Talk:Continued fraction

citation needed

"One theorem states that any real number k can be approximated by rational m/n " .. I'd like to see a reference for this, and the name of the theorem cited. —Preceding unsigned comment added by 99.226.209.46 (talk) 05:19, 6 January 2008 (UTC)

It's Theorem 193 in Hardy & Wright (5th ed.) - I've added a reference in the article. Gandalf61 (talk) 17:23, 6 January 2008 (UTC)

Rational induction

As an additional point of interest, the fact that all rational numbers can be represented as a finite continued fraction allows mathematical induction to be extended to the positive rationals. This can be done by proving:

1. $n=1$ is true
2. If $n=k$ is true, then $n=k+1$ is true
3. If $n=k$ is true, then $n=1/k$ is true

Using these operations, it is possible to create all finite continued fractions and hence all rational numbers. Perhaps this deserves a mention on either this page or the page on induction?

--124.189.100.175 (talk) 03:03, 12 January 2008 (UTC)

That can be done without continued fractions. E.g., consider a grid of points in a coordinate system, e.g. {..., -3, -2, -1, 0, 1, 2, 3, ...} X {1, 2, 3, ...}. Define a sequence going through all these points, beginning at (0, 1), spiralling or zigzagging outwards as you please. Let each point (p, q) represent the fraction p/q, and skip it if it is not to lowest terms. Of course, this skipping in some contexts may makes this sequence inferior to the one suggested by 124.189.100.175, so I guess what I'm saying is, if we include 124.189.100.175's sequence in the article, we should indicate what it might be good for.--Niels Ø (noe) (talk) 11:23, 12 January 2008 (UTC)

representation of 1

Every finite continued fraction represents a rational number, and every rational number (except 1) can be represented in precisely two different ways as a finite continued fraction.

Why except 1? 1 can also be represented in two different ways : 1 = [1] = [0;1]. Indeed any integer n = [n] = [n-1;1].--143.248.181.36 (talk) 06:48, 28 January 2008 (UTC)

Theorem 2

Is Theorem 2 really a Theorem? All it is, is the definition of h and k, which is a given. TechnoFaye Kane 03:28, 8 February 2008 (UTC)

As the article stands, h_n and k_n are defined at the beginning of the Some useful theorems section as integer sequences with recurrence relations that use the a_n as parameters. So, yes, it is then necessary to prove that h_n/k_n is the n^th convergent of the continued fraction. An alternative approach is to reverse the order, defining h_n and k_n as the numerator and denominator of the n^th convergent, and then showing that h_n and k_n satisfy the recurrence relations. Gandalf61 (talk) 11:17, 8 February 2008 (UTC)

THANK YOU!!!! TechnoFaye Kane 20:46, 8 February 2008 (UTC)

infinite continued fraction

Let's discuss about the formula of this picture:

${\frac {a_{0}}{1}},\qquad {\frac {a_{1}a_{0}+1}{a_{1}}},\qquad {\frac {a_{2}(a_{1}a_{0}+1)+a_{0}}{a_{2}a_{1}+1}},\qquad {\frac {a_{3}(a_{2}(a_{1}a_{0}+1)+a_{0})+(a_{1}a_{0}+1)}{a_{3}(a_{2}a_{1}+1)+a_{1}}}.$

from infinite continued fraction of continued fraction. Continued_fraction

please compare with this.

$x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+{\cfrac {1}{\ddots \,}}}}}}}}$

and here is my proof to show that conflicts.

Answer:Covergents:

$x_{1}=[a_{0}]=a_{0}\qquad$

$x_{2}=[a_{0};a_{1}]=a_{0}+{\cfrac {1}{a_{1}}}={\frac {a_{1}a_{0}+1}{a_{1}}}\qquad$

$x_{3}=[a_{0};a_{1},a_{2}]=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}}}}}=a_{0}+{\cfrac {1}{\cfrac {a_{1}a_{2}+1}{a_{2}}}}=a_{0}+{\cfrac {a_{2}}{a_{1}a_{2}+1}}={\cfrac {a_{0}(a_{1}a_{2}+1)+a_{2}}{a_{1}a_{2}+1}}\qquad$

$x_{4}=[a_{0};a_{1},a_{2},a_{3}]=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}}}}}}}\qquad$

$=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {a_{3}}{a_{2}a_{3}+1}}}}\qquad$

$=a_{0}+{\cfrac {a_{2}a_{3}+1}{a1(a_{2}a_{3}+1)+a_{3}}}\qquad$

$={\cfrac {a_{0}(a_{1}(a_{2}a_{3}+1)+a_{3})+a_{2}a_{3}+1}{a_{1}(a_{2}a_{3}+1)+a_{3}}}\qquad$--Meavel (talk) 00:26, 17 March 2008 (UTC)

As far as I can see the formulae are only seemingly different, but in fact the same Hkleinnl —Preceding comment was added at 20:00, 19 March 2008 (UTC)

Nerd Bias

I saw this at the beginning of the first section, "Motivation", and it made me laugh:

Most people are familiar with the decimal representation of real numbers:

$r=\sum _{i=0}^{\infty }a_{i}10^{-i}$

It looks as if it's being said that most people understand the above mathematical representation. Freaked me out.

(This entry edited once for my own reasons. New sig below)

- Misha Vargas

216.254.12.114 (talk) 04:08, 21 November 2008 (UTC)

This is only true. Most people are familiar with decimal representation, even if not in the rigorous form. Scineram (talk) 11:38, 17 December 2008 (UTC)