Proof for a theorem about relations beween convergents in continued fractions

Question

Upon reading about some properties of numerators and denominators in a textbook called Continued Fractions (here, chapter 2.3), I was unable to understand the following transmutation of the expression (circled in red in the image link), and highlighted red here:

2.3 Relations between convergents In this section, we see some properties of the simple continued fractions in terms of the numerators and denominators appearing in the convergents.

Theorem 2.4. If $ p_{n} $ and $ q_{n} $ are defined by $ \begin{array}{l} p_{0}=a_{0}, p_{1}=a_{1} a_{0}+1, p_{n}=a_{n} p_{n-1}+p_{n-2} \text { for } 2 \leq n \\ q_{0}=1, q_{1}=a_{1}, q_{n}=a_{n} q_{n-1}+q_{n-2} \text { for } 2 \leq n \end{array} $

then $ \left[a_{0}, a_{1}, \ldots, a_{n}\right]=\frac{p_{n}}{q_{n}} $

Proof. The proof proceeds by induction. The base cases are seen to be true by the assumptions given for $ n=0, n=1 $. Let us assume the statement to be true for some $ m $. Then

$ \left[a_{0}, a_{1}, \ldots a_{m-1}, a_{m}\right]=\frac{p_{m}}{q_{m}}=\frac{a_{m} p_{m-1}+p_{m-2}}{a_{m} q_{m-1}+q_{m-2}} $

Hence, we get

$ \left[a_{0}, a_{1}, \ldots a_{m-1}, a_{m},\color{red}{a_{m+1}}\right]=\left[a_{0}, a_{1}, \ldots a_{m-1}, \color{red}{a_{m}+\frac{1}{a_{m+1}}}\right]$

In addition to not seeing how the equality in the last line was achieved, I was also under the impression that convergents of continued fractions (the quotients in square bracket notation) must by definition always be integers. The marked transmutation makes the last quotient a fraction.

Answer 1

This is an interpretation question, rather than a request for a problem to be solved. Therefore, I personally see no problem answering it, even though the OP has shown no work. If I get downvoted, okay.

Because of the difficulty displaying long continued fractions, I am going to illustrate the OP's question, under the assumption that $m = 3$.

$a_0 +\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3 + \cfrac{1}{a_4}}}}$

can be equivalently interpeted as

$a_0 +\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{\{a_3 + \frac{1}{a_4}\}}}}$

I was also under the impression that convergents of continued fractions (the quotients in square bracket notation) must by definition always be integers.

I am assuming that you intend the coefficients of continued fractions, which are normally expressed as integers, rather than the convergents of continued fractions, which normally have form $\frac{p_n}{q_n}.$

While it's true that the coefficients are normally computed to be integers, you specifically asked how a specific line in a proof can be algebraically justified. As indicated in my continued fraction examples above, the algebra is justified.

In my examples, what this means is that

$$[a_0, a_1, a_2, a_3, a_4]$$

is algebraically equivalent to

$$\left[a_0, a_1, a_2, \left(a_3 + \frac{1}{a_4}\right)\right].$$

Answer 2

here's an example, including the Bezout equation at the end

$$ \gcd( 479, 231 ) = ??? $$

$$ \frac{ 479 }{ 231 } = 2 + \frac{ 17 }{ 231 } $$ $$ \frac{ 231 }{ 17 } = 13 + \frac{ 10 }{ 17 } $$ $$ \frac{ 17 }{ 10 } = 1 + \frac{ 7 }{ 10 } $$ $$ \frac{ 10 }{ 7 } = 1 + \frac{ 3 }{ 7 } $$ $$ \frac{ 7 }{ 3 } = 2 + \frac{ 1 }{ 3 } $$ $$ \frac{ 3 }{ 1 } = 3 + \frac{ 0 }{ 1 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccc} & & 2 & & 13 & & 1 & & 1 & & 2 & & 3 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 2 }{ 1 } & & \frac{ 27 }{ 13 } & & \frac{ 29 }{ 14 } & & \frac{ 56 }{ 27 } & & \frac{ 141 }{ 68 } & & \frac{ 479 }{ 231 } \end{array} $$ $$ $$ $$ 479 \cdot 68 - 231 \cdot 141 = 1 $$

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different, infinite but periodic continued fraction, here for $\sqrt {13}$

$$ \sqrt { 13} = 3 + \frac{ \sqrt {13} - 3 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {13} - 3 } = \frac{ \sqrt {13} + 3 }{4 } = 1 + \frac{ \sqrt {13} - 1 }{4 } $$ $$ \frac{ 4 }{ \sqrt {13} - 1 } = \frac{ \sqrt {13} + 1 }{3 } = 1 + \frac{ \sqrt {13} - 2 }{3 } $$ $$ \frac{ 3 }{ \sqrt {13} - 2 } = \frac{ \sqrt {13} + 2 }{3 } = 1 + \frac{ \sqrt {13} - 1 }{3 } $$ $$ \frac{ 3 }{ \sqrt {13} - 1 } = \frac{ \sqrt {13} + 1 }{4 } = 1 + \frac{ \sqrt {13} - 3 }{4 } $$ $$ \frac{ 4 }{ \sqrt {13} - 3 } = \frac{ \sqrt {13} + 3 }{1 } = 6 + \frac{ \sqrt {13} - 3 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccc} & & 3 & & 1 & & 1 & & 1 & & 1 & & 6 & & 1 & & 1 & & 1 & & 1 & & 6 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 3 }{ 1 } & & \frac{ 4 }{ 1 } & & \frac{ 7 }{ 2 } & & \frac{ 11 }{ 3 } & & \frac{ 18 }{ 5 } & & \frac{ 119 }{ 33 } & & \frac{ 137 }{ 38 } & & \frac{ 256 }{ 71 } & & \frac{ 393 }{ 109 } & & \frac{ 649 }{ 180 } \\ \\ & 1 & & -4 & & 3 & & -3 & & 4 & & -1 & & 4 & & -3 & & 3 & & -4 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 13 \cdot 0^2 = 1 & \mbox{digit} & 3 \\ \frac{ 3 }{ 1 } & 3^2 - 13 \cdot 1^2 = -4 & \mbox{digit} & 1 \\ \frac{ 4 }{ 1 } & 4^2 - 13 \cdot 1^2 = 3 & \mbox{digit} & 1 \\ \frac{ 7 }{ 2 } & 7^2 - 13 \cdot 2^2 = -3 & \mbox{digit} & 1 \\ \frac{ 11 }{ 3 } & 11^2 - 13 \cdot 3^2 = 4 & \mbox{digit} & 1 \\ \frac{ 18 }{ 5 } & 18^2 - 13 \cdot 5^2 = -1 & \mbox{digit} & 6 \\ \frac{ 119 }{ 33 } & 119^2 - 13 \cdot 33^2 = 4 & \mbox{digit} & 1 \\ \frac{ 137 }{ 38 } & 137^2 - 13 \cdot 38^2 = -3 & \mbox{digit} & 1 \\ \frac{ 256 }{ 71 } & 256^2 - 13 \cdot 71^2 = 3 & \mbox{digit} & 1 \\ \frac{ 393 }{ 109 } & 393^2 - 13 \cdot 109^2 = -4 & \mbox{digit} & 1 \\ \frac{ 649 }{ 180 } & 649^2 - 13 \cdot 180^2 = 1 & \mbox{digit} & 6 \\ \end{array} $$

Answer 3

Something I programmed recently; this is the Gauss-Lagrange method of chains of reduced forms, here I use the left neighbors. The Pell equation deals with $x^2 - n y^2,$ the continued fraction being for $\sqrt n.$ With little extra effort we get the continued fraction for $\frac{B + \sqrt D}{2A},$ where $D=B^2 -4AC.$

I also have it print some 2 by 2 matrices in proper format for pari-gp; in each case below, the product pari calls rt * h * r is something specific related to the original Hessian matrix h

The "partial quotients" of the continued fraction are the absolute values of the "digits" I write at the right hand side of each line. The output below shows $\frac{5 + \sqrt {597}}{22}$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycleLeft 11 5 -13

0  form   11 5 -13   epsilon  1
1  form   -7 17 11   Epsilon  -2
2  form   17 11 -7   Epsilon  1
3  form   -1 23 17   Epsilon  -23     ambiguous  
4  form   17 23 -1   Epsilon  1     ambiguous  
5  form   -7 11 17   Epsilon  -2
6  form   11 17 -7   Epsilon  1
7  form   -13 5 11   Epsilon  -1 opposite  
8  form   3 21 -13   Epsilon  7     ambiguous  
9  form   -13 21 3   Epsilon  -1     ambiguous  
10  form   11 5 -13


  form   11 x^2  + 5 x y  -13 y^2 

minimum was   1rep   x = -4   y = 3 disc 597 dSqrt 24
Automorph, written on right of Gram matrix:  
-5872  5187
4389  -3877
  for   Pari/gp: rt =  [ -5872 , 4389 ; 5187 , -3877 ] ;    h =  [ 22 , 5 ; 5 , -26 ] ;    r =  [ -5872 , 5187 ; 4389 , -3877 ] ; 


 opposite Pari/gp: rt =  [ -392 , 293 ; -293 , 219 ] ;    h =  [ 22 , 5 ; 5 , -26 ] ;    r =  [ -392 , -293 ; 293 , 219 ] ; 

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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$