Infinite Continued Fractions and Irrational Numbers

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! :smile:

Reply 1

Lord of the Flies

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).

(edited 11 years ago)

Reply 2

Magu1re

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?

Reply 3

Lord of the Flies

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?

Reply 4

Magu1re

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. :smile:

Reply 5

Lord of the Flies

Original post by Magu1re

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

u_n

alternates between being

>x

and

<x

. Hence

x

is the centre of the oscillation and therefore the limit.

Reply 6

Mladenov

Somewhat detailed proof.

Spoiler

Reply 7

Magu1re

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

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. :smile:

Thank-you for persevering with me through this question, I appreciate it!