$N in mathbb{N}$ is not a square, show that the continued fraction expansion of $sqrt N/lfloorsqrt Nrfloor$...












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$begingroup$



Let $N in mathbb{N}$ not a square, show that the continued fraction expansion of $sqrt{N}/lfloorsqrt{N}rfloor$ is $[1,overline{a_1,a_2,dots,2}]$.




My notations: the fractional part of $a$ is denoted by ${a}$. Let $N_1 < N < N_2$, where $N_1$ and $N_2$ are squares the closest to $N$.



$sqrt{N}/lfloorsqrt{N}rfloor = (sqrt{N} - lfloorsqrt{N}rfloor+lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor = 1 + (sqrt{N} - lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor$.



Now I'm stuck at the further steps.






number-theory elementary-number-theory continued-fractions







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    $begingroup$



    Let $N in mathbb{N}$ not a square, show that the continued fraction expansion of $sqrt{N}/lfloorsqrt{N}rfloor$ is $[1,overline{a_1,a_2,dots,2}]$.




    My notations: the fractional part of $a$ is denoted by ${a}$. Let $N_1 < N < N_2$, where $N_1$ and $N_2$ are squares the closest to $N$.



    $sqrt{N}/lfloorsqrt{N}rfloor = (sqrt{N} - lfloorsqrt{N}rfloor+lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor = 1 + (sqrt{N} - lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor$.



    Now I'm stuck at the further steps.






    number-theory elementary-number-theory continued-fractions







    share|cite|improve this question











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      $begingroup$



      Let $N in mathbb{N}$ not a square, show that the continued fraction expansion of $sqrt{N}/lfloorsqrt{N}rfloor$ is $[1,overline{a_1,a_2,dots,2}]$.




      My notations: the fractional part of $a$ is denoted by ${a}$. Let $N_1 < N < N_2$, where $N_1$ and $N_2$ are squares the closest to $N$.



      $sqrt{N}/lfloorsqrt{N}rfloor = (sqrt{N} - lfloorsqrt{N}rfloor+lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor = 1 + (sqrt{N} - lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor$.



      Now I'm stuck at the further steps.






      number-theory elementary-number-theory continued-fractions







      share|cite|improve this question











      $endgroup$





      Let $N in mathbb{N}$ not a square, show that the continued fraction expansion of $sqrt{N}/lfloorsqrt{N}rfloor$ is $[1,overline{a_1,a_2,dots,2}]$.




      My notations: the fractional part of $a$ is denoted by ${a}$. Let $N_1 < N < N_2$, where $N_1$ and $N_2$ are squares the closest to $N$.



      $sqrt{N}/lfloorsqrt{N}rfloor = (sqrt{N} - lfloorsqrt{N}rfloor+lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor = 1 + (sqrt{N} - lfloorsqrt{N}rfloor)/lfloorsqrt{N}rfloor$.



      Now I'm stuck at the further steps.






      number-theory elementary-number-theory continued-fractions




      number-theory elementary-number-theory continued-fractions






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 15 '18 at 16:45









      Namaste

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      asked Dec 15 '18 at 11:29









      Jens WagemakerJens Wagemaker

      554312




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          1 Answer
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          $begingroup$

          In the algorithm of writing a continued fraction $[a_0,overline{a_1,ldots,2a_0}]$ of any square root $sqrt{N}$, we have that $a_0=lfloorsqrt{N}rfloor$. Now the first part is writing
          $$begin{align*}frac{sqrt{N}}{lfloorsqrt{N}rfloor}&=frac{a_0+frac{1}{a_1+frac{1}{ddots}}}{a_0}\
          &=1+frac{1}{a_0a_1+frac{a_0}{a_2+frac{1}{ddots}}}\
          &=1+frac{1}{a_0a_1+frac{1}{frac{1}{a_0}left(a_2+frac{1}{ddots}right)}}\
          &=ldots\
          &=1+frac{1}{a_0a_1+frac{1}{frac{ddots}{frac{1}{a_0}left(a_n+frac{1}{ddots}right)}}}.end{align*}$$
          Since $a_n=2a_0$, we get that $a_0$ gets divided by itself at the "end" of the repetition shown above. Hence, the continued fraction is now $[1,overline{b_1,ldots,b_{n-1},2}]$ with $b_iinmathbb{Z}$.



          Note that the case where eventually somewhere in the continued fraction $a_0left(a_k+frac{1}{ddots}right)$ can't appear because then that part would be $a_0a_k+frac{a_0}{ddots}$ which is not part of the partial fraction for we need the "$1$ over something" in each "layer" of the continued fraction.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:52










          • $begingroup$
            That's correct, typo
            $endgroup$
            – Algebear
            Dec 15 '18 at 18:54










          • $begingroup$
            But also in the lines below, and I think that changes the story
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:54










          • $begingroup$
            The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
            $endgroup$
            – Algebear
            Dec 15 '18 at 19:00











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          $begingroup$

          In the algorithm of writing a continued fraction $[a_0,overline{a_1,ldots,2a_0}]$ of any square root $sqrt{N}$, we have that $a_0=lfloorsqrt{N}rfloor$. Now the first part is writing
          $$begin{align*}frac{sqrt{N}}{lfloorsqrt{N}rfloor}&=frac{a_0+frac{1}{a_1+frac{1}{ddots}}}{a_0}\
          &=1+frac{1}{a_0a_1+frac{a_0}{a_2+frac{1}{ddots}}}\
          &=1+frac{1}{a_0a_1+frac{1}{frac{1}{a_0}left(a_2+frac{1}{ddots}right)}}\
          &=ldots\
          &=1+frac{1}{a_0a_1+frac{1}{frac{ddots}{frac{1}{a_0}left(a_n+frac{1}{ddots}right)}}}.end{align*}$$
          Since $a_n=2a_0$, we get that $a_0$ gets divided by itself at the "end" of the repetition shown above. Hence, the continued fraction is now $[1,overline{b_1,ldots,b_{n-1},2}]$ with $b_iinmathbb{Z}$.



          Note that the case where eventually somewhere in the continued fraction $a_0left(a_k+frac{1}{ddots}right)$ can't appear because then that part would be $a_0a_k+frac{a_0}{ddots}$ which is not part of the partial fraction for we need the "$1$ over something" in each "layer" of the continued fraction.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:52










          • $begingroup$
            That's correct, typo
            $endgroup$
            – Algebear
            Dec 15 '18 at 18:54










          • $begingroup$
            But also in the lines below, and I think that changes the story
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:54










          • $begingroup$
            The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
            $endgroup$
            – Algebear
            Dec 15 '18 at 19:00
















          0












          $begingroup$

          In the algorithm of writing a continued fraction $[a_0,overline{a_1,ldots,2a_0}]$ of any square root $sqrt{N}$, we have that $a_0=lfloorsqrt{N}rfloor$. Now the first part is writing
          $$begin{align*}frac{sqrt{N}}{lfloorsqrt{N}rfloor}&=frac{a_0+frac{1}{a_1+frac{1}{ddots}}}{a_0}\
          &=1+frac{1}{a_0a_1+frac{a_0}{a_2+frac{1}{ddots}}}\
          &=1+frac{1}{a_0a_1+frac{1}{frac{1}{a_0}left(a_2+frac{1}{ddots}right)}}\
          &=ldots\
          &=1+frac{1}{a_0a_1+frac{1}{frac{ddots}{frac{1}{a_0}left(a_n+frac{1}{ddots}right)}}}.end{align*}$$
          Since $a_n=2a_0$, we get that $a_0$ gets divided by itself at the "end" of the repetition shown above. Hence, the continued fraction is now $[1,overline{b_1,ldots,b_{n-1},2}]$ with $b_iinmathbb{Z}$.



          Note that the case where eventually somewhere in the continued fraction $a_0left(a_k+frac{1}{ddots}right)$ can't appear because then that part would be $a_0a_k+frac{a_0}{ddots}$ which is not part of the partial fraction for we need the "$1$ over something" in each "layer" of the continued fraction.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:52










          • $begingroup$
            That's correct, typo
            $endgroup$
            – Algebear
            Dec 15 '18 at 18:54










          • $begingroup$
            But also in the lines below, and I think that changes the story
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:54










          • $begingroup$
            The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
            $endgroup$
            – Algebear
            Dec 15 '18 at 19:00














          0












          0








          0





          $begingroup$

          In the algorithm of writing a continued fraction $[a_0,overline{a_1,ldots,2a_0}]$ of any square root $sqrt{N}$, we have that $a_0=lfloorsqrt{N}rfloor$. Now the first part is writing
          $$begin{align*}frac{sqrt{N}}{lfloorsqrt{N}rfloor}&=frac{a_0+frac{1}{a_1+frac{1}{ddots}}}{a_0}\
          &=1+frac{1}{a_0a_1+frac{a_0}{a_2+frac{1}{ddots}}}\
          &=1+frac{1}{a_0a_1+frac{1}{frac{1}{a_0}left(a_2+frac{1}{ddots}right)}}\
          &=ldots\
          &=1+frac{1}{a_0a_1+frac{1}{frac{ddots}{frac{1}{a_0}left(a_n+frac{1}{ddots}right)}}}.end{align*}$$
          Since $a_n=2a_0$, we get that $a_0$ gets divided by itself at the "end" of the repetition shown above. Hence, the continued fraction is now $[1,overline{b_1,ldots,b_{n-1},2}]$ with $b_iinmathbb{Z}$.



          Note that the case where eventually somewhere in the continued fraction $a_0left(a_k+frac{1}{ddots}right)$ can't appear because then that part would be $a_0a_k+frac{a_0}{ddots}$ which is not part of the partial fraction for we need the "$1$ over something" in each "layer" of the continued fraction.






          share|cite|improve this answer











          $endgroup$



          In the algorithm of writing a continued fraction $[a_0,overline{a_1,ldots,2a_0}]$ of any square root $sqrt{N}$, we have that $a_0=lfloorsqrt{N}rfloor$. Now the first part is writing
          $$begin{align*}frac{sqrt{N}}{lfloorsqrt{N}rfloor}&=frac{a_0+frac{1}{a_1+frac{1}{ddots}}}{a_0}\
          &=1+frac{1}{a_0a_1+frac{a_0}{a_2+frac{1}{ddots}}}\
          &=1+frac{1}{a_0a_1+frac{1}{frac{1}{a_0}left(a_2+frac{1}{ddots}right)}}\
          &=ldots\
          &=1+frac{1}{a_0a_1+frac{1}{frac{ddots}{frac{1}{a_0}left(a_n+frac{1}{ddots}right)}}}.end{align*}$$
          Since $a_n=2a_0$, we get that $a_0$ gets divided by itself at the "end" of the repetition shown above. Hence, the continued fraction is now $[1,overline{b_1,ldots,b_{n-1},2}]$ with $b_iinmathbb{Z}$.



          Note that the case where eventually somewhere in the continued fraction $a_0left(a_k+frac{1}{ddots}right)$ can't appear because then that part would be $a_0a_k+frac{a_0}{ddots}$ which is not part of the partial fraction for we need the "$1$ over something" in each "layer" of the continued fraction.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 19:27

























          answered Dec 15 '18 at 16:35









          AlgebearAlgebear

          704419




          704419








          • 1




            $begingroup$
            You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:52










          • $begingroup$
            That's correct, typo
            $endgroup$
            – Algebear
            Dec 15 '18 at 18:54










          • $begingroup$
            But also in the lines below, and I think that changes the story
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:54










          • $begingroup$
            The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
            $endgroup$
            – Algebear
            Dec 15 '18 at 19:00














          • 1




            $begingroup$
            You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:52










          • $begingroup$
            That's correct, typo
            $endgroup$
            – Algebear
            Dec 15 '18 at 18:54










          • $begingroup$
            But also in the lines below, and I think that changes the story
            $endgroup$
            – Jens Wagemaker
            Dec 15 '18 at 18:54










          • $begingroup$
            The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
            $endgroup$
            – Algebear
            Dec 15 '18 at 19:00








          1




          1




          $begingroup$
          You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
          $endgroup$
          – Jens Wagemaker
          Dec 15 '18 at 18:52




          $begingroup$
          You write: $1+frac{1}{a_1+frac{a_0}{a_2+frac{1}{ddots}}}$ in the second line, shouldn't it be $1+frac{1}{a_0 a_1+frac{a_0}{a_2+frac{1}{ddots}}}$?
          $endgroup$
          – Jens Wagemaker
          Dec 15 '18 at 18:52












          $begingroup$
          That's correct, typo
          $endgroup$
          – Algebear
          Dec 15 '18 at 18:54




          $begingroup$
          That's correct, typo
          $endgroup$
          – Algebear
          Dec 15 '18 at 18:54












          $begingroup$
          But also in the lines below, and I think that changes the story
          $endgroup$
          – Jens Wagemaker
          Dec 15 '18 at 18:54




          $begingroup$
          But also in the lines below, and I think that changes the story
          $endgroup$
          – Jens Wagemaker
          Dec 15 '18 at 18:54












          $begingroup$
          The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
          $endgroup$
          – Algebear
          Dec 15 '18 at 19:00




          $begingroup$
          The terms $b_i$ are $a_0cdot a_i$, so they are still in $mathbb{Z}$ for both $a_0$ and $a_i$ are integers.
          $endgroup$
          – Algebear
          Dec 15 '18 at 19:00


















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