more on continued fractions

[richardt]

Greetings:

In a previous post, I had expressed a desire to prove that every simple continued fraction (CF) converges to some real number. Suppose now that I seek to prove the converse. That is, for every real number, r, there exists some CF that converges to r. Further suppose I have already shown this to be true for r rational (*). In proving for the irrationals, is it enough to show that for any interval containing r, there exists some continued fraction within that interval per (*)? Because the interval can be made arbitrarily small, it seems reasonable to me that in the limiting case, there must exist some CF whose value converges upon r.

As an alternative, one construct of the irrationals is by way of the decimal representations of the rationals. For instance, suppose the sequence of rational numbers {a_n} = {1, 1.4, 1.41, 1.414, ..., a_n} converges to sqrt(2) as n increases without bound. As there exists some continued fraction, F_i, such that F_i = a_i for each a_i in {a_n}, then there exists a sequence of continued fractions, {F_n}, such that {a_n} = {F_n}. Thus {F_n} converges to sqrt(2). As the same argument can be applied for any irrational number, it seems that we a basis for valid proof.

What do you think?

Rich B.

 

[daon2]

In proving for the irrationals, is it enough to show that for any interval containing r, there exists some continued fraction within that interval per (*)? Because the interval can be made arbitrarily small, it seems reasonable to me that in the limiting case, there must exist some CF whose value converges upon r.

Assuming you mean open interval, then no, that is not sufficient, because you already know that every open interval contains a rational number (in fact an infinite number). So there are an infinite number of rational numbers as close as you want to r with distinct continued fraction representations (by (*)).

As an alternative, one construct of the irrationals is by way of the decimal representations of the rationals. For instance, suppose the sequence of rational numbers {a_n} = {1, 1.4, 1.41, 1.414, ..., a_n} converges to sqrt(2) as n increases without bound. As there exists some continued fraction, F_i, such that F_i = a_i for each a_i in {a_n}, then there exists a sequence of continued fractions, {F_n}, such that {a_n} = {F_n}. Thus {F_n} converges to sqrt(2). As the same argument can be applied for any irrational number, it seems that we a basis for valid proof.

What do you think?

Rich B.

I'm not sure what you mean here. Are you attempting to say that a sequence of numbers expressible by continued fractions, which converges, must converge to a continued fraction? That seems a bit more difficult than just showing irrational numbers have continued fraction representations. Or am I misreading?

 

[richardt]

daon2:

To clarify, I am trying to prove that every irrational number can be represented by some continued fraction (CF). To that end, I thought that since, for any irrational number, r, there exists a sequence of decimal approximations that converges to r, namely, {a_n}, where a_n = r to n places, and since each such approximation has a CF equivalent, then we must have a sequence of continued fractions that converges to r.

For instance, suppose {a_n} = {2.2, 2.23, 2.236, 2.2360, 2.23606, ...}, where a_n is the n^th decimal approximation of sqrt(5). Now let F_n be the continued fraction representation of a_n. E.g., F₂ = 2.23 = [2; 4, 2, 1, 7] = 2 + 1/(4 + 1/(2 + 1/(1 + 1/(7)))). Then {a_n} = {2.2, 2.23, 2.236, 2.2360, 2.23606, ...} = {F_n} = {F₁, F₂, F₃, ...}. Since the two sequences are identical, and since {a_n} converges to sqrt(5), then {F_n} must converge to sqrt(5). I was hoping that such argument would succeed in proving that sqrt(5) can indeed be represented via continued fraction, namely, the limiting case of F_n.

Please excuse my less than elegant approach; everything I know about real analysis is self taught.

The aforesaid argument notwithstanding, can you offer a simpler way to prove that every irrational number can be represented by a CF?

Thank you kindly.

Rich B.

 

[daon2]

I do not have a proof, no, and I have never studied the subject. I only pointed out that while you may prove that a sequence of continued fractions converges to every irrational number, that does not imply the irrational number has representation as a continued fraction.

Criticize the following proof:

The sequence defined by

\(\displaystyle \displaystyle a_n = \sum_{i=0}^n \dfrac{4(-1)^{k}}{2k+1}\)

is a sequence of rational numbers. Since \(\displaystyle \{a_n\}\) converges to \(\displaystyle \pi\), that means \(\displaystyle \pi\) is a rational number.

With additional facts your proof may become valid, but I am examining your attempt solely on what you have presented.

 

[richardt]

Deon2:

Thank you for your response. I must confess that my experience with real analysis is limited indeed. I do believe I will take a course or two. Thanks again.

Rich