Proof of q n q n − 1 = [ a n , a n − 1 , a n − 2 , . . . , a 2 , a 1 ]?Proof of continued fractions axiom.Let c = [ a 0 , a 1 , a 2 , … , a n , … ] = a 0 + 1 a 1 + 1 a 2 + ⋱ be a continued fraction which could be finite or infinite.By p n q n we denote the n-th convergent of the continued fraction, i.e. p n q n = [ a 0 , a 1 , a 2 , … , a n ] .How to prove the formula formula: 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 p n p n − 1 = [ a n , a n − 1 , … , a 1 , a 0 ] .

Question

Proof of             q      n              q              n        −        1              =  [      a    n    ,      a          n      −      1        ,      a          n      −      2        ,  .  .  .  ,      a    2    ,      a    1    ]?Proof of continued fractions axiom.Let   c  =  [      a    0    ,      a    1    ,      a    2    ,  …  ,      a    n    ,  …  ]  =      a    0    +                                                  1                                                                              a            1                    +                                                                                                                  1                                                                                                                                                                      a                    2                                    +                  ⋱                                                                         be a continued fraction which could be finite or infinite.By             p      n              q      n       we denote the   n-th convergent of the continued fraction, i.e.            p      n              q      n        =  [      a    0    ,      a    1    ,      a    2    ,  …  ,      a    n    ]  .How to prove the formula formula:            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            p      n              p              n        −        1              =  [      a    n    ,      a          n      −      1        ,  …  ,      a    1    ,      a    0    ]  .

floygdarvn · Accepted Answer

Ok, I guess it is by far simple answer, sorry for posting silly questions :(.            q      n              q              n        −        1             =                     a        n            ∗              q                  n          −          1                    +              q                  n          −          2                            q              n        −        1             =                     a        n            ∗              q                  n          −          1                            q              n        −        1             +             q              n        −        2                    q              n        −        1             =       a    n   +             q              n        −        2                    q              n        −        1             =       a    n   +       1                  q                  n          −          1                            q                  n          −          2                    So             q      n              q              n        −        1             =       a    n   +       1                  q                  n          −          1                            q                  n          −          2                    and             q              n        −        1                    q              n        −        2             =       a          n      −      1       +       1                  q                  n          −          2                            q                  n          −          3                    and             q              n        −        2                    q              n        −        3             =       a          n      −      2       +       1                  q                  n          −          3                            q                  n          −          4                      ⋮So by induction you will have:            q      n              q              n        −        1             =       a    n   +       1                  a                  n          −          1                    +              1                              a                          n              −              2                                +                      1                                          q                                  n                  −                  3                                            ⋱                                           =   [      a          n        ,      a          n      −      1        ,      a          n      −      2        ,  …  ,      a          1        ]

Proof of q n </msub> q <mrow class="MJX-TeXAtom-ORD">

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