Proof of $frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},…,a_2,a_1]$?












1












$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







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  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45
















1












$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45














1












1








1





$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







share|cite|improve this question











$endgroup$




Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions




fractions continued-fractions






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edited Jun 19 '16 at 14:39









Martin Sleziak

44.8k9118272




44.8k9118272










asked Jun 18 '16 at 16:22









qwerryqwerry

314




314








  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45














  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45








1




1




$begingroup$
I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
$endgroup$
– Martin Sleziak
Jun 19 '16 at 14:45




$begingroup$
I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
$endgroup$
– Martin Sleziak
Jun 19 '16 at 14:45










1 Answer
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$begingroup$

Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



$frac{q_n}{q_{n-1}}$ =
$frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
$frac{a_n*q_{n-1}}{q_{n-1}}$ +
$frac{q_{n-2}}{q_{n-1}}$ =
$a_n$ +
$frac{q_{n-2}}{q_{n-1}}$ =
$a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



$vdots$



So by induction you will have:



$frac{q_n}{q_{n-1}}$ =
$a_n$ + $frac{1}
{a_{n-1} + frac{1}
{a_{n-2} + frac{1}
{frac{q_{n-3}}{
ddots}}}}$ = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    2












    $begingroup$

    Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



    $frac{q_n}{q_{n-1}}$ =
    $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
    $frac{a_n*q_{n-1}}{q_{n-1}}$ +
    $frac{q_{n-2}}{q_{n-1}}$ =
    $a_n$ +
    $frac{q_{n-2}}{q_{n-1}}$ =
    $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



    So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



    and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



    and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



    $vdots$



    So by induction you will have:



    $frac{q_n}{q_{n-1}}$ =
    $a_n$ + $frac{1}
    {a_{n-1} + frac{1}
    {a_{n-2} + frac{1}
    {frac{q_{n-3}}{
    ddots}}}}$     = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



      $frac{q_n}{q_{n-1}}$ =
      $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
      $frac{a_n*q_{n-1}}{q_{n-1}}$ +
      $frac{q_{n-2}}{q_{n-1}}$ =
      $a_n$ +
      $frac{q_{n-2}}{q_{n-1}}$ =
      $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



      So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



      and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



      and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



      $vdots$



      So by induction you will have:



      $frac{q_n}{q_{n-1}}$ =
      $a_n$ + $frac{1}
      {a_{n-1} + frac{1}
      {a_{n-2} + frac{1}
      {frac{q_{n-3}}{
      ddots}}}}$       = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



        $frac{q_n}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}}{q_{n-1}}$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



        and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



        $vdots$



        So by induction you will have:



        $frac{q_n}{q_{n-1}}$ =
        $a_n$ + $frac{1}
        {a_{n-1} + frac{1}
        {a_{n-2} + frac{1}
        {frac{q_{n-3}}{
        ddots}}}}$         = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






        share|cite|improve this answer











        $endgroup$



        Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



        $frac{q_n}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}}{q_{n-1}}$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



        and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



        $vdots$



        So by induction you will have:



        $frac{q_n}{q_{n-1}}$ =
        $a_n$ + $frac{1}
        {a_{n-1} + frac{1}
        {a_{n-2} + frac{1}
        {frac{q_{n-3}}{
        ddots}}}}$         = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 14:28









        Algebear

        616319




        616319










        answered Jun 19 '16 at 21:17









        qwerryqwerry

        314




        314






























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