that a little bit in order to digest all step! Bye
            – gimusi
            Nov 27 '18 at 23:00










          • Fell free to ask if anything is unclear.
            – Ingix
            Nov 27 '18 at 23:01










          • Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
            – gimusi
            Nov 27 '18 at 23:07










          • Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
            – gimusi
            Nov 27 '18 at 23:12










          • Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
            – Ingix
            Nov 27 '18 at 23:27
















          Thanks! I need to study that a little bit in order to digest all step! Bye
          – gimusi
          Nov 27 '18 at 23:00




          Thanks! I need to study that a little bit in order to digest all step! Bye
          – gimusi
          Nov 27 '18 at 23:00












          Fell free to ask if anything is unclear.
          – Ingix
          Nov 27 '18 at 23:01




          Fell free to ask if anything is unclear.
          – Ingix
          Nov 27 '18 at 23:01












          Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
          – gimusi
          Nov 27 '18 at 23:07




          Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
          – gimusi
          Nov 27 '18 at 23:07












          Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
          – gimusi
          Nov 27 '18 at 23:12




          Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
          – gimusi
          Nov 27 '18 at 23:12












          Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
          – Ingix
          Nov 27 '18 at 23:27




          Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
          – Ingix
          Nov 27 '18 at 23:27


















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