Talk:Liouville number

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Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. —Cronholm144 22:57, 12 November 2007 (UTC)

Could someone clarify the definition? I don't understand what a Liouville number is. Also, the Liouville constant should be explained better; how is it defined? AxelBoldt 04:43 Dec 14, 2002 (UTC)

Hope that helps; the proof doesn't seem super dense if someone wants to tackle it. Chas zzz brown 10:41 Dec 17, 2002 (UTC)

Yes, thanks, that clarifies it. The proof would be nice, since the Liouville constant seems to be one of the few numbers that can be proven to be transcendental rather easily.

I still don't understand the irrationality measure. Is it the supremum over all approximating sequences? Are Liouville numbers precisely the numbers with infinite irrationality measure?

On http://mathworld.wolfram.com/LiouvilleNumber.html, they require infinitely many rational p/q for any n. Is that equivalent? AxelBoldt 02:17 Dec 18, 2002 (UTC)

Using the article definition, given a Liouville number x, there exists an infinite sequence of integer pairs {(p_i, q_i)} which satisfy |x - p_i/q_i| < 1/(q_i)ⁱ. Then for all m > n,

|x - p_m/q_m| < 1/(q_m)^m < 1/(q_m)ⁿ;

so there are an infinite number of such integer pairs for a given n. Conversely, starting with the Mathworld definition, all we require is a single example for each n to satisfy the article definition, so the two statements are equivalent.

I'm not sure that the given value really is Liouville's constant; for example, some seem to prefer ∑ 2^-j! instead of ∑ 10^-j! (both are Liouville numbers). I don't understand the irrationality measure either; one would assume that it would want to be over all sequences - or over all rationals with positive denominator?

The proof is paraphrased from the given link - what is the copyright issue here (if any)?

My understanding (or opinion) on this question would be as follows:

1. There are certain proofs which are either so old, so famous, or so important, that they are a part of "folk mathematics", i.e. they are part of the common intellectual body of mathematics (e.g. the proof of Lagrange's theorem, Heine-Borel theorem, Cantor's diagonal argument, Erdos's proof of Bertrand's postulate, and most "basic" results in a particular area) and "belong" to no one.

2. There are other more recent proofs, which are attributable to particular individuals. These may appear in journal articles, class notes, etc. These "belong" to someone, in the sense that they may or may not be legally copyrighted.

3. In any case, it seems to me that proofs are not really "copyrightable"...it is probably best not to copy verbatim someone's wording. But the essence of a proof can be rephrased without verbatim copying. In this case, unless it's obviously a "folk proof", credit should be given (e.g. "The following proof is due to Niven, see...")

4. Unless you are copying entire papers, articles, notes, or chapters, I doubt that anyone would object to a single proof of theirs being paraphrased here. I would guess (at least, I would feel this way) that it would be something of an honour to think that one's proof warranted specific mention in a general encyclopedia, given the bulk of math that's published every year.

Revolver

Finally, I'll be SOOOO glad when the TeX support is added; in my browser, I can't see either ≤ or ≥, so the proof looks a bit like gobbely-dy-gook to me :(. Chas zzz brown 09:12 Dec 18, 2002 (UTC)

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In re: the irrationality measure. The article defines it as...

the limit superior of -ln(|x-pi/qi|)/ln(qi) for a sequence of rational approximations {pi/qi} to x.

I think this statement relates to, for example, mathworld's definition http://mathworld.wolfram.com/IrrationalityMeasure.html, as follows. Assume we have a sequence {(p_k, q_k)} approximating x, and define the sequence {μ_k} as the largest value (handwaving... lim sup) such that

0 < |x - p_k/q_k| < 1/q_k^μ_k

Note that, if we furthermore order the sequence so that μ_j > μ_k iff j > k, then again we get (as noted above) that there are an infinite number of pairs (p,q) which satisfy

0 < |x - p/q| < 1/q^μ_k

for each μ_k. The article definition then seeks the largest value of μ_k which appears in any approximating sequence, via the following:

|x - p_k/q_k| < 1/q_k^μ_k

(q_k^μ_k)|x - p_k/q_k| < 1

log_qk ((q_k^μ_k)|x - p_k/q_k|) < 0

μ_k + log_qk(|x - p_k/q_k|) < 0

μ_k < -log_qk(|x - p_k/q_k|)

μ_k < -ln(|x - p_k/q_k|) / ln(q_k)

Now, if we take the lim sup of {μ_k} over all rational sequences which approximate x, we get the article's definition of the measure.

Mathworld's definition in this case seems simpler; they take the infinum of the set of all μ for which

0 < |x - p/q| < 1/q^μ

has at most a finite number of solutions (p, q), which is equivalent to the article definition, but more clear I think. Chas zzz brown 22:59 Dec 18, 2002 (UTC)

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This part of the proof:

Then, since x is a Liouville number, there exists integers a, b > 1 such that

|x - a/b| < 1/b^r+n

seems unclear to me. Do you choose b to be a power of 10? AxelBoldt 05:11 Jan 28, 2003 (UTC)

The recent edit hasn't clarified it for me: why do a and b exist? AxelBoldt 17:21 Jan 28, 2003 (UTC)

Ermmm, by the assumption that x is Liouville. By the definition of Liouville number, for any integer n, there exist p and q with q > 1 and

|x - p/q| < 1/qⁿ

Substitute m = (r+n) for n, a for p and b for q in the above, and you get that, since x is Liouville number, and given m = (r+n), there exist a and b > 1 such that

|x - a/b| < 1/b^m = 1/b^r+n = 1/(b^rbⁿ)

Since b ≥ 2, and then by our choice of r, 1/b^r ≤ 1/2^r < A. Thus 1/(b^rbⁿ) = (1/b^r) (1/bⁿ) < A/bⁿ. Thus, given A and n, exists a, b such that

|x - a/b| < A/bⁿ

and the existence of a and b contradicts the lemma based on the assumption that x is algebraic. Maybe I should have stuck with p and q instead of introducing a and b, but I thought that would be more confusing! Chas zzz brown 12:05 Jan 29, 2003 (UTC)

Oh, now I get it. I thought you were proving that the Liouville constant is transcendental, but your proof is for Liouville numbers. Sure, then everything is fine. What I was missing was the argument that the Liouville constant is indeed a Liouville number I guess. AxelBoldt 17:57 Jan 29, 2003 (UTC)

I was wondering if you quit coffee or something! I added a section showing that the L. constant is indeed a L. number Chas zzz brown 21:34 Jan 30, 2003 (UTC)

Shouldn't we convert the formulas involved in the proof to PNG? Looking at the HTML gives me a headache.Scythe33 30 June 2005 02:06 (UTC)

[edit] Continued fraction expansion

The terms in the continued fraction expansion of every Liouville number are unbounded...

Is the converse of that true? That is, if the terms in a continued fraction expansion are unbounded, does it represent a Liouville number? -GTBacchus^(talk) 18:44, 29 April 2006 (UTC)

No; counterexample is e. AxelBoldt 00:25, 30 April 2006 (UTC)

[edit] Irrationality measure of logarithms

I'm curious about the irrationality measure of some of the irrational numbers I run into regularly. Does anybody know a good source for that? In particular, I'd like to know about calculating irrationality measure for integer-base logarithms of integers, such as log₂3. Thanks in advance for any pointers. -GTBacchus^(talk) 17:28, 19 March 2007 (UTC)

[edit] What does the irrationality measure measure?

Since Liouville numbers have μ = ∞, one would say that approximation is harder when μ is lower. This cannot easily be seen by the measure of rational or algebraic numbers, which I hear is 1 and 2 respectively. Obviously, rational numbers are easy to approximate...

Approximating x to n+1 decimal digits, one may get

  except for at most a finite number of "lucky" pairs (p, q)

hence

  for n large enough.

For example, what does μ(π) ~ 8 imply? Brute force approach, where q=10ⁿ, can be used to approximate π with, say, 314159265358979323846/10²⁰. In this case I may divide by 2, and then, maybe, simplify more. 10²⁰ >= 10^20/8 ~ 316.2 is not really sharp in this case, is it?

How does one characterize that measure? —Preceding unsigned comment added by Ale2006 (talk • contribs) 19:28, 14 February 2008 (UTC)

Is there a flaw with the irrationality proof? I should be able to select any positive integer n. However, the n chosen is one such that 2^{n − 1} > d. If I use n=1 then 1 > d which is false. Unmasked (talk) 19:47, 8 May 2008 (UTC)