Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hey,

    I am currently learning about infinite continued fractions, having studied finite continued fractions.

    I understand why all infinite continued fractions do indeed have convergents which tend to a limit and that this limit must be irrational. However I do not understand why all irrationals can be represented as an infinite continued fraction.

    By attempting to write a general irrational number as a continued fraction I can see how a single infinite continued fraction is the only possible option. As far as I can tell, this does not necessarily mean that the infinite continued fraction one produces has to be equal to the original irrational.

    For example: One is forced to write root(2) as [1; 2, 2, 2, ... ]

    But what is to say [1; 2, 2, 2, ... ]?

    I hope this makes sense!
    Offline

    18
    ReputationRep:
    (Original post by Magu1re)
    I do not understand why all irrationals can be represented as an infinite continued fraction.
    A brief summary of how it works:

    Denote an arbitrary real as x. Then there is an integer a_0 and a non-negative real r_0<1 such that x=a_0+r_0.

    Now if r_0\neq 0 apply the reasoning to \frac{1}{r_0}:\;\frac{1}{r_0}=a_  1+r_1 for some positive integer a_1 and non-negative real r_1<1. Pluggin this in to the 1st expression:

    x=a_0+\dfrac{1}{a_1+r_1}

    Going with the same idea for \frac{1}{r_1}:

    x=a_0+\dfrac{1}{a_1+\dfrac{1}{a_  2+r_2}}

    \text{etc.}

    Now the key here, is to note that whenever x is rational, there is an n for which r_n=0 (i.e. the process stops - can you see why?). Hence if it does not then x is not rational (and since x was arbitrary, this process works for any x; hence any irrational can be expressed in this form).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Lord of the Flies)
    A brief summary of how it works:

    Denote an arbitrary real as x. Then there is an integer a_0 and a non-negative real r_0<1 such that x=a_0+r_0.

    Now if r_0\neq 0 apply the reasoning to \frac{1}{r_0}:\;\frac{1}{r_0}=a_  1+r_1 for some positive integer a_1 and non-negative real r_1<1. Pluggin this in to the 1st expression:

    x=a_0+\dfrac{1}{a_1+r_1}

    Going with the same idea for \frac{1}{r_1}:

    x=a_0+\dfrac{1}{a_1+\dfrac{1}{a_  2+r_2}}

    \text{etc.}

    Now the key here, is to note that whenever x is rational, there is an n for which r_n=0 (i.e. the process stops - can you see why?). Hence if it does not then x is not rational (and since x was arbitrary, this process works for any x; hence any irrational can be expressed in this form).
    I do see this method of deriving the infinite continued fraction.

    Perhaps I did not make my misgivings clear: I am not currently convinced this infinite continued fraction does actually equal the initial number.

    I can see the equivalence at each step, but at each step we are essentially dealing with a finite continued fraction. Why does the limiting result also hold? Why could this infinite continued fraction not equal another irrational number for example?
    Offline

    18
    ReputationRep:
    (Original post by Magu1re)
    I am not currently convinced this infinite continued fraction does actually equal the initial number.

    I can see the equivalence at each step, but at each step we are essentially dealing with a finite continued fraction. Why does the limiting result also hold? Why could this infinite continued fraction not equal another irrational number for example?
    Consider the process above and let u_n be the rational number determined by the finite fraction up to a_n\;(setting r_n=0). You agree that u_n converges. Can you see why this sequence alternates between being smaller than and greater than x? And if so, can you see why this implies that the sequence cannot converge to anything but x?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Lord of the Flies)
    Consider the process above and let u_n be the rational number determined by the finite fraction up to a_n\;(setting r_n=0). You agree that u_n converges. Can you see why this sequence alternates between being smaller than and greater than x? And if so, can you see why this implies that the sequence cannot converge to anything but x?
    I understand why the convergents oscillate with an ever-decreasing amplitude, so to be speak, and thus the infinite continued fraction converges. I do not see why x must be in the "centre" of this oscillation and thus the limit.
    Offline

    18
    ReputationRep:
    (Original post by Magu1re)
    I understand why the convergents oscillate with an ever-decreasing amplitude, so to be speak, and thus the infinite continued fraction converges. I do not see why x must be in the "centre" of this oscillation and thus the limit.
    Perhaps I wasn't clear. Using the sequence described above:

    \begin{aligned} a_0+r_0=x &\Rightarrow u_0<x\\ a_0+\dfrac{1}{a_1+r_1}=x &\Rightarrow u_1>x\\ a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+  r_2}}=x &\Rightarrow u_2<x\end{aligned}

    \text{etc.}

    Spoiler:
    Show
    In case it wasn't clear, u_n is simply the LHS with r_n=0

    u_n alternates between being >x and <x. Hence x is the centre of the oscillation and therefore the limit.
    Offline

    1
    ReputationRep:
    Somewhat detailed proof.
    Spoiler:
    Show
    Let \displaystyle [x]=p_{1} and write \displaystyle x=p_{1}+\frac{1}{a_{1}}. We have \displaystyle a_{1}>1 and \displaystyle [a_{1}]=p_{2}. If \displaystyle a_{1} is an integer, then \displaystyle x=p_{1}+\frac{1}{p_{2}}. Otherwise we have  \displaystyle x=p_{1}+\frac{1}{p_{2}+\frac{1}{  a_{2}}}. Continuing in this way, we can write \displaystyle x=[p_{1},p_{2},...,p_{n},a_{n}]. If there is \displaystyle n such that \displaystyle a_{n}, then we are done; otherwise we obtain an infinite continued fraction. Let us denote its limit by \displaystyle x', and suppose\dispalystyle x \neq x'.
    Let \displaystyle K_{m} be a sequence of best rational approximations to \displaystyle x'. Then \displaystyle K_{1}, K_{2}, ... K_{n} are best rational approximation fractions to \displaystyle K_{n}. We can write \displaystyle x=\frac{p_{n+1}P_{n}+P_{n-1}}{p_{n+1}Q_{n}+Q_{n-1}}, where \displaystyle P_{n}=p_{n}P_{n-1}+P_{n-2}, Q_{n}=p_{n}Q{n-1}+Q_{n-2}, P_{0}=1, P_{1}=p_{1}, Q_{0}=0, Q_{1}=1. Since \displaystyle p_{n}>1 for each \displaystyle n \in \mathbb{Z^{+}}, the inequality \displaystyle | \frac{P_{n}}{Q_{n}}-x|<\frac{1}{{Q_{n}}^2} leads to \displaystyle x'=\lim_{n \to \infty}\frac{P_{n}}{Q_{n}}=x - contradiction.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Lord of the Flies)
    Perhaps I wasn't clear. Using the sequence described above:

    \begin{aligned} a_0+r_0=x &\Rightarrow u_0<x\\ a_0+\dfrac{1}{a_1+r_1}=x &\Rightarrow u_1>x\\ a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+  r_2}}=x &\Rightarrow u_2<x\end{aligned}

    \text{etc.}

    Spoiler:
    Show
    In case it wasn't clear, u_n is simply the LHS with r_n=0

    u_n alternates between being >x and <x. Hence x is the centre of the oscillation and therefore the limit.
    This does make sense. Only one number can be less than all the odd convergents and greater than all the even convergents, namely the limit. x satisfies this property, therefore x is the limit.

    Thank-you for persevering with me through this question, I appreciate it!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.