How can we show these two relations?












0












$begingroup$


I want to show that $$frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]
(ngeq 1)$$ and that $$p_nq_{n-3}-q_np_{n-3}=(-1)^{n-1}(a_na_{n-1}+1)$$ but I don't really have an idea how to start.



Could you give me a hint?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Induction on $n$?
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:07










  • $begingroup$
    $p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 20:09






  • 2




    $begingroup$
    In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:10






  • 1




    $begingroup$
    Let us know if you need further hints.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:22








  • 1




    $begingroup$
    Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 21:09
















0












$begingroup$


I want to show that $$frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]
(ngeq 1)$$ and that $$p_nq_{n-3}-q_np_{n-3}=(-1)^{n-1}(a_na_{n-1}+1)$$ but I don't really have an idea how to start.



Could you give me a hint?






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Induction on $n$?
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:07










  • $begingroup$
    $p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 20:09






  • 2




    $begingroup$
    In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:10






  • 1




    $begingroup$
    Let us know if you need further hints.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:22








  • 1




    $begingroup$
    Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 21:09














0












0








0


0



$begingroup$


I want to show that $$frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]
(ngeq 1)$$ and that $$p_nq_{n-3}-q_np_{n-3}=(-1)^{n-1}(a_na_{n-1}+1)$$ but I don't really have an idea how to start.



Could you give me a hint?






elementary-number-theory







share|cite|improve this question











$endgroup$




I want to show that $$frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]
(ngeq 1)$$ and that $$p_nq_{n-3}-q_np_{n-3}=(-1)^{n-1}(a_na_{n-1}+1)$$ but I don't really have an idea how to start.



Could you give me a hint?






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 20:12







Mary Star

















asked Dec 17 '18 at 20:07









Mary StarMary Star

3,08982369




3,08982369








  • 3




    $begingroup$
    Induction on $n$?
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:07










  • $begingroup$
    $p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 20:09






  • 2




    $begingroup$
    In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:10






  • 1




    $begingroup$
    Let us know if you need further hints.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:22








  • 1




    $begingroup$
    Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 21:09














  • 3




    $begingroup$
    Induction on $n$?
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:07










  • $begingroup$
    $p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 20:09






  • 2




    $begingroup$
    In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:10






  • 1




    $begingroup$
    Let us know if you need further hints.
    $endgroup$
    – MisterRiemann
    Dec 17 '18 at 20:22








  • 1




    $begingroup$
    Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
    $endgroup$
    – Mary Star
    Dec 17 '18 at 21:09








3




3




$begingroup$
Induction on $n$?
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:07




$begingroup$
Induction on $n$?
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:07












$begingroup$
$p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
$endgroup$
– Mary Star
Dec 17 '18 at 20:09




$begingroup$
$p_n$ is the $n$th prime, isn't it? What exactly is $q_n$ ? @MisterRiemann
$endgroup$
– Mary Star
Dec 17 '18 at 20:09




2




2




$begingroup$
In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:10




$begingroup$
In the context of continued fractions, $p_n$ and $q_n$ are respectively the numerator and the denominator of the $n$-th convergent $C_n$.
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:10




1




1




$begingroup$
Let us know if you need further hints.
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:22






$begingroup$
Let us know if you need further hints.
$endgroup$
– MisterRiemann
Dec 17 '18 at 20:22






1




1




$begingroup$
Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
$endgroup$
– Mary Star
Dec 17 '18 at 21:09




$begingroup$
Base Case: For $n=1$ we have $frac{p_1}{p_0}=frac{a_1a_0+1}{1}=a_1a_0+1$. Is this equal to $[a_1, a_0]$ ? $$$$ Inductive Hypothesis: We suppose that the statement holds for some $n$, i.e. $frac{p_n}{p_{n-1}}=[a_n, a_{n-1}, ldots , a_1, a_0]$. $$$$ Inductive Step: We want to show that it holds for $n+1$: $$frac{p_{n+1}}{p_n}=frac{a_{n+1}p_{n}+p_{n-1}}{p_n}=a_{n+1}+frac{p_{n-1}}{p_n}=a_{n+1}+frac{1}{frac{p_n}{p_{n-1}}}$$ Is this correct so far? @MisterRiemann
$endgroup$
– Mary Star
Dec 17 '18 at 21:09










1 Answer
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Hint for the second question!!



Use that $p_{n-3}=p_{n-1}-a_{n-1}p_{n-2}$ and $q_{n-3}=q_{n-1}-a_{n-1}q_{n-2}$.



Doing some math you will get
$(p_{n}q_{n-1} - q_{n}p_{n-1}) + a_{n-1} ( q_{n}p_{n-2} - p_{n}q_{n-2} ) $



Use induction for the 2 brackets.






share|cite|improve this answer









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    $begingroup$

    Hint for the second question!!



    Use that $p_{n-3}=p_{n-1}-a_{n-1}p_{n-2}$ and $q_{n-3}=q_{n-1}-a_{n-1}q_{n-2}$.



    Doing some math you will get
    $(p_{n}q_{n-1} - q_{n}p_{n-1}) + a_{n-1} ( q_{n}p_{n-2} - p_{n}q_{n-2} ) $



    Use induction for the 2 brackets.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint for the second question!!



      Use that $p_{n-3}=p_{n-1}-a_{n-1}p_{n-2}$ and $q_{n-3}=q_{n-1}-a_{n-1}q_{n-2}$.



      Doing some math you will get
      $(p_{n}q_{n-1} - q_{n}p_{n-1}) + a_{n-1} ( q_{n}p_{n-2} - p_{n}q_{n-2} ) $



      Use induction for the 2 brackets.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint for the second question!!



        Use that $p_{n-3}=p_{n-1}-a_{n-1}p_{n-2}$ and $q_{n-3}=q_{n-1}-a_{n-1}q_{n-2}$.



        Doing some math you will get
        $(p_{n}q_{n-1} - q_{n}p_{n-1}) + a_{n-1} ( q_{n}p_{n-2} - p_{n}q_{n-2} ) $



        Use induction for the 2 brackets.






        share|cite|improve this answer









        $endgroup$



        Hint for the second question!!



        Use that $p_{n-3}=p_{n-1}-a_{n-1}p_{n-2}$ and $q_{n-3}=q_{n-1}-a_{n-1}q_{n-2}$.



        Doing some math you will get
        $(p_{n}q_{n-1} - q_{n}p_{n-1}) + a_{n-1} ( q_{n}p_{n-2} - p_{n}q_{n-2} ) $



        Use induction for the 2 brackets.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 0:08









        Dr.MathematicsDr.Mathematics

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