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# Infinite Continued Fractions and Irrational Numbers Watch

1. 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!
2. (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 . Then there is an integer and a non-negative real such that .

Now if apply the reasoning to for some positive integer and non-negative real . Pluggin this in to the 1st expression:

Going with the same idea for

Now the key here, is to note that whenever is rational, there is an n for which (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).
3. (Original post by Lord of the Flies)
A brief summary of how it works:

Denote an arbitrary real as . Then there is an integer and a non-negative real such that .

Now if apply the reasoning to for some positive integer and non-negative real . Pluggin this in to the 1st expression:

Going with the same idea for

Now the key here, is to note that whenever is rational, there is an n for which (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?
4. (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 be the rational number determined by the finite fraction up to setting . You agree that 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?
5. (Original post by Lord of the Flies)
Consider the process above and let be the rational number determined by the finite fraction up to setting . You agree that 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.
6. (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:

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

alternates between being and . Hence is the centre of the oscillation and therefore the limit.
7. Somewhat detailed proof.
Spoiler:
Show
Let and write . We have and . If is an integer, then . Otherwise we have . Continuing in this way, we can write . If there is such that , then we are done; otherwise we obtain an infinite continued fraction. Let us denote its limit by , and suppose.
Let be a sequence of best rational approximations to . Then are best rational approximation fractions to . We can write , where . Since for each , the inequality leads to - contradiction.
8. (Original post by Lord of the Flies)
Perhaps I wasn't clear. Using the sequence described above:

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

alternates between being and . Hence 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!

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Updated: April 6, 2013
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